Math 440

An Introduction to Vector Analysis

Steven A. Chapin

c©2010

ii

Contents

Preface ix

1 Vectors 11.1 Basic Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 121.2 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 181.3 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . 20

1.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 241.4 Lines and Planes in Space . . . . . . . . . . . . . . . . . . . . 27

1.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2 Curves in Space 372.1 Vector Functions . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 442.2 Space Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

2.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 552.3 More About Curves . . . . . . . . . . . . . . . . . . . . . . . . 58

2.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 642.4 Plotting Curves . . . . . . . . . . . . . . . . . . . . . . . . . . 65

2.4.1 Project . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3 Scalar Fields and Vector Fields 693.1 Scalar Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

3.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 773.2 Vector Fields and Flow Curves . . . . . . . . . . . . . . . . . . 80

3.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 903.3 Plotting Functions of Two Variables . . . . . . . . . . . . . . . 94

iii

iv CONTENTS

3.3.1 Project . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

4 Line Integrals and Surface Integrals 974.1 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

4.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 1024.2 Conservative Fields . . . . . . . . . . . . . . . . . . . . . . . . 103

4.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 1154.3 Area Integrals and Volume Integrals . . . . . . . . . . . . . . . 117

4.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 1264.4 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . 129

4.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 1434.5 Plotting Parametric Surfaces . . . . . . . . . . . . . . . . . . . 146

4.5.1 Project . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

5 Stokes-type Theorems 1495.1 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 149

5.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 1535.2 The Divergence Theorem . . . . . . . . . . . . . . . . . . . . . 155

5.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 1595.3 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 160

5.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 1645.4 Cylindrical and Spherical Coordinates . . . . . . . . . . . . . . 165

5.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 1745.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

5.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 182

A The Modern Stokes’ Theorem 185A.1 Differential Forms . . . . . . . . . . . . . . . . . . . . . . . . . 187

A.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 190A.2 The Modern Stokes’ Theorem . . . . . . . . . . . . . . . . . . 190

A.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 196A.3 Stokes-Type Theorems, Revisited . . . . . . . . . . . . . . . . 197

A.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 203A.4 Project . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

B Planar Motion in Polar Coordinates: Kepler’s Laws 207B.1 Planar Motion in Polar Coordinates . . . . . . . . . . . . . . . 207

B.1.1 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . 208

CONTENTS v

B.2 Kepler’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . 209B.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 214

C Solutions to Selected Exercises 215

vi CONTENTS

List of Figures

1.1 The xy-axes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 The xyz-axes . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 The vector−→PQ . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.4−→PQ =

−→RS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.5−→PQ+

−→QR =

−→PR . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.6 Multiplication of a vector by scalar . . . . . . . . . . . . . . . 91.7 Example 1.1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.8 The angle between two vectors . . . . . . . . . . . . . . . . . . 151.9 The decomposition of a vector . . . . . . . . . . . . . . . . . . 161.10 The cross product of two vectors . . . . . . . . . . . . . . . . 221.11 The parallelogram generated by a and b . . . . . . . . . . . . 231.12 The parallelepiped generated by a, b, and c . . . . . . . . . . 251.13 A line in space . . . . . . . . . . . . . . . . . . . . . . . . . . 281.14 A plane in space . . . . . . . . . . . . . . . . . . . . . . . . . 301.15 Example 1.4.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.1 A space curve . . . . . . . . . . . . . . . . . . . . . . . . . . . 462.2 A helix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.3 The velocity vector . . . . . . . . . . . . . . . . . . . . . . . . 492.4 An oriented path . . . . . . . . . . . . . . . . . . . . . . . . . 512.5 An oriented closed path . . . . . . . . . . . . . . . . . . . . . 522.6 T, N, and B . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3.1 A region . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 703.2 A gradient vector and tangent plane to a level surface . . . . . 763.3 A vector field . . . . . . . . . . . . . . . . . . . . . . . . . . . 813.4 Flow curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 823.5 The volume of a fluid passing through a surface . . . . . . . . 85

vii

viii LIST OF FIGURES

3.6 The flux through the surface of a box . . . . . . . . . . . . . . 863.7 The curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

4.1 The line integral . . . . . . . . . . . . . . . . . . . . . . . . . 984.2 A torus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1054.3 Example 4.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1194.4 Polar coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 1204.5 Cylindrical coordinates . . . . . . . . . . . . . . . . . . . . . . 1234.6 Spherical coordinates . . . . . . . . . . . . . . . . . . . . . . . 1244.7 A compact smooth surface with boundary . . . . . . . . . . . 1314.8 A Mobius strip . . . . . . . . . . . . . . . . . . . . . . . . . . 1334.9 The induced orientation . . . . . . . . . . . . . . . . . . . . . 1344.10 A closed surface with several outward unit normal vectors . . 1354.11 A piecewise smooth surface with boundary . . . . . . . . . . . 1354.12 A surface that is not piecewise smooth . . . . . . . . . . . . . 1364.13 Surface area . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

5.1 Green’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 1515.2 A domain on which Green’s theorem holds . . . . . . . . . . . 1525.3 Exercise 1, Section 5.1.1 . . . . . . . . . . . . . . . . . . . . . 1545.4 The divergence theorem . . . . . . . . . . . . . . . . . . . . . 1575.5 A domain on which the divergence theorem holds . . . . . . . 1585.6 A cylindrical rectangular solid . . . . . . . . . . . . . . . . . . 1685.7 A cylindrical rectangle . . . . . . . . . . . . . . . . . . . . . . 169

Preface

This text is designed for a one semester or one quarter course in vectoranalysis (sometimes called vector calculus) for undergraduates majoring inone of the sciences, engineering, or mathematics. The prerequisite is theusual calculus sequence taught in most universities in the United States. Ihave written this book in an informal style, so I hope that it will also proveuseful for self-study and for review.

I considered titling this book A Brief Introduction to Vector Analysis,since it is considerably shorter than most modern-day mathematics texts. Ihope that many will consider its length a desirable feature. In my experience,from teaching this subject several times, this text contains more than enoughmaterial for a one quarter or one semester course, especially if Section 2.3,Appendix A, and Appendix B are covered. I have long been of the opinionthat almost all mathematics texts nowadays are much too long (and muchtoo expensive!).

Because of the level of this text, many of the derivations and argumentsare not totally rigorous in the modern mathematical sense. In many casesan appeal is made to physical or geometric intuition or to formal manipula-tions. In other instances, I have given proofs that rely on assumptions thataren’t strictly necessary, if this resulted in a simpler and/or more instructiveargument. I expect that few students will lose any sleep over this approach,and the references include sources that cover all of this material in a rigorousfashion. I’ve made every effort not to be too technical without sacrificingveracity.

My aim in writing this text has been to give a straightforward treatmentof the basics of vector analysis at a level appropriate for the majority ofstudents likely to take such a course in the United States.

Perhaps an overview of the material covered in this text is in order. Mostof Chapter 1 deals with basic facts about vectors and vector operations,

ix

x PREFACE

including the dot product and the cross product. Vectors are treated bothgeometrically and algebraically. In particular, two directed line segments areconsidered to be equal as vectors if and only if they have the same lengthand the same direction. In the final section in Chapter 1 lines and planesin three dimensional euclidean space are discussed. Since most students willhave seen the material in Chapter 1 in their elementary calculus courses, thetreatment is fairly brief. In particular, for the proofs of some results studentsare referred to their favorite elementary calculus text, while the proofs of afew results are left as exercises. Chapter 2 concerns vector-valued functionsand curves in three dimensional space. Section 2.3 (More About Curves) isnot required in the sequel and may be treated as optional.

Chapter 3 introduces scalar fields and vector fields, including the gradi-ent, the divergence, and the curl which are crucial to the remainder of thetext. Chapter 4 concerns integrals (both oriented and unoriented) as well asconservative vector fields and their relationship to line integrals. Althoughthe main focus in Chapter 4 is line integrals and surface integrals, a sectionon area and volume integrals (Section 4.3) is included in the nature of a re-view. This section may also be omitted by those who are proficient in theseareas. In the development of surfaces, I have tried very hard to strike theright balance between intuition and rigor.

This all leads up to the major theorems of vector analysis — Green’stheorem, the divergence theorem, and Stokes’ theorem — as well as some oftheir applications, which are taken up in Chapter 5. Section 5.4 covers thegradient, the divergence, and the curl in cylindrical and spherical coordinates.With a few very minor modifications, this section could be covered any timeafter Section 3.2.

The final sections of Chapters 2, 3, and 4 are devoted to using computersoftware to plot some of the geometric objects considered in those chapters.I have chosen Matlab for this purpose, but any other computer algebra sys-tem or general purpose graphing software could be substituted for Matlab.These sections could be omitted by those so inclined, but I hope that moststudents will find them interesting and instructive.

In addition, I’ve included a few appendices. In Appendix A I’ve at-tempted to give a treatment of differential forms and the modern versionof Stokes’ theorem that is accessible to students with modest mathematicalbackgrounds. In this I have generally followed Spivak [21], but I have stuckto one, two, and three dimensions and have tried to give a much simpler, andless abstract, treatment. (I have, though, added a project in which the reader

xi

is asked to extend some of these results to four dimensional space, and to usethis to derive the volumes of balls and spheres in four dimensional space.)My hope is that students who study this material later in their careers willbenefit from having done some hands-on calculations.

Appendix B is on planar motion in polar coordinates and contains aderivation of Kepler’s laws of planetary motion.

This text contains over 350 exercises of varying degrees of difficulty. Thesolutions to almost all of the non-proof exercises are given in Appendix C.It would be hard to overstate the importance of working a large number ofexercises. As a professor of mine once said, “If you can’t do the problems,then you don’t understand the material.”

I would like to thank Ms. Joan Seder for giving me a copy of the article[3] referred to in the beginning of Section 5.5. And I would like to thankMr. Vusi Mpendulo Magagula for checking almost all of the solutions in theback of the text. Of course, any remaining errors are my responsibility. Mythanks also go to Mr. Bob Sims at Zip Publishing for all of his assistancein getting this book into print. In addition, I would like to thank all of thestudents who have taken vector analysis from me over the years.

Finally, I would very much appreciate any suggestions or comments thatyou, the reader, might have. Please feel free to write or send e-mail to oneof the addresses given below.

Steven A. ChapinDepartment of Mathematics

Ohio UniversityAthens, OH 45701

[email protected]

xii PREFACE

Chapter 1

Vectors

1.1 Basic Concepts

A nonzero vector, at least in a finite dimensional euclidean space, can becharacterized by its magnitude (or length) and its direction. The vector 0has magnitude 0, but we do not assign it a direction. Vectors have beenproven to be very successful in modeling various quantities in many differentfields. Displacement, velocity, momentum, acceleration, and force are but afew examples.

In this text vectors will be indicated by boldfaced type in contrast toscalars (i.e., real numbers) which will be printed in ordinary type.1 Whenwriting vectors by hand the usual practice is to place an arrow over the sym-bol or to underline the symbol. It is very important to distinguish betweenquantities that are scalars and those that are vectors. In particular, thevector 0 is not the same thing as the number 0.

We will be concerned mostly with vectors, and other objects, in threedimensional euclidean space, which we will refer to simply as space. Forthis reason, and to avoid a lot of unnecessary repetition, we will often statedefinitions and results only in space. However, most of the material in thischapter, and in the remainder of the text, can be easily generalized to theplane (two dimensional euclidean space) and to higher (finite) dimensionaleuclidean spaces.2 Thus, we will often leave it to the reader to formulate

1We will actually drop this convention in Appendix A.2An exception to this is the cross product. Our definition makes sense only for vectors

in space.

1

2 CHAPTER 1. VECTORS

x

y

Figure 1.1: The xy-axes

definitions and results for the plane, and for higher dimensional euclideanspaces.

How to Draw Coordinate Axes

Before we proceed with our discussion of vectors, we would like to make afew remarks concerning the xy-axes in the plane and the xyz-axes in space.

Much of this text concerns oriented curves in the plane and in space andoriented surfaces in space. To be consistent in how we define and picture theorientation of such objects we need to depict the xy-axes in the plane andthe xyz-axes in space in a particular standard fashion.

The xy-axes in the plane are perpendicular and almost always drawnwith the positive x-axis pointing to the right and the positive y-axis pointingupward. See Figure 1.1. Rotating this picture in the plane is allowed. Thus,we could draw the xy-axes with the positive y-axis pointing to the rightand the positive x-axis pointing downward; although, this is seldom done.Other depictions, for example, the positive y-axis pointing to the right andthe positive x-axis pointing upward, are not compatible with the usual waycurves are oriented in the plane.

The xyz-axes are perpendicular to each other and are usually depictedin a few different ways. In this text we will most often draw the xyz-axesso that the positive x-axis points out of the page toward the reader, the

1.1. BASIC CONCEPTS 3

x

y

z

Figure 1.2: The xyz-axes

positive y-axis points to the right, and the positive z-axis points upward.See Figure 1.2. Another common way of depicting the xyz-axes is so thatthe positive x-axis points to the right, the positive y-axis points into the pageaway from the reader, and the positive z-axis points upward. The generalrule is referred to as the right-hand rule: Point the fingers of your right handin the direction of the positive x-axis and rotate them in the direction ofthe positive y-axis through the smaller angle, then your extended thumb willpoint in the direction of the positive z-axis. Other depictions, for example,the positive x-axis pointing to the right, the positive y-axis pointing out ofthe page toward the reader, and the positive z-axis pointing upward, are notcompatible with the usual way curves and surfaces are oriented in space.

We assume that the reader is familiar with cartesian coordinates and howto locate points in the plane and in space. We also assume familiarity withthe graphs of certain basic functions and equations in both the plane and inspace.


Vectors in Terms of Components

We will usually express vectors in terms of their cartesian components.A vector u in space can be written

u = 〈u1, u2, u3〉,

where the real numbers u1, u2, and u3 are called the components of u.In terms of components, addition of vectors is defined by

〈u1, u2, u3〉+ 〈v1, v2, v3〉 = 〈u1 + v1, u2 + v2, u3 + v3〉;

scalar multiplication is defined by

t〈u1, u2, u3〉 = 〈tu1, tu2, tu3〉;

and the magnitude of a vector is defined by

|〈u1, u2, u3〉| =√u2

1 + u22 + u2

3.

Two vectors are parallel if and only if each is a scalar multiple of theother. A vector of magnitude 1 is called a unit vector.

We also define:

0 = 〈0, 0, 0〉 and − u = (−1)u.

In addition, we will use the following notation (definitions):

ut = tu, u− v = u + (−v), and u/t = (1/t)u.

It follows easily that

〈u1, u2, u3〉 − 〈v1, v2, v3〉 = 〈u1 − v1, u2 − v2, u3 − v3〉.

Example 1.1.1. Let a = 〈2,−2, 1〉 and b = 〈−5, 4, 7〉. Compute |a| and2a− 3b

Solution.

|a| =√

22 + (−2)2 + 12 = 3

2a− 3b = 2〈2,−2, 1〉 − 3〈−5, 4, 7〉 = 〈4,−4, 2〉 − 〈−15, 12, 21〉= 〈19,−16,−19〉


It turns out to be convenient to let

i = 〈1, 0, 0〉, j = 〈0, 1, 0〉, and k = 〈0, 0, 1〉.

Then we have

〈u1, u2, u3〉 = u1i + u2j + u3k.

(According to the next theorem, vector addition is associative so it doesn’tmatter how one groups the terms.) So, for example, one can write

〈4,−5, 7〉 = 4i− 5j + 7k.

It is important to note that two vectors are equal if and only if each oftheir corresponding components are equal.

The following theorem says that, with this structure (i.e., addition andscalar multiplication), the set of vectors in space is a vector space over thereal numbers.

Theorem 1.1.1. For any vectors a, b, c and any scalars r, s each of thefollowing is true.

1. (a + b) + c = a + (b + c)

2. a + b = b + a

3. a + 0 = a

4. a + (−a) = 0

5. (r + s)a = ra + sa

6. (rs)a = r(sa)

7. r(a + b) = ra + rb

8. 1a = a

We will prove the first of these and leave the remainder as an exercise.


Proof of (a + b) + c = a + (b + c). Write a = 〈a1, a2, a3〉, b = 〈b1, b2, b3〉,and c = 〈c1, c2, c3〉. Then

(a + b) + c = (〈a1, a2, a3〉+ 〈b1, b2, b3〉) + 〈c1, c2, c3〉= 〈a1 + b1, a2 + b2, a3 + b3〉+ 〈c1, c2, c3〉= 〈(a1 + b1) + c1, (a2 + b2) + c2, (a3 + b3) + c3〉= 〈a1 + (b1 + c1), a2 + (b2 + c2), a3 + (b3 + c3)〉= 〈a1, a2, a3〉+ 〈b1 + c1, b2 + c2, b3 + c3〉= 〈a1, a2, a3〉+ (〈b1, b2, b3〉+ 〈c1, c2, c3〉)= a + (b + c).

Theorem 1.1.1 implies that certain simple equations involving vectors canbe manipulated in basically the same way as similar equations involving justreal numbers. For example, if

su + tv = w, s 6= 0, then

u =1

s(w − tv).

We will feel free to use such manipulations in the sequel without furthercomment. Remember, however, that you cannot divide by a vector.

Vectors as Directed Line Segments

Geometrically, a nonzero vector can be represented by a directed line seg-

ment. For distinct points P and Q, we write−→PQ for the directed line segment

from the point P to the point Q. In terms of components, given P (x1, y1, z1)

and Q(x2, y2, z2) the vector u represented by−→PQ is given by

u = 〈x2 − x1, y2 − y1, z2 − z1〉.

Example 1.1.2. Suppose u is represented by−→PQ for the points P (2,−1, 1)

and Q(−5, 3, 4). Find u.

Solution.u = 〈−5− 2, 3− (−1), 4− 1〉 = 〈−7, 4, 3〉


P

Q

Figure 1.3: The vector−→PQ

The magnitude of a vector is the length of any directed line segmentwhich represents it, and the direction of a nonzero vector is the direction ofany directed line segment which represents it. If P = Q, then we interpret−→PQ as the point P which represents 0. If v is represented by

−→PQ we will

write v =−→PQ, and feel free to refer to

−→PQ itself as a vector. See Figure 1.3.

If O is the origin, then the directed line segment−→OP is often referred to as

the position vector of the point P . Directed line segments that do not startat the origin are sometimes referred to as displacement vectors, especially inphysics.

In light of these considerations, every directed line segment of a particularlength and direction represents the same vector. For example, given thepoints P (−1,−2, 1), Q(1,−1, 1), R(1, 1, 0), and S(3, 2, 0) the directed line

segments−→PQ and

−→RS have the same magnitude and the same direction.

Therefore, even though−→PQ and

−→RS are different directed line segments they

represent the same vector, and with this understanding, we write−→PQ =−→

RS. See Figure 1.4. This means that we are free to locate or place vectorswherever we choose: the magnitude and direction are what determine thevector.

In conformity with previous notation, the magnitude (or length) of−→PQ

will be denoted∣∣∣−→PQ∣∣∣.

Addition of two vectors can be defined, geometrically, by (see Figure 1.5):

−→PQ+

−→QR =

−→PR.


P

Q

R

S

Figure 1.4:−→PQ =

−→RS

It is probably worth mentioning that this implies that

−→PR−

−→PQ =

−→QR.

Again, see Figure 1.5.One can define scalar multiplication, geometrically, as follows (see Fig-

ure 1.6): Given a scalar t and a vector−→PQ,∣∣∣t(−→PQ)∣∣∣ = |t|∣∣∣−→PQ∣∣∣ .

If t > 0, then t(−→PQ)

has the same direction as−→PQ; if t < 0, then t

(−→PQ)

has the direction opposite of−→PQ.

The reader should be able to see that all of these definitions are consistentwith the definitions, in terms of components, given earlier.

Remark. A somewhat more formal approach to defining vectors geometricallyis to define a vector as the set of all directed line segments that have the samelength and the same direction. Here we consider a point to be a degenerate

directed line segment. In this approach we write v =[−→PQ], where

[−→PQ]

denotes the set of directed line segments with the same length and the same

direction as−→PQ. Thus, if

−→PQ and

−→RS have the same length and the same

direction, then[−→PQ]

=[−→RS]. The set consisting of all points is denoted by

0.


P

Q

R

Figure 1.5:−→PQ+

−→QR =

−→PR

PQ t PQt>1

t PQ-1<t<0

→ →

→

Figure 1.6: Multiplication of a vector by scalar


These sets partition the set of all directed line segments (each directed linesegment belongs to one and only one of these sets) and are called equivalenceclasses.

We then define addition and scalar multiplication for directed line seg-ments as above. This is used to define addition and scalar multiplicationfor equivalence classes, that is, vectors. One needs to observe that these

are well-defined: If−−−→P1Q1 ∈

[−→PQ]

and−−−→Q1R1 ∈

[−→QR], then

−−−→P1R1 ∈

[−→PR].

Moreover, if c is a scalar, then c(−−−→P1Q1

)∈[c(−→PQ)]

. Likewise, any other

vector operation that is defined in terms of directed line segments has to beshown to be independent of the directed line segments chosen to representthe vectors.

Once these things been established it is customary to remove the brackets

and write, say,−→PQ for

[−→PQ].

Plane Vectors

Vectors in the plane can be written in terms of their two cartesian compo-nents. So, a vector in the plane can be written

u = 〈u1, u2〉,

where u1 and u2 are real numbers. When we are considering vectors in theplane, we let

i = 〈1, 0〉 and j = 〈0, 1〉,

so we can write

〈u1, u2〉 = u1i + u2j.

All of the material in this section can be applied to vectors in the plane(and, indeed, to vectors in higher dimensional euclidean spaces) in the obvi-ous way.

It is sometimes useful to remember that the plane can, of course, beviewed as a subset of space. That is, we can identify the point (x, y) in theplane with the point (x, y, 0) in space and the vector 〈x, y〉 in the plane withthe vector 〈x, y, 0〉 in space, whenever this is helpful.


P Q

RS

T

Figure 1.7: Example 1.1.3

Application

Picturing vectors as directed line segments is a tremendous aid in the sciencesand engineering. We can also use this approach to prove various standardresults from plane geometry.

Example 1.1.3. Prove that the diagonals of a parallelogram bisect eachother.

Solution. Consider Figure 1.7. Write−→PT = s

(−→PR)

and−→QT = t

(−→QS)

. We

need to prove that s = t = 1/2.Using the geometric definition of addition of vectors, we have

−→PT = s

(−→PR)

= s(−→PQ+

−→QR)

−→QT = t

(−→QS)

= t(−→PS −

−→PQ).

Now,−→PT =

−→PQ +

−→QT and, since the figure represents a parallelogram,−→

QR =−→PS. Making these substitutions in the previous equations, we have

−→PQ+

−→QT = s

(−→PQ+

−→QR)

−→QT = t

(−→QR−

−→PQ).

Solving each of these equations for−→QT and equating the results, we obtain

(s− 1 + t)−→PQ = (t− s)

−→QR.


Since−→PQ and

−→QR are nonzero, nonparallel vectors, this implies

s− 1 + t = 0

t− s = 0.

The unique solution to this system of equations is s = t = 1/2, as desired.

1.1.1 Exercises

1. Given P (1, 1, 0), Q(3, 2, 0), and R(−1,−2, 1), find−→PQ+

−→PR.

2. Let a = 3i− j and b = i + 2j. Find 2a + b.

3. Let a = 2i− j + k and b = −i + 3j + 2k. Compute

(a) a− 2b

(b) |b|

4. Let a = 〈3, 1, 2〉 and b = 〈0,−4, 5〉. Compute

(a) 6a− 2b

(b) |6a− 2b|

5. Let a = 3i + 4j and b = 2i + 2j− k. Compute

(a) 2a− 3b

(b) |2a− 3b|

6. Find a unit vector in the direction of 〈1, 2,−1〉.

7. Find the vector of magnitude 2 with the same direction as −4i + 5j.

8. Find a vector of magnitude 2 whose direction is opposite that of a =−4i + 5j.

9. Find the vector of magnitude 3 whose direction is opposite 6i + 3j.

10. Find a vector of length 4 in the direction of i− j + 2k.

11. Prove, using components, that |tu| = |t||u| for vectors in space.

12. Prove that medians of a triangle intersect at a single point.

1.2. THE DOT PRODUCT 13

13. Complete the proof of Theorem 1.1.1 for vectors in space.

14. Using only the properties of a vector space given in Theorem 1.1.1,prove that, for every vector a, 0a = 0.

1.2 The Dot Product

In this section we define the dot product and study some of its properties.We will be working exclusively in space in this section. The reader shouldhave no problem adapting this material to the plane and higher dimensionalspaces.

Definition 1.2.1. Let a = 〈a1, a2, a3〉 and b = 〈b1, b2, b3〉. Then the dotproduct a · b is given by

a · b = a1b1 + a2b2 + a3b3.

Example 1.2.1. Let a = 〈2,−2, 1〉 and b = 〈−5, 4, 7〉. Compute a · b.

Solution.a · b = 2(−5) + (−2)(4) + 1 · 7 = −11

The dot product is also called the scalar product or the inner product. Itis very important to remember that the result of taking the dot product oftwo vectors is a scalar (i.e., a number).

The following basic properties can be easily proven using components.

Theorem 1.2.1. For any vectors a, b, c and any scalar r each of the fol-lowing is true.

1. a · 0 = 0

2. a · b = b · a

3. (a + b) · c = a · c + b · c

4. a · (b + c) = a · b + a · c

5. r(a · b) = (ra) · b


6. |a|2 = a · a

Again, we prove the first of these and leave the remainder as an exercise.

Proof of a · 0 = 0. Write a = 〈a1, a2, a3〉. Then

a · 0 = 〈a1, a2, a3〉 · 〈0, 0, 0〉= a1 · 0 + a2 · 0 + a3 · 0 = 0.

The following results are very important.

Theorem 1.2.2 (The Cauchy-Schwarz Inequality). For any vectors a andb,

|a · b| ≤ |a||b|.

Theorem 1.2.3 (The Triangle Inequality). For any vectors a and b,

|a + b| ≤ |a|+ |b|.

Theorem 1.2.4 (The Pythagorean Theorem).

|a + b|2 = |a|2 + |b|2

if and only if a · b = 0.

The Angle Between Two Vectors

We can use the dot product to compute the angle between two nonzerovectors. See Figure 1.8. When we talk about the angle between two vectorswe mean the smaller angle without regard to direction. So, the angle betweentwo vectors is always between 0 and π.

Theorem 1.2.5. If a 6= 0 and b 6= 0 and θ is the angle between a and b,then

θ = cos−1

(a · b|a||b|

),

where 0 ≤ θ ≤ π.


θ

Figure 1.8: The angle between two vectors

This result is proven in most standard calculus textbooks. See, for exam-ple, Stewart [22]. Notice that the Cauchy-Schwarz inequality implies that

−1 ≤ a · b|a||b|

≤ 1,

so the theorem at least makes sense.

Example 1.2.2. Find the angle θ between a = 〈2,−2, 1〉 and b = 〈−5, 4, 7〉.

Solution.

θ = cos−1

(2(−5) + (−2)(4) + 1 · 7√

22 + (−2)2 + 12√

(−5)2 + 42 + 72

)= cos−1

(−11

9√

10

)

If the angle between a and b is 0 or π, then a and b are parallel. If theangle between a and b is π/2, then a and b are perpendicular (or orthogonal).In particular, two nonzero vectors are perpendicular if and only if their dotproduct is 0.

The Decomposition of a Vector

Given two nonzero vectors a and b we can use the dot product to write

a = a‖ + a⊥,


b

a

a

a

⊥

II

Figure 1.9: The decomposition of a vector


where a‖ is parallel to b and a⊥ is perpendicular to b. See Figure 1.9.If θ is the angle between a and b, then

a‖ = |a| cos θb

|b|= |a|

(a · b|a||b|

)b

|b|=

a · b|b|2

b.

We can then use the equation

a⊥ = a− a‖,

to compute a⊥. We also note that, since a‖ and a⊥ are orthogonal,

|a|2 =∣∣a‖∣∣2 + |a⊥|2 .

Example 1.2.3. Let a = 〈2,−2, 1〉 and b = 〈−5, 4, 7〉. Write

a = a‖ + a⊥,

where a‖ is parallel to b and a⊥ is perpendicular to b.

Solution.

a‖ =a · b|b|2

b =−11

90〈−5, 4, 7〉 =

⟨11

18,−22

45,−77

90

⟩

a⊥ = a− a‖ =

⟨25

18,−68

45,167

90

⟩

In the above, a‖ is also called the projection of a onto b, and is oftenwritten projba. (This notation has the advantage of including b explicitly.)

The scalar

|a| cos θ =a · b|b|

is called the component of a in the direction of b, written compba.Note that

projba = compba

(b

|b|

).


Application

If a constant force F, acting in the direction of motion, moves an object alonga straight line from the point P to the point Q, then the work W is given by

W = |F|∣∣∣−→PQ∣∣∣ .

For the slightly more complicated situation where the constant force doesnot necessarily act in the direction of motion the work is given by

W =(

comp−−→PQ

F) ∣∣∣−→PQ∣∣∣ = F ·−→PQ.

Example 1.2.4. A girl is pulling a sled horizontally in a straight line adistance of 100 meters (m) by exerting a constant force of 50 Newtons (N).The rope she is pulling on is at an angle 45◦ above the horizontal. How muchwork is done?

Solution. Recall thata · b = |a||b| cos θ,

where θ is the angle between a and b.Therefore, the work is given by

W = (50N.)(100m.) cos (45◦) =5000√

2N-m ≈ 3535.5 N-m

(Note: 1 N-m = 1 joule = 1 J.)3

1.2.1 Exercises

1. Let a = 〈3, 1, 2〉 and b = 〈0,−4, 5〉. Compute a · b.

2. Let a = 3i + 4j and b = 2i + 2j− k. Compute a · b.

3. Let a = 2i− j + k and b = −i + 3j + 2k. Find a · b.

4. Find the angle between the vectors in Exercise 1.

3In the English System of units, the work done in moving an object in a straight linea distance of 1 foot (ft) by a constant force of 1 pound (lb), acting in the direction ofmotion, is 1 foot-pound (ft-lb).




7. Let a = 3i− j and b = i + 2j. Find cos θ, where θ is the angle betweena and b.

8. Given a = 〈1, 2,−1〉 and b = 〈2, 2, 1〉, find the cosine of the anglebetween a and b.

9. Find the cosine of the angle between 4i− 3j + k and i + 2j− 2k.

10. Given P (1, 2,−2), Q(3, 1,−2), and R(1,−1, 1), find the cosine of the

angle between−→PQ and

−→PR.

11. Find the angle between the vectors 〈1, 1, 1, 1〉 and 〈1, 2, 3, 4〉.

12. Let a = −4i + 5j and b = 6i + 3j. Find the component of b on a.

13. Given P (2,−1,−4), Q(1, 1, 3), R(−4, 5,−1), and S(−3, 4,−5), find the

component of−→PS on

−→QR.

14. Given P (−3, 0,−4), Q(−3, 1,−1), R(1, 1, 0), and S(4, 0,−5), find the

component of−→PS on

−→QR.

15. Write

a = a‖ + a⊥,

where a‖ is parallel to b and a⊥ is perpendicular to b for the vectorsin Exercise 1.

16. Write

a = a‖ + a⊥,

where a‖ is parallel to b and a⊥ is perpendicular to b for the vectorsin Exercise 2.

17. Given a = 〈4,−5, 3〉 and b = 〈2, 1,−2〉 find vectors a‖ and a⊥ suchthat a = a‖ + a⊥, where a‖ is parallel to b and a⊥ is perpendicular tob.


18. A child pulls a wagon along level ground in a straight line by exertinga force of 30 pounds on a handle that makes an angle of 30◦ with thehorizontal. Find the work done in pulling the wagon 80 feet.

19. For the triangle with vertices P (−3,−3), Q(4, 0), and R(0, 5), find allof its (internal) angles.

20. Prove the parallelogram law:

|a + b|2 + |a− b|2 = 2(|a|2 + |b|2

).

Interpret this geometrically.

21. Complete the proof of Theorem 1.2.1.

22. Prove Theorem 1.2.2. (Hint: If a and b are linearly independent, then|a−λb|2 > 0, for every real number λ. Do the linearly dependent caseseparately.)

23. Prove Theorem 1.2.3. (Hint: Use the Cauchy-Schwarz inequality.)

24. Prove Theorem 1.2.4.

1.3 The Cross Product

In this section we define the cross product and study some of its properties.

Definition 1.3.1. Let a = 〈a1, a2, a3〉 and b = 〈b1, b2, b3〉. Then the crossproduct a× b is given by

a× b = (a2b3 − a3b2)i + (a3b1 − a1b3)j + (a1b2 − a2b1)k.

Unlike all of the other binary operations we’ve discussed, we define thecross product only for vectors in space. The cross product is also called thevector product. It is very important to remember that the result of takingthe cross product of two vectors is a vector.

For most people the definition we’ve given for the cross product is difficultto remember. Fortunately, there is a much easier way to compute it. If weformally expand the symbolic determinant∣∣∣∣∣∣

i j ka1 a2 a3

b1 b2 b3

∣∣∣∣∣∣

1.3. THE CROSS PRODUCT 21

along the first row as we would an ordinary determinant, we obtain thecorrect expression for the cross product. We refer to the array above as asymbolic determinant since it is not an actual determinant, but simply acomputational aid.

Example 1.3.1. Let a = 〈2,−2, 1〉 and b = 〈−5, 4, 7〉. Compute a× b.

Solution.

a× b =

∣∣∣∣∣∣i j k2 −2 1−5 4 7

∣∣∣∣∣∣ = −18i− 19j− 2k = 〈−18,−19,−2〉

The following basic properties can be proven using the definition. How-ever, they can also be demonstrated using elementary properties of determi-nants.

Theorem 1.3.1. For any vectors a, b, c and any scalar r each of the fol-lowing is true.

1. a× b = −b× a

2. (a + b)× c = a× c + b× c

3. a× (b + c) = a× b + a× c

4. r(a× b) = (ra)× b

Note that the cross product is not commutative. It is also not associative.Geometrically,

|a× b| = |a||b| sin θ,

where θ is the angle between a and b. If a×b is nonzero, then the directionof a× b is perpendicular to both a and b. Now, this leaves two possibilitiesfor the direction a× b. It turns out that the correct choice is given by theright-hand rule: Point the fingers of your right hand in the direction of a androtate them in the direction of b through the smaller angle, your extendedthumb will then point in the direction of a × b. See Figure 1.10. This isthe direction that a right-handed screw will progress if it is rotated in thedirection just described. Note that this is consistent with the way we depict


a

b

a x b

Figure 1.10: The cross product of two vectors


a

b

Figure 1.11: The parallelogram generated by a and b

the xyz-axes, since i× j = k. Again, these facts are proven in most standardcalculus textbooks. See, for example, Stewart [22].

It follows that the cross product can be expressed as

a× b = (|a||b| sin θ) n,

where θ is the angle between a and b and n is the unit vector perpendicularto both a and b whose direction is given by the right-hand rule.

The Area of a Parallelogram

Given a parallelogram with adjacent sides PQ and PR, we say that the

parallelogram is spanned (or generated) by a =−→PQ and b =

−→PR. See

Figure 1.11.By trigonometry, it is easy to see that the area of this parallelogram is

|a||b| sin θ,

where θ is the angle between a and b. But this is precisely |a× b|.

Example 1.3.2. Find the area of the parallelogram spanned by a = 〈2,−2, 1〉and b = 〈−5, 4, 7〉.

Solution. In the last example we found that

a× b = 〈−18,−19,−2〉.

Therefore, the area of the parallelogram is

|a× b| = |〈−18,−19,−2〉| =√

689.


The Triple Scalar Product

The expression

a · b× c =

∣∣∣∣∣∣a1 a2 a3

b1 b2 b3

c1 c2 c3

∣∣∣∣∣∣is called the triple scalar product. Notice that there is only one way to groupthe three factors that makes sense, so we don’t need to use parentheses. Also,notice that the result of the triple scalar product is, indeed, a scalar.

Given a parallelepiped with adjacent sides PQ, PR, and PS we say that

the parallelepiped is spanned (or generated) by a =−→PQ, b =

−→PR, and

c =−→PS. See Figure 1.12. Again using trigonometry, we see that the volume

of this parallelepiped is

(area of the parallelogram spanned by b and c) |a|| cos θ|,

where θ is the angle between a and b× c. Using the expression above forthe area of a parallelogram, we see that this is precisely

|a · b× c|.

Notice that this provides a geometric interpretation of the determinant.That is, the determinant of a matrix is (up to sign) the volume of the par-allelepiped spanned by its rows. This is also the case in two dimensions andeven in higher dimensions.

Example 1.3.3. Find the volume of the parallelepiped spanned by a =〈1, 0, 1〉, b = 〈−2,−1, 1〉, and c = 〈0, 2, 3〉.

Solution. We compute ∣∣∣∣∣∣1 0 1−2 −1 10 2 3

∣∣∣∣∣∣ = −9.

So, the volume is | − 9| = 9.

1.3.1 Exercises

1. Let a = 〈3, 1, 2〉 and b = 〈0,−4, 5〉. Compute a× b.

2. Let a = 2i− j + k and b = −i + 3j + 2k. Find a× b.


b

c

a

b x c

Figure 1.12: The parallelepiped generated by a, b, and c


3. Let a = 3i + 4j and b = 2i + 2j− k. Compute b× a.

4. Find the area of the parallelogram spanned by the vectors a and b inExercise 1.



7. Find a vector perpendicular to the plane that contains the pointsP (−5, 4,−4), Q(2, 5,−4), and R(5,−5, 4).

8. Find the two unit vectors that are perpendicular to the plane thatcontains the points (3,−6, 4), (2, 1, 1), and (5, 0,−2).

9. Show, by example, that the cross product is not associative.

10. Find the area of the triangle with vertices (5, 4, 0), (−4, 3,−3), and(−1, 1,−4).

11. Find the area of the triangle with vertices (−5,−1,−4), (4,−1, 0), and(3, 4, 2).

12. Find the area of the triangle with vertices (5, 1, 6), (−1,−2,−3), and(4,−2, 1).

13. Find the area of the triangle with vertices (2, 3, 1), (1,−1, 2), and(−4, 3, 0).

14. Given P (1, 1, 0), Q(3, 2, 0), and R(−1,−2, 1) find the area of the trian-gle with vertices P , Q, and R.

15. There are two parallelograms with vertices P (1, 2,−2), Q(3, 1,−2), andR(1,−1, 1). Find the fourth vertex and the area of each of these par-allelograms.

16. Given P (2, 1,−1), Q(3, 0, 2), R(4,−2, 1), and S(5,−3, 0), find the vol-ume of the parallelepiped having adjacent sides PQ, PR, and PS.

17. Let O be the origin and P , Q, and R be as in Exercise 14. Find thevolume of tetrahedron with vertices O, P , Q, and R.

1.4. LINES AND PLANES IN SPACE 27

18. Find the volume of the parallelepiped in four dimensional euclideanspace spanned by the vectors: 〈1, 1, 0, 0〉, 〈0, 1, 1, 0〉, 〈0, 0, 1, 1〉, and〈0, 1, 0, 1〉.

19. Show that the area of the parallelogram in the plane spanned by a =〈a1, a2〉 and b = 〈b1, b2〉 is equal to the absolute value of∣∣∣∣a1 a2

b1 b2

∣∣∣∣ .20. Prove Theorem 1.3.1.

21. Use Theorem 1.3.1. to prove the following:

(a) a× 0 = 0

(b) If a and b are parallel, then a× b = 0.

1.4 Lines and Planes in Space

Much of this text is devoted to studying curves and surfaces in space. Thesimplest example of a curve is a line; the simplest example of a surface is aplane.

Lines

A line is determined by a point and a direction. If we choose a point withposition vector R0 = 〈x0, y0, z0〉 and a vector v = 〈a, b, c〉 which specifies thedirection, then

R = R0 + tv, −∞ < t <∞,

is called a vector equation of the given line. See Figure 1.13. Note that thehead of the vector R traces out the line as t ranges from −∞ to ∞.

Writing R = 〈x, y, z〉, and equating components, one obtains

x = x0 + ta, y = y0 + tb, z = z0 + tc, −∞ < t <∞.

These are called parametric equations of the given line and t is referred toas a parameter. (In the sequel, if we are considering an entire line, we willusually omit the domain −∞ < t <∞.)


O

R

R

0

v

Figure 1.13: A line in space

If a 6= 0, b 6= 0, and c 6= 0, then we can eliminate t, and obtain

x− x0

a=y − y0

b=z − z0

c.

These are called symmetric equations for the line.If one or more of a, b, and c are 0, then the symmetric equations take a

slightly different form. For example, if a 6= 0, b 6= 0, and c = 0, then we write

x− x0

a=y − y0

b, z = z0,

for the symmetric equations. We leave it to the reader to consider the otherpossibilities.

Example 1.4.1. Find equations of each type for the line through P (1, 4,−1)and Q(2, 2, 7).

Solution. We can take v =−→PQ = 〈1,−2, 8〉 and use P for the known point

on the line. Therefore,

R = 〈1, 4,−1〉+ t〈1,−2, 8〉


is a vector equation,

x = 1 + t, y = 4− 2t, z = −1 + 8t

are parametric equations, and

x− 1 =y − 4

−2=z + 1

8

are symmetric equations for the given line.

Example 1.4.2. Determine whether the lines given by

R = 3i + 2j + (2i + j + k)t and R = i− 2k + (j + k)t

intersect and, if they do, find the point(s) of intersection.

Solution. Parametrically, these can be written

x = 3 + 2t, y = 2 + t, z = t, and

x = 1, y = s, z = −2 + s.

(Note that we use different letters to denote the parameters, since the linescan intersect at a point where the values of the parameters are different.)This leads to three equations with two unknowns:

3 + 2t = 1, 2 + t = s, t = −2 + s.

Solving the first two equations gives t = −1 and s = 1. Since, this is alsoa solution to the third equation, we find that the system of equations isconsistent and t = −1, s = 1 is the only solution.

Substituting into the parametric equations gives (1, 1,−1) as the uniqueintersection point.

Example 1.4.3. Find the (smaller) angle between the two lines in the pre-vious example.

Solution. The (smaller) angle between the two lines is

θ = cos−1

(〈2, 1, 1〉 · 〈0, 1, 1〉|〈2, 1, 1〉||〈0, 1, 1〉|

)= cos−1

(2√6√

2

)= cos−1

(1√3

)≈ 0.955

This is around 54.7◦.


n

P Q

O

RR0

Figure 1.14: A plane in space

Planes

A plane is determined by a point and a direction normal (i.e., perpendicular)to the plane. If we choose a point with position vector R0 = 〈x0, y0, z0〉 anda vector n = 〈a, b, c〉 which specifies a normal direction, then

n · (R−R0) = 0

is a vector equation of the given plane. See Figure 1.14. Note that the headof the vector R lies in the plane if and only if it satisfies a vector equationof the plane.

Writing R = 〈x, y, z〉 one obtains

a(x− x0) + b(y − y0) + c(z − z0) = 0,


orax+ by + cz = d,

where d = ax0 + by0 + cz0.

Example 1.4.4. Find an equation for the plane through the following points:P (1, 0,−1), Q(0, 2, 0), and R(1, 2, 3).

Solution. We could solve this problem by substituting the given points intothe general equation for a plane, and then solving a system of three equationsin four unknowns. However, we can also employ the cross product.

Using the given points, we write−→PQ = 〈−1, 2, 1〉 and

−→PR = 〈0, 2, 4〉.

Now, according to the geometric interpretation of the cross product,−→PQ×−→

PR is a normal vector to the plane. So, we compute

−→PQ×−→PR =

∣∣∣∣∣∣i j k−1 2 10 2 4

∣∣∣∣∣∣ = 6i + 4j− 2k.

If 6i + 4j − 2k is normal to our plane, then so is 3i + 2j − k. Using thelatter as our normal vector to the plane and P as our known point on theplane, we obtain

3(x− 1) + 2y − (z + 1) = 0 or 3x+ 2y − z = 4

as equations for the desired plane.

With regard to this example, a normal vector can, of course, be found

using different vectors than we chose. For example, we could have used−→QP

and−→RQ to obtain a normal vector. Also, we could have used Q or R as our

known point on the plane.If a given plane intersects all three coordinate axes and none of the in-

tercepts are zero, then there is a particularly easy way to write its equation.Suppose that x0, y0, and z0 are the (nonzero) x-, y-, and z-intercepts, re-spectively, then the equation of the plane can be written

x

x0

+y

y0

+z

z0

= 1,

since this is obviously the equation of a plane and (x0, 0, 0), (0, y0, 0), and(0, 0, z0) all evidently satisfy the equation.


Example 1.4.5. Find an equation for the plane through P (1, 0, 0), Q(0, 2, 0),and R(0, 0, 3).

Solution. An equation for the desired plane is

x+y

2+z

3= 1 or 6x+ 3y + 2z = 6.

Application

Using the decomposition of a vector in the manner we have discussed previ-ously, we can determine the distance between certain geometric objects suchas two parallel planes or two skew lines, etc. Rather than trying to remem-ber a general formula for each type of problem, we suggest that the readerapproach each problem from scratch. The first step should always be to drawa picture.

Example 1.4.6. Find the distance between the plane 2x − 2y + z = 4 andthe point P (1, 2, 3).

Solution. We choose any point on the plane, say Q(2, 0, 0). Then we letn = 〈2,−2, 1〉 which is normal to the plane. The distance from the point tothe plane is then ∣∣∣compn

−→QP∣∣∣ =

∣∣∣∣∣−→QP · n|n|

∣∣∣∣∣ .See Figure 1.15. Using this expression, we obtain∣∣∣compn

−→QP∣∣∣ =

∣∣∣∣−2− 4 + 3

3

∣∣∣∣ = | − 1| = 1.

So, the distance is 1.

1.4.1 Exercises

1. Find parametric and symmetric equations for the line through thepoints P (−1, 2, 3) and Q(−2, 0, 7).


P

Q

n


2. Find parametric and symmetric equations for the line of intersection ofthe planes 2x+ y + z = 4 and 3x− y + z = 3.

3. Find parametric equations for the line containing the points (−2, 1, 5)and (6, 2,−3).

4. Find parametric and symmetric equations for the line through (2,−1, 0)and (−1, 2, 2).

5. Determine whether the following lines intersect and, if so, find the pointof intersection:

x = 3− 3t, y = 3− 3t, z = 2 + t

andx = 3 + t, y = 6 + 2t, z = −6− 2t.


x = 1− 6t, y = 3 + 2t, z = 1− 2t


and

x = 2 + 2t, y = 6 + t, z = 2 + t.


x = 3− 4t, y = −3− 2t, z = 2− 2t

and

x = 3 + 2t, y = −13− 4t, z = −6− 3t.

8. Find the (smaller) angle between the two planes in Exercise 2.

9. Find the distance between the origin and the line of Exercise 2.

10. Find an equation for the plane through P (1, 0,−1), Q(3, 3, 2), andR(4, 5, 1).

11. Find an equation of the plane through the points (5, 1,−1), (4, 3,−5),and (−2,−5,−2).

12. Find an equation of the plane through the points (−2, 4, 4), (−3,−3, 1),and (−4,−3, 4).

13. Find an equation of the plane through (2, 3, 1), (1,−1, 2), and (−4, 3, 0).

14. Find an equation of the plane containing the points (1, 2,−2), (1,−1, 1),and (3, 2,−1).

15. Find an equation of the plane through (1, 2,−1), (3, 1, 2), and (2, 3,−5).

16. Given P (2, 4, 1), Q(1,−2, 2), and R(3, 1, 2), find the equation of theplane through P , Q, and R.

17. Given the line with parametric equations

x = −4t+ 1, y = −2t− 5, z = −2t− 5,

find an equation for the plane containing this line and through thepoint (−3, 2,−2).


18. Show that the following planes are parallel and find the distance be-tween them:

5x+ y − z = −3 and 15x+ 3y − 3z = 6.

19. Show that the following planes are parallel and find the distance be-tween them:

−2x− 2y + 4z = 12 and 6x+ 6y − 12z = 24.

20. Find the distance between the origin and the plane of Exercise 10.

21. Find an equation of the plane containing the line x = y = z and thepoint P (1, 2, 3).

22. Redo Example 1.4.4 by substituting the given points into the generalequation ax+ by+ cz = d, and then solving the resulting system for a,b, c, and d.


Chapter 2

Curves in Space

2.1 Vector Functions

In this section we consider functions whose values are vectors in space. Byobvious modifications this material can be applied to functions whose valuesare vectors in the plane or in higher dimensional euclidean space (except forformulas involving the cross product).

A vector (or vector-valued) function (of one variable) can be written:

F(t) = F1(t)i + F2(t)j + F3(t)k,

where F1, F2, and F3 are real-valued functions of a real variable. Unlessotherwise stated, the domain of F is the set of real numbers for which F1, F2,and F3 are all defined. The functions F1, F2, and F3 are called the componentfunctions of F.

The sum, dot product, and cross product of vector functions are all de-fined pointwise. For example,

(F + G)(t) = F(t) + G(t).

The scalar product of a real-valued function and a vector function is alsodefined pointwise:

(sF)(t) = s(t)F(t).

Of course, if c is a constant, then

(cF)(t) = cF(t).

37

38 CHAPTER 2. CURVES IN SPACE

The development of the calculus for vector functions in large measureparallels the development for real-valued functions. Therefore, we will givea very brief treatment omitting almost all of the proofs.

Limits

Definition 2.1.1. Suppose t0 belongs to an open interval contained in thedomain of F and that limt→t0 Fi(t) = ai for i = 1, 2, 3. Then we define

limt→t0

F(t) = a1i + a2j + a3k.

We note that we could have given an “ε-δ definition” for the limit of avector function, but that turns out to be equivalent to our definition. Infact, a good exercise for the more motivated student is to formulate an “ε-δdefinition” for the limit of a vector function and prove that it is equivalentto Definition 2.1.1.

One-sided limits are defined similarly.The following basic limit laws can all be proven using components. They

are also valid for one sided limits.

Theorem 2.1.1. Suppose that limt→t0 F(t) and limt→t0 G(t) both exist. Also,suppose that limt→t0 s(t) exists. Then

• limt→t0

(cF(t)) = c limt→t0

F(t) (c a constant)

• limt→t0

(F(t) + G(t)) = limt→t0

F(t) + limt→t0

G(t)

• limt→t0

(F(t) ·G(t)) = limt→t0

F(t) · limt→t0

G(t)

• limt→t0

(F(t)×G(t)) = limt→t0

F(t)× limt→t0

G(t)

• limt→t0

(s(t)F(t)) = limt→t0

s(t) limt→t0

F(t)

Continuity

Definition 2.1.2. Suppose t0 belongs to an open interval contained in thedomain of F. F is continuous at t0 if

limt→t0

F(t) = F(t0).

2.1. VECTOR FUNCTIONS 39

We define continuity on an interval (of any type) in the same way as forreal-valued functions.

From these definitions it is easy to obtain the following:

Theorem 2.1.2. Suppose t0 belongs to an open interval contained in the do-main of F. F is continuous at t0 if and only if F1, F2, and F3 are continuousat t0.

Similarly, F is continuous on an interval (of any type) if and only if F1,F2, and F3 are.

Differentiation

Definition 2.1.3. Suppose t0 belongs to an open interval contained in thedomain of F. F is differentiable at t0 if

limh→0

1

h[F(t0 + h)− F(t0)] exists.

If this limit does exist, then we write F′(t0) for the limit (read: the derivativeof F at t0).

We say F is differentiable on an open interval if F is differentiable at everypoint in the interval. We will also use the notation dF/dt for the derivative.

Theorem 2.1.3. F is differentiable at t if and only if F1, F2, and F3 are. IfF is differentiable at t, then

F′(t) = F ′1(t)i + F ′2(t)j + F ′3(t)k.

The proof of this is quite straightforward, so we leave it as an exercise forthe reader.

If a real-valued function is differentiable at a point, then it is continuousat that point. Combining this fact with the results above, we easily obtainthe following:

Theorem 2.1.4. If F is differentiable at t, then F is continuous at t.

Example 2.1.1. Given

F(t) = sin ti + e−tj + 3k,

compute F′(t).


Solution. Differentiating each component function, we obtain

F′(t) = cos ti− e−tj.

Theorem 2.1.5. If F and G are differentiable at t, and c is a constant, thencF, F + G, F ·G, and F×G are all differentiable at t. Their derivativesare given as follows:

• (cF)′(t) = cF′(t)

• (F + G)′(t) = F′(t) + G′(t)

• (F ·G)′(t) = F′(t) ·G(t) + F(t) ·G′(t)

• (F×G)′(t) = F′(t)×G(t) + F(t)×G′(t)

If, in addition, s = s(t) is a real-valued function which is differentiable at t,then

(sF)′(t) = s′(t)F(t) + s(t)F′(t).

Note that in the formula for the derivative of the cross product of twovector functions the order of the factors is crucial, since the cross product isnot commutative.

The proof of these formulas can be given in the same way as their real-valued counterparts. (One can also use components.) We will prove theformula for the dot product, and leave the others to the reader.

Proof of (F ·G)′(t) = F′(t) ·G(t) + F(t) ·G′(t). We perform the usual trick


of subtracting and adding F(t) ·G(t+ h).

(F ·G)′(t) = limh→0

1

h[F(t+ h) ·G(t+ h)− F(t) ·G(t)]

= limh→0

1

h[F(t+ h) ·G(t+ h)− F(t) ·G(t+ h)

+F(t) ·G(t+ h)− F(t) ·G(t)]

= limh→0

1

h[(F(t+ h)− F(t)) ·G(t+ h) + F(t) · (G(t+ h)−G(t))]

= limh→0

1

h[(F(t+ h)− F(t)) ·G(t+ h)]

+ limh→0

1

h[F(t) · (G(t+ h)−G(t))]

= limh→0

1

h[F(t+ h)− F(t)] ·G(t+ h)

+ F(t) · limh→0

1

h[G(t+ h)−G(t)]

= F′(t) ·G(t) + F(t) ·G′(t)

Note that we have used some of the basic limit laws here. Also, we’veused the fact that G is continuous at t, since it is assumed to be differentiableat t. The reader should be able to fill in the details.

Example 2.1.2. Find F′(t) for

F(t) = (eti + j + t2k)× (t3i + j− k).

Solution. Using the formula for the derivative of the cross product, we obtain

F′(t) = (eti + 2tk)× (t3i + j− k) + (eti + j + t2k)× 3t2i

= −2ti +(et + 5t4

)j +(et − 3t2

)k.

The reader should check this result by first taking the cross product andthen differentiating.

The grandaddy of all differentiation rules is, of course, the chain rule.

Theorem 2.1.6 (The Chain Rule). If F is differentiable at h(t) and h isdifferentiable at t, then

d

dtF (h(t)) = h′(t)F′ (h(t)) .


This follows from the elementary calculus version using components, sowe leave the proof to the reader.

The development of higher order derivatives is essentially the same as forreal-valued functions, so, once again, we leave the details to the reader.

Integration

IfF(t) = F1(t)i + F2(t)j + F3(t)k,

then we define the definite integral of F from a to b by∫ b

a

F(t) dt =

[∫ b

a

F1(t) dt

]i +

[∫ b

a

F2(t) dt

]j +

[∫ b

a

F3(t) dt

]k,

provided all of the integrals on the right of the equal sign exist. If∫ ba

F(t) dtexists, then F is said to be integrable on [a, b].

The next theorem follows from the corresponding elementary calculusresult simply by considering components.

Theorem 2.1.7. If F and G are integrable on [a, b] and c is a constant, then∫ b

a

cF(t) dt = c

∫ b

a

F(t) dt, and∫ b

a

(F(t) + G(t)) dt =

∫ b

a

F(t) dt+

∫ b

a

G(t) dt.

By the fundamental theorem of calculus for real-valued functions, oneeasily obtains the following.

Theorem 2.1.8 (Fundamental Theorem of Calculus). If F is continuous on[a, b] and F = G′, then ∫ b

a

F(t) dt = G(b)−G(a).

Example 2.1.3. Evaluate∫ 2

0

(3t2i + etj + cos tk

)dt.


Solution. ∫ 2

0

(3t2i + etj + cos tk

)dt = t3i + etj + sin tk

∣∣∣20

= 8i + e2j + sin(2)k− j

= 8i +(e2 − 1

)j + sin(2)k

Change of variables or integration by substitution is also valid for vectorfunctions. The same proof as for the elementary calculus version, using thechain rule and the fundamental theorem of calculus, works here. One canalso give a proof using components.

Theorem 2.1.9 (Change of Variables). Suppose h is continuously differen-tiable on [a, b] and F is continuous on the image of h. Then∫ b

a

h′(t)F(h(t)) dt =

∫ h(b)

h(a)

F(t) dt.

Indefinite Integrals

If F = G′, then G is called an antiderivative of F. If G1 and G2 areantiderivatives of F defined on the same interval, say I, then

G2(t) = G1(t) + C, t ∈ I,

where C is a constant vector. This follows from the corresponding elementarycalculus result simply by considering components.

Thus, if G is an antiderivative of F, we write∫F(t) dt = G(t) + C.

The expression∫

F(t) dt is referred to as the indefinite integral of F, and Cis referred to as an arbitrary constant.

In terms of components, this can be written∫F(t) dt =

[∫F1(t) dt

]i +

[∫F2(t) dt

]j +

[∫F3(t) dt

]k.


Example 2.1.4. Suppose

F′(t) = 3t2i + etj + cos tk and F(0) = i + 2j + 3k.

Find F(t).

Solution.

F(t) =

∫ (3t2i + etj + cos tk

)dt = t3i + etj + sin tk + C,

where C satisfiesF(0) = j + C = i + 2j + 3k.

This givesC = i + j + 3k,

soF(t) = (t3 + 1)i + (et + 1)j + (sin t+ 3)k.

2.1.1 Exercises

1. GivenF(t) = sin t cos ti + eπtj + 3t5k,

compute F′(t).

2. GivenF(t) = 4 ln(5− t)i + 12

√t− 3j + 3t4k.

Find the domain of F. Also, find F′(t) and F′′(t).

3. Find the domain of F, F′(t), and F′′(t) for

F(t) = 3 ln(2 + t)i + 12√t+ 1j + 2t6k.

4. Evaluate ∫ 9

1

(3ti− 5

√tj)dt

5. Evaluate ∫ 9

4

(5t2i− 3

√tj + 2tk

)dt.

2.2. SPACE CURVES 45

6. If

u′′(t) = −6ti− 12t2j + 6k,

u′(0) = −7i− 3j + 2k, and

u(0) = 5i + 7j + 4k,

find u(t).

7. Let I be an open interval and c be a constant. Prove: If F is differ-entiable and |F(t)| = c, for all t ∈ I, then F(t) · F′(t) = 0, for allt ∈ I.

8. Let f(t) = |u(t)|n, where u is differentiable. Show:

f ′(t) = n|u(t)|n−2u(t) · u′(t).

(Hint: |u(t)|n = (u(t) · u(t))n/2.)










2.2 Space Curves

We begin this section with the definition of a curve.

Definition 2.2.1. A continuous, nonconstant, vector function R = R(t)defined on an interval (of any type) is called a curve.


O

R(t)

Figure 2.1: A space curve

In this context, we always picture R(t) with its tail (initial point) at theorigin, that is, as a position vector. See Figure 2.1. Since it is natural toidentify the position vector of a point with the point itself, we can view theimage of a curve as a set of points in space (as opposed to a collection ofdirected line segments whose tails are all at the origin). In applications, weoften think of a moving object whose position is given by R(t) at time t.

It is important to note that a curve is a vector function and that this isnot the same thing as the image of a curve.

Example 2.2.1. Describe the curve R(t) = R0 + tv.

Solution. As we’ve already seen, this represents the straight line through thepoint with position vector R0 and parallel to v.

Example 2.2.2. Describe the curve R(t) = a cosωti+a sinωtj, where a andω are positive constants.

Solution. The image of this curve is the circle in the xy-plane centered at theorigin of radius a. The point corresponding to the position vector of R(t)travels around the circle in the counterclockwise direction as t increases. Thepoint takes 2π/ω to traverse a complete circle.


−1

−0.5

0

0.5

1

−1

−0.5

0

0.5

10

5

10

15

20

xy

z

Figure 2.2: A helix

Example 2.2.3. Describe the curve R(t) = ρ cos ti + ρ sin tj + btk, where ρand b are positive constants.

Solution. The image of this curve looks like a spring and is called a (circular)helix.1 Figure 2.2 shows the helix with ρ = b = 1

If R(t) = f(t)i + g(t)j + h(t)k is a curve, then

x = f(t), y = g(t), z = h(t)

are called the parametric equations for the curve.

Example 2.2.4. Write the parametric equations for

R(t) = a cosωti + a sinωtj,

1Despite what some writers and editors of crossword puzzles seem to think, a helix isnot the same thing as a spiral, which is a planar curve.


where a and ω are positive constants.

Solution.x = a cosωt, y = a sinωt, z = 0

Example 2.2.5. Write the parametric equations for

R(t) = ρ cos ti + ρ sin tj + btk,

where ρ and b are positive constants.

Solution.x = ρ cos t, y = ρ sin t, z = bt

It’s often useful to remember that if a curve in the xy-plane is given byy = f(x), then it can be expressed as parametric equations:

x = t, y = f(t), z = 0.

Tangent Vectors

Definition 2.2.2. Let R = R(t) be a curve. If R is differentiable at t, thenv(t) = R′(t) is the velocity of R at t and v(t) = |v(t)| is the speed of R at t.

The reader should note that velocity is not the same thing as speed. Inparticular, the velocity is a vector, while the speed is a scalar.

Suppose R = R(t) is a curve and that R(t0) =−→OP , where O is the origin.

If v(t0) 6= 0, then v(t0) and any nonzero scalar multiple of v(t0) are said tobe tangent to the curve at the point P . Figure 2.3 shows why we adopt thisterminology. When we depict a tangent vector to a curve at a point P , as inFigure 2.3, we always show the vector with its tail at P .

Definition 2.2.3. Let R = R(t) be a curve. Assume that v(t) exists and isnonzero. Then the unit tangent vector T(t) is given by

T(t) =1

v(t)v(t).


v(t0)

O

R(t0 )

R(t0+h)

P

Figure 2.3: The velocity vector

Example 2.2.6. Consider the helix R(t) = ρ cos ti + ρ sin tj + btk, where ρand b are positive constants. Find v(t), v(t), and T(t).

Solution. Differentiating, we obtain

v(t) = −ρ sin ti + ρ cos tj + bk.

Using the basic identity sin2 t+ cos2 t = 1, we obtain

v(t) =√

(−ρ sin t)2 + (ρ cos t)2 + b2 =√ρ2 + b2.

So,

T(t) =1√

ρ2 + b2(−ρ sin ti + ρ cos tj + bk).

Smooth Curves

A general curve can be quite exotic. Nowhere-differentiable curves (i.e., func-tions that are continuous at every point in an interval, but not differentiableat any point in the interval) were constructed as early as 1875. See Singerand Thorpe [20]. At least partially motivated by this discovery, much has


been written in the past several decades about extremely irregular curvescalled fractals. A fractal curve is a curve for which the dimension of its im-age is strictly between 1 and 2 according to some reasonable definition ofdimension. The classic work on fractals is [13]. In addition, it is also possibleto prove the existence of so-called space-filling curves. A space-filling curve isone whose image intersects every point on a (two dimensional) surface. SeeApostol [1].

Therefore, it turns out that the requirement that a vector function simplybe continuous is not restrictive enough for our purposes, and we will limitour attention to curves that satisfy some additional properties.

Definition 2.2.4. A curve R = R(t) defined on an interval I is said to besimple if I is not a closed bounded interval and R is 1-to-1, or I = [a, b] andR is 1-to-1 on the interval [a, b) and on the interval (a, b].

Roughly speaking, a simple curve cannot cross itself, but it may join upat the ends.

If the domain of R is [a, b] and R(a) = R(b), then R is said to be a closedcurve.

Definition 2.2.5. A curve R = R(t) is said to be smooth if:

1. R is continuously differentiable on its domain.2

2. R is simple.

3. R′(t) 6= 0 for any t.

We note that a particular set of points in space can be the image ofboth a smooth curve and a nonsmooth curve. As an example, considerR(t) = t3i+ t3j and S(t) = ti+ tj. Both of these curves have the same image:the line with slope equal 1 through the origin and lying in the xy-plane.However, S is a smooth curve and R is not a smooth curve. (Why?)

We will call a set of points a [smooth] path if it is the image of some[smooth] curve. If C is a [smooth] path and R is a [smooth] curve whoseimage is C, then R is referred to as a [smooth] parametrization of C.

2If the domain of R is not an open interval, then this means that R can be extended toa continuously differentiable function defined on an open interval containing its domain.


Figure 2.4: An oriented path

Oriented Paths

In this text we will need to consider oriented paths. If C is a non-closedsmooth path, then any smooth parametrization R = R(t) can be used todefine an ordering on C in the obvious way:

R(t1) > R(t2) if t1 > t2.

By the definition of smoothness, it follows that there are exactly two orderingsor orientations (or directions) of C that can be defined in this manner. Toorient a closed path C one can divide the path into two overlapping, non-closed, paths C1 and C2, and then orient C1 and C2 separately requiring thatthe orientations coincide on C1 ∩C2. Again, it follows that there are exactlytwo orientations (or directions) of C.

If a particular orientation of a smooth path C has been chosen, then wewill refer to C as an oriented (or directed) path. The same path with the otherorientation will then be denoted −C. Pictorially, we indicate the orientation(or direction) of an oriented path by arrows pointing in the direction ofincrease along the path. See Figures 2.4 and 2.5.

The velocity vector obviously depends on the particular parametriza-tion. However, the unit tangent vector depends only on the path and itsorientation. That is, if R1 and R2 are smooth parametrizations of the


Figure 2.5: An oriented closed path


same path that determine the same orientation and R1(t1) = R2(t2), thenT1(t1) = T2(t2). Thus, we can view T as a function defined on the orientedsmooth path itself.

Arc Length

Next we discuss arc length and parametrization by arc length.

Definition 2.2.6. Suppose R(t) = f(t)i + g(t)j + h(t)k, a ≤ t ≤ b, is asmooth curve. Then the arc length of R from t = a to t = b is given by∫ b

a

[(f ′(t))2 + (g′(t))2 + (h′(t))2

]1/2dt =

∫ b

a

v(t) dt.

Notice that, since the velocity of a smooth curve is continuous, the integralin Definition 2.2.6 exists as a finite number.

We leave it as an exercise to show that the arc length of a curve is in-dependent of (smooth) parametrization. That is, if R1(t), a ≤ x ≤ b, andR2(t), c ≤ x ≤ d, are both smooth parametrizations of the same path, then∫ b

a

|R′1(t)| dt =

∫ d

c

|R′2(t)| dt.

It therefore makes sense to talk about the arc length of a smooth path.

Example 2.2.7. Find the arc length of the helix

R(t) = ρ cos ti + ρ sin tj + btk,

where ρ and b are positive constants, from (ρ, 0, 0) to (ρ, 0, 2πb).

Solution. First note that (ρ, 0, 0) corresponds to t = 0 and (ρ, 0, 2πb) cor-responds to t = 2π. In the previous example we found that the speed isv(t) =

√ρ2 + b2. Therefore, the desired arc length is∫ 2π

0

√ρ2 + b2 dt = 2π

√ρ2 + b2.


Suppose that R(t) is a smooth curve defined at t0. Then the arc lengthfrom t0 to t is given by

s = s(t) =

∫ t

t0

v(τ) dτ.

Now, by the fundamental theorem of calculus and the definition of smooth-ness s′(t) = v(t) > 0. Therefore, s(t) is strictly increasing and thus invertible.This means that one can (at least theoretically) solve the equation s = s(t)for t = t(s). Thus, a smooth curve can always be “parametrized by arclength.”

Henceforth, when we use “s” as a parameter for a curve we will alwaysinterpret it as arc length.

We trust that the following example will help clarify these ideas.

Example 2.2.8. Reparametrize the curve

x = sin t, y = cos t, z = t, 0 ≤ t ≤ 2π,

by arc length.

Solution. For this curve v(t) =√

2, so

s = s(t) =

∫ t

0

v(τ) dτ = t√

2.

Therefore, t = s/√

2.Substituting, we obtain

x = sin

(s√2

), y = cos

(s√2

), z =

s√2, 0 ≤ s ≤ 2π

√2.

Note that we can choose the lower limit of integration in the integralabove to be any convenient number in the domain of the curve. In fact, wecan even allow the integral to be improper, as long as it is convergent.

In what follows it will be useful to remember that

ds =ds

dtdt = v(t) dt.


ds is referred to as the element of arc length. Also, we have

dR

ds=dR/dt

ds/dt=

v(t)

v(t)= T(t),

so dR/ds is always a unit vector. For this reason a curve that is parametrizedby arc length is often called a unit speed curve.

2.2.1 Exercises

1. Sketch the image of the curve R(t) = t2i + t−2j, t > 0, in the plane.On the same plot, sketch R(1) and v(1).

2. SupposeR(t) = −4t2i + 4

√tj + (t2 − 1)k

is the position vector of a moving particle. Find the velocity and thespeed at time t = 4.

3. Given R(t) = t2i + 23t3j + tk, find (a) T(t) and (b) the value of T at

the point(1, 2

3, 1).

4. The coordinates of a moving point are given as follows:

x = 3t cos t, y = 3t sin t, z = 4t.

Find T(t).

5. Find the unit tangent vector for the circle

x2 + y2 = a2, z = 0, (a > 0 constant),

oriented counterclockwise. Write the answer in terms of x and y.

6. Find symmetric equations for the line tangent to the curve

x = t2, y = t3, z = 1− t

through the point (1,−1, 2).

7. Find symmetric equations of the tangent line to the path parametrizedby

R(t) = 2 cos ti + 6 sin tj + tk at t = π/3.


8. Find parametric equations for the tangent line to the curve

x = −t− 8, y = t2 − 3, z = 2t− 5

at the point (−9,−2,−3).


x = 4√t, y = t2 − 10, z =

4

t

at the point (8, 6, 1).


x = t+ 3, y = −t3 − 7, z = −3t− 4

at the point (1, 1, 2).

11. For the curve defined by

x = 2 cos t, y = −t, z = 2 sin t,

find (a) the velocity at any time t and (b) the arc length of the curvefrom t = 0 to t = 2π.

12. For the spiralx = t cos t, y = t sin t, t ≥ 0,

find an equation for the tangent line through (0, π/2).

13. Let C be the path parametrized by R(t) = 〈t, cos 2t, sin 2t〉.

(a) Find the arc length of C from t = 0 to t = π.

(b) Find parametric equations of the line tangent to C at the point(π/2,−1, 0).

14. Find the arc length of the curve defined by

x = t3, y = 3t2, z = 6t

from t = 2 to t = 4.


15. Find the arc length of the path parametrized by

x =2

3t3, y = 2t2, z = 4t, 0 ≤ t ≤ 1.

16. Find the arc length of the spiral

x = t cos t, y = t sin t, t ≥ 0,

from t = 0 to t = π/2. (Hint. You may want to look the integral up ina table or use computer software.)

17. Find the arc length of the hypocycloid with four cusps (or astroid)

x = a cos3 t, y = a sin3 t, 0 ≤ t ≤ 2π.

18. Reparametrize the curve

x = et cos t, y = et sin t, z = et

by arc length.

19. For the curve

x = sin t− t cos t, y = cos t+ t sin t, z = t2

find T(t), t 6= 0.

20. Find the arc length from (π,−1, π2) to (−2π, 1, 4π2) for the curve inExercise 19.

21. Suppose a curve in the xy-plane is given by y = f(x). Show that thearc length from x = a to x = b is given by∫ b

a

[1 + [f ′(x)]

2]1/2

dx.

Find the arc length of the curve given by y = ln(secx) from x = 0 tox = π/4.


22. Prove that the arc length of a curve is independent of (smooth) pa-rametrization. (Hint: If R1(t), a ≤ x ≤ b, and R2(t), c ≤ x ≤ d,are both smooth parametrizations of the same path, then there exits acontinuously differentiable function h from [c, d] onto [a, b] such that h′

is nonvanishing and R2(t) = R1(h(t)).)

23. Prove: If R1 and R2 are smooth parametrizations of the same path thatdetermine the same orientation and R1(t1) = R2(t2), then T1(t1) =T2(t2). (Hint: See the hint for Exercise 22. In this case, h′ is positive.)

2.3 More About Curves

In this section we study curves in somewhat more detail.3 Throughout thissection, R = R(t) will be a smooth curve that is twice continuously differ-entiable. Also, as previously stated, we reserve the letter “s” for arc lengthalong the curve.

Acceleration and Curvature

Recall that the velocity is given by v(t) = R′(t) and that the speed is givenby v(t) = |v(t)|. Also recall that the unit tangent vector is given by

T(t) =v(t)

v(t)or T(s) =

dR

ds.

We define the curvature by

k =

∣∣∣∣dTds∣∣∣∣ .

In terms of t this can be written

k =

∣∣∣∣dTdt /dsdt∣∣∣∣ =

1

v(t)

∣∣∣∣dTdt∣∣∣∣ .

One can show that the curvature of a straight line is 0, and that the curvatureof a circle is 1/a, where a is the radius of the circle.

3This section may be omitted without loss of continuity.

2.3. MORE ABOUT CURVES 59

By Exercise 7, Section 2.1.1, dT/dt is always perpendicular to T. There-fore, we define N, the principle unit normal vector, by

N =dT/dt

|dT/dt|

or

N =dT/ds

|dT/ds|=

1

k

dT

ds.

The acceleration is defined by

a(t) = v′(t) = R′′(t).

If we write v = vT and differentiate both sides of this equation withrespect to t, we obtain

a =dv

dtT + v

dT

dt

=dv

dtT + kv2N.

Thus, the tangential component of the acceleration is given by

aT =dv

dt,

and the normal component of the acceleration is given by

aN = kv2.

Since T and N are perpendicular unit vectors, it follows that

|a|2 = a2T + a2

N .

Example 2.3.1. Suppose the coordinates of a moving point are given asfollows:

x = t, y =1√2t2, z =

1

3t3.

Find T, N, k, aT , and aN .


Solution. First, we compute

dx

dt= 1,

dy

dt= t√

2,dz

dt= t2.

So,

v =(1 + 2t2 + t4

)1/2= 1 + t2,

and we obtain

T =1

1 + t2

(i + t√

2j + t2k).

Differentiating, we find

dT

dt=

1

(1 + t2)2

(−2ti +

√2(1− t2

)j + 2tk

).

A straightforward calculation then shows that∣∣∣∣dTdt∣∣∣∣ =

√2

1 + t2.

It follows that

N =1

1 + t2

(−t√

2i +(1− t2

)j + t√

2k)

and

k =

√2

(1 + t2)2 .

Finally,

aT =dv

dt= 2t

andaN = kv2 =

√2.

Example 2.3.2. Derive the following alternative formula for the curvature:

k =|a× v|v3

.


Solution. We start out with

a =dv

dtT + kv2N.

Using the fact that T×T = 0, we have

a×T = kv2N×T.

N×T is a unit vector, so

|a×T| = kv2.

Substituting T = v/v and solving for k gives us

k =|a× v|v3

,

as desired.

The Frenet Equations

Assume for now that the smooth path in question is parametrized by arclength and that the curvature k is nonvanishing.

Recall thatdT

ds= kN.

We define B = T×N. B is a unit vector that is normal to T and to N andis called the binormal vector. See Figure 2.6. (In Figure 2.6, T and N bothlie on the page and B points directly into the page.)

We leave it as an exercise to show that

dN

ds+ kT

is also perpendicular to T and to N. It follows that we can define τ by theequation

dN

ds= −kT + τB.

τ is called the torsion.Differentiating both sides of the equation B = T×N with respect to s,

one can show thatdB

ds= −τN.


T

N

B

Figure 2.6: T, N, and B

So,

τ = −dBds·N.

The following equations from above are referred to as the Frenet equa-tions:

dT

ds= kN

dN

ds= −kT + τB

dB

ds= −τN

These equations can be used to prove the fundamental theorem of spacecurves. This says, roughly, that a smooth path with nonvanishing curvatureis completely determined up to position by its curvature and its torsion. See[16].

Finally, we note that although we have for simplicity assumed that thegiven path is parametrized by arc length, we could, using the chain rule, writeformulas for N, B, and τ in terms of an arbitrary parameter t. In any event,each of the functions T, N, B, k, and τ depend only on the smooth pathand possibly the orientation (k does not depend on the orientation), and not


on the particular parametrization. The reader should recall the discussion inthe previous section concerning independence of parametrization for T.

Example 2.3.3. Find T, N, B, k, and τ for the circular helix

x = 3 cos t, y = 3 sin t, z = 4t.

Solution. From Example 2.2.6, with ρ = 3 and b = 4, we have

T =1

5(−3 sin ti + 3 cos tj + 4k)

and v = 5. Differentiating, we obtain

dT

dt= −3

5(cos ti + sin tj)

and |dT/dt| = 3/5. It follows that

N = − cos ti− sin tj

and

B = T×N =1

5(4 sin ti− 4 cos tj + 3k).

Next, we have

k =1

v

∣∣∣∣dTdt∣∣∣∣ =

3

25.

Finally,

dB

ds=dB/dt

ds/dt=

1

v

dB

dt=

4

25(cos ti + sin tj),

so, comparing dB/ds with N, we obtain

τ =4

25.


2.3.1 Exercises

1. The coordinates of a moving point are given as follows:

x = 3t cos t, y = 3t sin t, z = 4t.

Find aT and aN .

2. Suppose a curve in the xy-plane is given by y = f(x). Derive theformula

k =|f ′′(x)|[

1 + [f ′(x)]2]3/2 .

Use this to find the curvature of the curve y = x2.

3. Find k, N, B, and τ for the circle

x2 + y2 = a2, z = 0 (a > 0 constant)

oriented counterclockwise. Write the answers in terms of x and y. (Youwere asked to find T in Exercise 5, Section 2.2.1.)

4. Show that the curve

x =3

5cos s, y = 1 + sin s, z =

4

5cos s, 0 ≤ s ≤ 2π,

is a unit speed curve and compute k, τ , T, N, and B.

5. For the curve

x = sin t− t cos t, y = cos t+ t sin t, z = t2

find k = k(t), t 6= 0.

6. The osculating plane to a smooth curve R = R(t) at the point (with po-sition vector) R(t0) is the plane through R(t0) perpendicular to B(t0).The point (with position vector)

mc = R(t0) +1

k(t0)N(t0)

is called the center of curvature of R at t0.

2.4. PLOTTING CURVES 65

ρ(t0) = 1/k(t0) is called the radius of curvature at t0. The circle of ra-dius ρ(t0), center (with position vector) mc, and lying in the osculatingplane is called the osculating circle of R at t0.

Find an equation for the osculating circle of

R(t) = cosh ti + sinh tj + tk

at the point (1, 0, 0). (Hint: cosh2 t− sinh2 t = 1.)

7. Prove that|a|2 = a2

T + a2N .

8. Show that the curvature of a straight line is 0.

9. Show thatdN

ds+ kT

is perpendicular to T and to N.

10. Prove thatdB

ds= −τN.

11. Let R = R(s) be a smooth curve with k nonzero. Show that if τ = 0for all s, then R(s) is a plane curve (i.e., the image of R(s) lies in aplane). Hint: Assume τ = 0 for all s. This implies that B is constant.(Why?) Let R0 = R(s0) for some s0 in the domain of R and consider

d

ds[(R(s)−R0) ·B] .

2.4 Plotting Curves

Matlab (with Symbolic Math Toolbox) makes it easy to plot space curvesthat are given parametrically. If you have access to Matlab try the follow-ing. You should pay careful attention to the syntax. Entering a command ora mathematical expression incorrectly is a very common mistake. You canuse the up-arrow key to edit the previous line. If you are using some othersoftware, then you will need to modify these commands. (We use “>” for theprompt. This may be different on your computer. Do not type the symbol“>”.)


> syms t

> ezplot3(cos(t)+t*sin(t),sin(t)-t*cos(t),2*t^2,[-2*pi,2*pi])

We list here the curves that have appeared in this chapter along withsuggested domains. (In some cases we have specified or changed some of theconstants.) We encourage the reader to experiment with different domains.

Note that you only need to type “syms t” once, at the beginning of, eachsession.

• x = t, y = t2, z = t3, [−2, 2]

• x = cos t, y = sin t, z = 1, [0, 2π]

• x = cos t, y = sin t, z = t, [−2π, 2π]

• x = t cos t, y = t sin t, z = t, [−2π, 2π]

• x = et cos t, y = et sin t, z = et, [−2π, 2π]

• x = 3 cos t, y = 1 + 5 sin t, z = 4 cos t, [0, 2π]

• x = sin t− t cos t, y = cos t+ t sin t, z = t2, [−2π, 2π]

• x = cosh t, y = sinh t, z = t, [−2, 2]

Planar Curves

We can also use Matlab (with Symbolic Math Toolbox) to plot planarcurves using the command “ezplot.” Try the following commands:

> syms t,x

> ezplot(cos(t)+t*sin(t),sin(t)-t*cos(t),[-2*pi,2*pi])

> ezplot(cos(2*x),[-2*pi,2*pi])

References

Matlab comes with extensive on-line help. For a general overview of Mat-lab we recommend [10].

2.4. PLOTTING CURVES 67

2.4.1 Project

1. Use Matlab or some other software to plot all of the curves listedin this section. Identify as many curves as you can. Experiment withdifferent domains.

2. Try plotting the function given by y = sin(x5) on the interval [0, 5] usingMatlab or some other software. Is the result a faithful representationof the graph of this function? (Hint: If you used the Matlab commandezplot, then the answer is: “No.”) Discuss some of the potentialpitfalls of using graphing software. In particular, what causes problemsin this example?


Chapter 3

Scalar Fields and Vector Fields

3.1 Scalar Fields

Before we consider scalar fields, we want to briefly discuss certain importanttypes of subsets of euclidean space. As usual, we will focus our attention onspace.

Regions

By a region in space we mean any nonempty set of points in space. The openball of radius ε > 0 centered at P (x0, y0, z0) is given by

Bε(P ) ={

(x, y, z) : (x− x0)2 + (y − y0)2 + (z − z0)2 < ε2}.

Given any region R a point P belongs to the interior of R, or is an interiorpoint of R, if there is an open ball centered at P that is contained in R. Pbelongs to the boundary of R, or is a boundary point of R, if every open ballcentered at P contains points in R and points not in R. Finally, P belongsto the exterior of R, or is an exterior point of R, if there is an open ballcentered at P that does not contain any of the points in R. In Figure 3.1,A is an interior point, B and C are boundary points, and D is an exteriorpoint.

A region R is said to be open if every point in R is an interior point.Since a region cannot contain any of its exterior points, a region is open ifand only if it contains none of its boundary points. A region that containsall of its boundary points is said to be closed. A region that contains some,

69

70 CHAPTER 3. SCALAR FIELDS AND VECTOR FIELDS

∙

∙

∙

∙

A

B

CD

R

Figure 3.1: A region

3.1. SCALAR FIELDS 71

but not all, of its boundary points is neither open nor closed. Note that theregion consisting of all points in space has no boundary points, so it is bothopen and closed. (The empty set is also both open and closed.)1 A region isbounded if it is contained in some open ball. A region is compact if it closedand bounded.

The boundary of the open ball of radius ε > 0 centered at P (x0, y0, z0) iscalled the sphere of radius ε > 0 centered at P (x0, y0, z0) and is given by

Sε(P ) ={

(x, y, z) : (x− x0)2 + (y − y0)2 + (z − z0)2 = ε2}.

Note that comparable definitions can be made for subsets of euclideanspace of any dimension. In the plane an open ball is called an open disk.The open disk of radius ε > 0 centered at P (x0, y0) is given by

Dε(P ) ={

(x, y) : (x− x0)2 + (y − y0)2 < ε2}.

The boundary of an open disk is called a circle.We caution, however, that even though a given region in the plane can be

viewed as a region in space, the characterization of the region (open, closed,or neither) may depend on whether it is viewed as a region in the plane oras a region in space. For example, a nonempty open subset of the plane isnever open as a subset of space. See also Exercises 4 and 5 in Section 3.3.1.

Scalar Fields

A scalar field is a real-valued function defined on a region of euclidean space,that is, a real-valued function of several variables. We assume that the readerhas studied scalar fields, including continuity, partial derivatives, and thechain rule. (Again, the reader is referred to his or her favorite calculus text.)We do, however, want to discuss the directional derivative, the gradient, andsome applications thereof, since the gradient, in particular, will play a keyrole in the sequel. As usual, we will, for the most part, restrict our attentionto space, and leave considerations in the plane and higher dimensional spacesto the reader.

Recall that a scalar field f is continuously differentiable on an open setA if f and all of its first partial derivatives are continuous on A. “Twicecontinuously differentiable,” and so forth are defined similarly.

1In any finite dimensional euclidean space the only sets that are both open and closedare the empty set and the entire space.


The Directional Derivative and the Gradient

Let u be a given nonzero vector and P (x0, y0, z0) be a given point. Take theline through P in the direction of u parametrized by arc length (so u pointsin the direction of increasing s) and with (x0, y0, z0) corresponding to s = 0:

x = x(s), y = y(s), z = z(s).

Suppose that the scalar field f is continuously differentiable on some openset containing P . The directional derivative of f at P in the direction of uis given by

Duf(P ) =d

dsf (x(s), y(s), z(s))

∣∣s=0

.

This is the rate of change of f at P in the direction of u per unit distance.Notice that Duf(P ) is not defined for u = 0 and depends only on the

direction of u.By the chain rule, we can write this as

Duf(P ) =∂f

∂x

dx

ds+∂f

∂y

dy

ds+∂f

∂z

dz

ds

=

(∂f

∂xi +

∂f

∂yj +

∂f

∂zk

)·(dx

dsi +

dy

dsj +

dz

dsk

),

where ∂f/∂x, ∂f/∂y, and ∂f/∂z are all evaluated at P , and dx/ds, dy/ds,and dz/ds are all constant. If we define

grad f =∂f

∂xi +

∂f

∂yj +

∂f

∂zk

and note thatu

|u|=dx

dsi +

dy

dsj +

dz

dsk,

then we can writeDuf(P ) = grad f(P ) · u

|u|.

grad f is called the gradient of f . The reader should always keep in mindthat the directional derivative is a scalar and that grad f(P ) is a vector.

Example 3.1.1. Find Duf(P ) for f(x, y, z) = x+ xyz at P (1, 3,−2) in thedirection of u = 〈−1, 2, 2〉.



grad f(x, y, z) = (1 + yz)i + xzj + xyk.

So,

grad f(1, 3,−2) = −5i− 2j + 3k.

Therefore,

Duf(P ) = (−5i− 2j + 3k) · 〈−1, 2, 2〉|〈−1, 2, 2〉|

=7

3.

Note that

Duf(P ) = grad f(P ) · u

|u|= |grad f(P )| cos θ,

where θ is the angle between grad f(P ) and u.

We can restate the above as follows:

Theorem 3.1.1. Suppose that the scalar field f is continuously differentiableon some open set containing P . Then the component of grad f(P ) in thedirection of u is Duf(P ).

Clearly, Duf(P ) is a maximum for f at a given point P when θ = 0or, equivalently, when u is in the direction of grad f(P ). So, we have thefollowing:

Theorem 3.1.2. Suppose that the scalar field f is continuously differentiableon some open set containing P . The maximum rate of increase of f at P(with respect to distance) is |grad f(P )|. This maximum rate of increaseoccurs in the direction of grad f(P ).

The reader should be able to formulate results concerning the maximumrate of decrease of f at P .

Example 3.1.2. Find the maximum rate of increase of f(x, y, z) = x+ xyzat (1, 3,−2). In what direction does this occur?


Solution. In the last example, we found that

grad f(1, 3,−2) = −5i− 2j + 3k.

So, the maximum rate of increase is

| − 5i− 2j + 3k| =√

38,

and this maximum rate of increase is in the direction of the vector

−5i− 2j + 3k.

Again, all of these results are valid in the plane and in higher dimensionaleuclidean spaces with the obvious modifications.

Level Sets and Level Surfaces

If f is a scalar field, then the set

Sk = {(x, y, z) : f(x, y, z) = k} ,

where k is a constant, is called a level set of f .If A is any closed set, then it can be shown that there exists a continuous

function f , defined on all of space, such that A is a level set of f . Therefore,in our definition of a smooth level surface, we shall impose conditions on fthat are stronger than mere continuity.

Definition 3.1.1. Let Sk be a level set of a scalar field f . Suppose that f iscontinuously differentiable in an open set containing Sk. Also, suppose thatgrad f 6= 0 for all points in Sk. Then Sk is a smooth level surface.

Note that a smooth level surface may be “disconnected.”2 Consider, forexample, f(x, y, z) = x2 − y2 − z2 for k > 0.

Example 3.1.3. Describe the level sets of f(x, y, z) = ax + by + cz, wherea, b, and c are constants.

2Beginning with Section 4.4, we will work with only “connected” surfaces.


Solution. These are all of the planes that have 〈a, b, c〉 as a normal vector.These are all, in fact, smooth level surfaces.

Example 3.1.4. Describe the level sets of f(x, y, z) = x2 + y2 + z2.

Solution. If k < 0, then the level set is the empty set. If k = 0, then thelevel set is the origin. If k > 0, then the level set is the sphere of radius

√k

centered at the origin. A sphere is, of course, a smooth surface.

Example 3.1.5. Describe the level sets of f(x, y, z) = |x|+ |y|+ |z|.

Solution. If k < 0, then the level set is the empty set. If k = 0, then thelevel set is the origin. If k > 0, then the level set is the surface of the cubewith vertices (±k, 0, 0), (0,±k, 0), and (0, 0,±k). The surface of a cube isa surface; however, it is not a smooth surface, because of the edges and thevertices.

We say a vector n is normal (or perpendicular) to a given smooth surfaceat (x0, y0, z0) if whenever R is a smooth curve whose image is contained inthe smooth surface and R(t0) = 〈x0, y0, z0〉, then n is perpendicular to v(t0).The plane perpendicular to such a vector n through (x0, y0, z0) is called thetangent plane to the smooth surface through (x0, y0, z0).

We then also have the following important property of the gradient:

Theorem 3.1.3. Let Sk be a level set of a scalar field f . Suppose that f iscontinuously differentiable in an open set containing Sk. Also, suppose thatgrad f 6= 0 for all points in Sk. If (x0, y0, z0) ∈ Sk, then grad f(x0, y0, z0) isperpendicular to the smooth level surface Sk at (x0, y0, z0).

Proof. Suppose that

x = x(s), y = y(s) z = z(s)

is a smooth curve (which we can assume is parametrized by arc length)through (x(0), y(0), z(0)) = (x0, y0, z0) and contained in the level surface Sk.Let u = v(0). Then f(x(s), y(s), z(s)) = k, so Duf(x0, y0, z0) = 0.

But

Duf(x0, y0, z0) = grad f(x0, y0, z0) · u

|u|,

where u/|u| is tangent to the given curve.


Figure 3.2: A gradient vector and tangent plane to a level surface

Figure 3.2 shows a gradient vector and the tangent plane through thesame point for a smooth level surface.

Thus, if f(x, y, z) = c is a smooth level surface through (x0, y0, z0), thenthe equation of the tangent plane to the surface through (x0, y0, z0) is givenby

grad f(x0, y0, z0) · 〈x− x0, y − y0, z − z0〉 = 0

or

∂f

∂x(x0, y0, z0)(x− x0) +

∂f

∂y(x0, y0, z0)(y − y0) +

∂f

∂z(x0, y0, z0)(z − z0) = 0.

Example 3.1.6. Find an equation for the tangent plane to the level surface

x2 + yz = 5 at (2, 1, 1).

Solution. Write f(x, y, z) = x2 + yz, then

grad f(x, y, z) = 〈2x, z, y〉

and

grad f(2, 1, 1) = 〈4, 1, 1〉.


It follows that4(x− 2) + (y − 1) + (z − 1) = 0

is an equation of the tangent plane.

A surface can also be given by a function z = g(x, y). If g is continuouslydifferentiable, then the surface z = g(x, y) can be viewed as a level surfaceby writing

f(x, y, z) = g(x, y)− z = 0.

It follows that grad f is nonvanishing and the tangent plane through thepoint (x0, y0, z0) is given by

∂g

∂x(x0, y0)(x− x0) +

∂g

∂y(x0, y0)(y − y0)− (z − z0) = 0

or

z = z0 +∂g

∂x(x0, y0)(x− x0) +

∂g

∂y(x0, y0)(y − y0).

3.1.1 Exercises

In Exercises 1–6, for the given region R, find: (a) the interior of R, (b) theboundary of R, and (c) the exterior of R. Also, (d) determine whether R isopen or closed or neither.

1. R = {(x, y, z) : x2 + y2 + z2 < 1}

2. R = {(x, y, z) : x2 + y2 + z2 ≤ 1}

3. R = {(x, y, z) : x2 + y2 + z2 = 1}

4. R = {(x, y, z) : x2 + y2 < 1, z = 0}

5. R = {(x, y) : x2 + y2 < 1}

6. R = {(x, y, z) : x, y, z are rational numbers}

7. Find grad f(1, 2,−1) for f(x, y, z) = y2 sin(xz).

8. Find the directional derivative of

f(x, y) = tan−1(3xy)

at the point (4, 2) in the direction of the vector u =√

32

i− 12j.



f(x, y) = 3x2 − y2 + 5xy

at the point (2,−1) in the direction of −3i− 4j.

10. Let f(x, y) = x3 + y3− 3xy. Find the directional derivative of f at thepoint (2, 1) in the direction of 〈−4, 3〉.

11. Find the directional derivative of f(x, y) = x2 − 2y2 at (3, 3) in thedirection of 2i− j.

12. In what direction does f(x, y) = x2 − 2y2 increase most rapidly at(3, 3)? What is the value of this maximum rate of increase?

13. Find the directional derivative of f(x, y) = −x2y2 + 2x2 at P in thedirection from P (−3, 1) to Q(−6,−1). Also, find a vector in the direc-tion in which f increases most rapidly at P and find the rate of changeof f in that direction.

14. Let f(x, y) = x2 cos y.

(a) Find Duf(P ) at P (√

2, π/2) in the direction of u = 〈√

3/2, 1/2〉.(b) What is the maximum value of Duf(P ) at P (

√2, π/2)?

(c) In what direction does Duf(P ) at P (√

2, π/2) attain its maximumvalue?


f(x, y, z) = 2xz − 3yz − xy

at the point (−1, 2,−1) in the direction of −i + 3j + k.

16. Find the directional derivative Duf(P ) of f(x, y, z) = x + xyz atP (1,−2, 2) in the direction of the vector u = 2i + 2j + k.

17. Find the maximum rate of increase of f(x, y, z) = x2 + 2yz − sin z at(3, 2, 0). In what direction does this occur?

18. Describe the level sets of f(x, y, z) = max{|x|, |y|, |z|}.


19. Find an equation of the tangent plane to the surface

z = x2 + 2xy − y2

through the point (1,−2,−7).

20. Find an equation of the tangent plane to z = (2x− y)2 at (2, 3, 1).

21. Find an equation of the tangent plane to the graph of

4x2 − 2y2 − 7z = 0 at (−2,−1, 2).

22. Find an equation of the tangent plane to the ellipsoid 9x2 +4y2 +9z2 =34 at (1, 2,−1).

23. Find an equation of the tangent plane to the hyperboloid z = xy at(2,−3, 6).

24. Find an equation of the tangent plane to the sphere x2 + y2 + z2 = 21at (2, 4,−1).

25. (a) Find an equation for the tangent plane to the ellipsoid

x2 + 2y2 + 3z2 = 6 at the point (1, 1, 1).

(b) Find parametric equations for the normal line to

x2 + 2y2 + 3z2 = 6 through the point (1, 1, 1).

26. (a) Find an equation for the tangent plane to the surface

z = cos(x) sin(y) at the point (π/3, π/6, 1/4).

(b) Find parametric equations for the normal line to

z = cos(x) sin(y) at the point (π/3, π/6, 1/4).

27. Find an equation for the tangent plane to the paraboloid

z = x2 + y2 at the point (1, 2, 5).


28. Find a point P on the surface z = (2x− y)2 with the property that theplane tangent to the surface at P is parallel to 12x− 6y − z = 0.

29. (Lagrange Multipliers) Assume f and g are scalar fields and grad g isnonvanishing. Then one can show that the maximum and minimumvalues of f subject to the constraint (or side condition) g(x, y, z) = 0(if they exist) occur at those points P for which

grad f(P ) = λgrad g(P ),

for some constant (scalar) λ. (λ is called a Lagrange multiplier.)

Use this result to find the point(s) on the surface z = xy + 5 closest tothe origin. (Hint: Minimize the square of the distance. Also rememberthat the point(s) must satisfy the constraint equation as well.)

30. Given a region R, the closure of R is given by

closure R = R ∪ boundary R.

Show that the closure of a region is a closed region.

3.2 Vector Fields and Flow Curves

By a vector field in space we mean a function F defined on a region of spacesuch that F(x, y, z) is a vector in space, that is, a space-vector-valued functionof three variables. We picture the directed line segment corresponding toF(x, y, z) with its initial point or tail located at (x, y, z). One can considervector fields in different dimensional spaces, but, once again, we are focusedhere on three dimensional space. Figure 3.3 depicts a vector field in space.

We can always write

F(x, y, z) = F1(x, y, z)i + F2(x, y, z)j + F3(x, y, z)k,

where F1, F2, and F3 are scalar fields. F is continuous [continuously differ-entiable, etc.] on an open set A if and only if F1, F2, and F3 are continuous[continuously differentiable, etc.] on A.

If f is a scalar field, then grad f is a vector field. Examples from physicsinclude gravitational fields, electric fields, and magnetic fields.

3.2. VECTOR FIELDS AND FLOW CURVES 81

Figure 3.3: A vector field

Flow Curves

A smooth curve is called a flow curve (or flow line) of a vector field F if thetangent to the curve at every point is parallel to F at that point. Figure 3.4shows a vector field in the plane together with several flow curves.

The condition that R = R(t) be a flow curve of F is equivalent to

T = βF, where β = β(x, y, z),

and T is the unit tangent vector for R. Since

T =dR

ds=dx

dsi +

dy

dsj +

dz

dsk,

we obtain

βF1 =dx

ds, βF2 =

dy

ds, βF3 =

dz

ds,

ordx

F1

=dy

F2

=dz

F3

,

if F1, F2, and F3 are all nonvanishing.As an example, in the magnetic field created by one or more small bar

magnets, iron filings will line up along flow curves.

Example 3.2.1. For F(x, y, z) = x2i + y2j + zk find the flow curve throughthe point (2, 2, 1).


Figure 3.4: Flow curves

Solution. We need to solve

dx

x2=dy

y2=dz

z.

This leads us todx

x2=dz

zand

dy

y2=dz

z.

Integrating, we obtain

−1

x= ln |z|+ c1 and − 1

y= ln |z|+ c2,

where c1 and c2 are constants. Upon substituting x = y = 2 and z = 1, weobtain c1 = c2 = −1/2. Thus, the flow curve through the point (2, 2, 1) isgiven by

x = y =2

1− 2 ln z.


The Divergence

If F = F1i + F2j + F3k is a continuously differentiable vector field we definethe divergence of F by

div F =∂F1

∂x+∂F2

∂y+∂F3

∂z.

Note that div F is a scalar field.

Example 3.2.2. Find div F for F(x, y, z) = xy2i + xyj + xyk.

Solution.

div F(x, y, z) = y2 + x

Example 3.2.3. Find div F for

F(x, y, z) =xi + yj + zk

(x2 + y2 + z2)3/2.


∂F1

∂x=

(x2 + y2 + z2)3/2 − 3x2 (x2 + y2 + z2)

1/2

(x2 + y2 + z2)3

=−2x2 + y2 + z2

(x2 + y2 + z2)5/2.

By symmetry, we see that

∂F2

∂y=

x2 − 2y2 + z2

(x2 + y2 + z2)5/2and

∂F3

∂z=

x2 + y2 − 2z2

(x2 + y2 + z2)5/2.

Adding these together we obtain

div F(x, y, z) = 0, (x, y, z) 6= (0, 0, 0).


Roughly speaking, the divergence of a vector field at a point is a measureof how much the vector field diverges or expands at that point.

More precisely, the divergence of F at a point P is the flux per unitvolume at P . The flux of a vector field through a surface is a measure of howmuch the vector field “flows” through the surface. (“Flux” means “flow” inLatin.)

To make this still more precise and concrete, consider the case where v isthe velocity field of a fluid. If we take a small planar surface and approximatethe volume of the fluid flowing through the surface we obtain

v · n∆S∆t,

where v is evaluated at any point on the surface, n is a unit normal to thesurface, ∆S is the area of the surface, and ∆t is the time increment. SeeFigure 3.5. In this case, the flux is the mass of the fluid passing through thesurface per unit time. Therefore, if we let δ = δ(x, y, z) be the density of thefluid and F = δv, we obtain:

flux ≈ F · n∆S.

Note that in Figure 3.5 we have chosen the unit normal vector to be in thedirection which makes the flux positive. The opposite direction would simplychange the sign of the flux.

Now, consider a small rectangular box centered at (x, y, z) with sidesparallel to the coordinate axes and with lengths ∆x, ∆y, and ∆z, as shownin Figure 3.6. We want to use the formula above to approximate the outwardflux through the six faces of the box.

Through face I the flux is approximately

F · n1∆S = −F1(x−∆x/2, y, z)∆y∆z.

Note that here we’ve taken n1 to be the outward unit normal, since we areapproximating the flux out of the box.

Similarly, through face II the flux is approximately

F · n2∆S = F1(x+ ∆x/2, y, z)∆y∆z.

If we add these we get

[F1(x+ ∆x/2, y, z)− F1(x−∆x/2, y, z)] ∆y∆z ≈ ∂F1

∂x(x, y, z)∆x∆y∆z.


v

v

n

vΔt

∆S

vΔt

Figure 3.5: The volume of a fluid passing through a surface


(x+Δx/2,y-Δy/2,z-Δz/2)

n1

n2

I

II

(x-Δx/2,y-Δy/2,z+Δz/2)

Figure 3.6: The flux through the surface of a box

Similarly, we can approximate the flux through the other four faces. Addingeverything together we obtain(

∂F1

∂x+∂F2

∂y+∂F3

∂z

)∆x∆y∆z,

where the partial derivatives are all evaluated at (x, y, z).If we divide this expression by the volume of the rectangular box (i.e.,

∆x∆y∆z) and let that volume go to zero, then we obtain

∂F1

∂x+∂F2

∂y+∂F3

∂z,

where the partial derivatives are all evaluated at (x, y, z). But this is preciselydiv F(x, y, z).

We note that these approximations are valid up to first order if all of thecomponent functions of F are continuously differentiable.

To summarize: If F represents the density of a fluid multiplied by thevelocity field of the fluid and we imagine a small rectangular solid centeredat P (with sides parallel to the coordinate axes), then div F(P ) is approxi-mately the mass of the fluid flowing out of the rectangular solid per unit time


divided by the volume of the rectangular solid. Taking smaller and smallerrectangular solids one obtains (in the limit, as all of the sides shrink to 0)div F(P ).

A vector field for which the divergence is everywhere 0 is called solenoidal.

The Curl

We next define the curl of a vector field. If F = F1i + F2j + F3k, then

curl F =

(∂F3

∂y− ∂F2

∂z

)i +

(∂F1

∂z− ∂F3

∂x

)j +

(∂F2

∂x− ∂F1

∂y

)k.

Note that the curl of a vector field is a vector field.

We can write this as another symbolic determinant:

curl F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

F1 F2 F3

∣∣∣∣∣∣ .Example 3.2.4. Find curl F for F(x, y, z) = xy2i + xyj + xyk.

Solution.

curl F(x, y, z) =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

xy2 xy xy

∣∣∣∣∣∣ = xi− yj + (y − 2xy)k

Roughly speaking, the curl of a vector field at a point gives a measure ofthe rotation of the vector field around the point.

Once again, to make this more precise and concrete, we consider the caseof the velocity field of a fluid. Here we just take F to be the velocity field.First, we consider the rotation of the fluid in the plane z = c, where c is aconstant, about a point (x, y, z). Referring to Figure 3.7, note that er andeθ are unit vectors perpendicular and tangent, respectively, to a small circlecentered at (x, y, z). The clockwise component of F at (x+ ∆x, y+ ∆y, z) isF · eθ, while the angular rate of rotation is F · eθ/r.


x

y

(x,y,z)

r

θ

e

e

r

θ

Figure 3.7: The curl


Using the linearization of F1, we can write

F1(x+ ∆x, y + ∆y, z) ≈ F1 +∂F1

∂x∆x+

∂F1

∂y∆y

F2(x+ ∆x, y + ∆y, z) ≈ F2 +∂F2

∂x∆x+

∂F2

∂y∆y,

where all the functions on the right are evaluated at (x, y, z).

Using these expressions and noting that

eθ = − sin θi + cos θj at (x+ ∆x, y + ∆y, z)

and

∆x = r cos θ and ∆y = r sin θ,

we obtain

F · uθ ≈ −(F1 +

∂F1

∂xr cos θ +

∂F1

∂yr sin θ

)sin θ

+

(F2 +

∂F2

∂xr cos θ +

∂F2

∂yr sin θ

)cos θ,

where all the functions are evaluated at (x, y, z).

The average of this around the circle is

1

2π

∫ 2π

0

F · uθ dθ =1

2r

(∂F2

∂x− ∂F1

∂y

).

Dividing by r we obtain the average angular velocity of the fluid about theaxis through (x, y, z) and perpendicular to z = c:

1

2

(∂F2

∂x− ∂F1

∂y

).

Neglecting the factor of 1/2 we obtain the component of the curl of F inthe k direction. Similar arguments give the other components.

A vector field for which the curl is everywhere 0 is said to be irrotational.


The ∇-Operator and the Laplacian

A standard way to express the gradient, divergence, and curl is using theoperator ∇ (read: del). This is written

∇ =∂

∂xi +

∂

∂yj +

∂

∂zk.

This enables us to write

grad f =∇f, div F =∇ · F, and curl F =∇× F.

If f is a scalar field, we define the Laplacian of f by

div (grad f) =∇ · (∇f) = ∇2f = ∆f =∂2f

∂x2+∂2f

∂y2+∂2f

∂z2.

For this reason one writes

∇2 = ∆ =∂2

∂x2+

∂2

∂y2+

∂2

∂z2.

This ubiquitous operator plays a key role in many areas of mathematicsincluding complex analysis and partial differential equations. All of the clas-sical partial differential equations (the heat equation, the wave equation, andLaplace’s equation) involve the Laplacian.

The equation ∇2f = 0 is called Laplace’s equation and the solutionsare called harmonic functions. The equation ∇2f = g is called Poisson’sequation. Both Laplace’s equation and Poisson’s equation play a key role inelectrostatics.

Analogously, we can define the vector Laplacian:

∇2F =∂2F

∂x2+∂2F

∂y2+∂2F

∂z2,

where F is a vector field.

3.2.1 Exercises

1. For F(x, y) = yi− xj find the flow curve through the point (3,−4).

2. For F(x, y) = xi− yj find the flow curve through the point (−3, 4).


3. For F(x, y, z) = xi + yj + zk find the flow curve through the point(1, 2, 3).

4. For F(x, y, z) = i+yj+zk find the flow curve through the point (1, 2, 3).

5. Let F(x, y, z) = xzi− xyj + yzk. Find

(a) div F

(b) curl F

6. Let F(x, y, z) = x2i + j + yzk. Find

(a) div F

(b) curl F

7. Let F(x, y, z) = x2zi + y2xj + (y + 2z)k. Find

(a) div F

(b) curl F

8. Let F(x, y, z) = 2xyi + (x2 + z) j + y3k. Find

(a) div F

(b) curl F

9. Let F(x, y, z) = cos(xy)i + sin(xy)j. Find

(a) div F

(b) curl F

10. Let F(x, y, z) = exyz(i + j + k). Find

(a) div F

(b) curl F

11. Let G(x, y, z) = xyi + y2j + yzk. Find

(a) ∇ ·G(b) ∇×G

(c) ∇2G


12. Let f(x, y, z) = e−xy sin yz. Find

(a) ∇f(b) 4f =∇2f =∇ ·∇f

13. Let F(x, y, z) =−xi− yj− zk

(x2 + y2 + z2)3/2.

Find a scalar field φ such that F = −∇φ. (Hint: Make an intelligentguess.)

14. The density of water is very close to 1000 kg/m3. Suppose water isflowing through a water main of diameter 1/2 m at a speed of 2 m/s.What is the flux through a cross-section of the water main? Giventhat 1 m3 is approximately 264 gallons, find the flow rate in gallonsper minute.

15. Let φ be a scalar field and F a vector field both twice continuouslydifferentiable. Prove (by direct computation) that

(a) curl (gradφ) = 0

(b) div (curl F) = 0

16. Let φ be a scalar field and F a vector field both continuously differen-tiable. Prove (by direct computation) that

∇ · (φF) = F · ∇φ+ φ∇ · F.

17. Let us write u = (x, y, z) and u = xi+yj+zk. A vector field F is calleda force field if the vector F(u) is interpreted as the force acting on aparticle placed at u. A force field is called conservative if F = −gradV ,for some scalar field V . The scalar field V is called a potential for F.If u(t) is the trajectory (or orbit) of a particle of mass m moving in aconservative force field, then the total energy is

E(t) =1

2m|u′(t)|2 + V (u(t)).

(m|u′(t)|2/2 is called the kinetic energy and V (u(t)) is called the po-tential energy.)


Prove the law of conservation of energy. That is, prove that if u(t) isthe trajectory of a particle of mass m moving in a conservative forcefield, then E(t) is a constant, independent of t. (Hint: It suffices toshow that E ′(t) = 0. Use Newton’s (second) law: mu′′(t) = F(u(t)).)

18. Let x represent the distance east, y the distance north, and z the heightof an object relative to a given spot on Earth. If |x|, |y|, and z > 0 arenot too large, and we ignore air resistance, then the force field due togravity can be modeled by F = −mgk, where m is the mass and g is apositive constant.

(a) Show that V = mgz is a potential for F in the sense of Exercise 17.

(b) Suppose an object is thrown with initial speed v0 ≥ 0 from aheight h ≥ 0. Ignoring air resistance, use conservation of energy,to show that the object hits the ground with speed

vf =√v2

0 + 2gh.

19. Consider an object of mass m attached to a “linear” spring and movingalong the x-axis. Assuming the equilibrium position is at the origin andneglecting any resistance, the force acting on the object can be modeledby F(x, y, z) = −kxi, where k is a positive constant. Suppose that theobject’s speed is v0 when it passes through the origin. Show that theamplitude of the motion is v0

√m/k. (Hint: Consider Exercises 17 and

18.)

20. A complex-valued function of a complex variable

f(x+ iy) = u(x, y) + iv(x, y)

is analytic on an open set G if and only if u and v are infinitely differ-entiable on G and satisfy the Cauchy-Riemann equations:

∂u

∂x=∂v

∂yand

∂u

∂y= −∂v

∂x.

Show that if f is analytic on G, then u and v satisfy Laplace’s equationin two-variables:

∂2u

∂x2+∂2u

∂y2= 0.

(That is, the real and imaginary parts of an analytic function are har-monic functions.)


3.3 Plotting Functions of Two Variables

Matlab (with Symbolic Math Toolbox) makes it easy to plot surfaces thatare given by z = f(x, y). (To plot a level surface that cannot be written asa function of two variables it is usually easier to express the surface in para-metric form.) If you have access to Matlab try the following. You shouldpay careful attention to the syntax. Entering a command or a mathematicalexpression incorrectly is a very common mistake. If you are using some othersoftware, then you will need to modify these commands. (We use “>” for theprompt. This may be different on your computer. Do not type the symbol“>”.)

> syms x y

> ezmesh(sqrt(x^2+y^2),[-2,2],[-2,2])

You should be able to rotate the surface and view it from different per-spectives.

Note here that “sqrt” stands for the square root. Also, in Matlab,“abs” stands for absolute value, “log” stands for the natural logarithm, and“exp” stands for the exponential function. The inverse tangent is given by“atan” and similarly for other inverse trigonometric functions.

You can use “ezsurf” in place of “ezmesh” for a different looking plot.We list below some surfaces along with suggested domains. We encourage

the reader to experiment by considering different domains.Note that you only need to type “syms x y” once, at the beginning of,

each session.

• z = x2 + y2, [−2, 2], [−2, 2]

• z = x2 − y2, [−2, 2], [−2, 2]

• z = xy, [−2, 2], [−2, 2]

• z = xy2 − x3, [−2, 2], [−2, 2]

• z = xy3 − x3y, [−2, 2], [−2, 2]

• z = |x+ y|, [−2, 2], [−2, 2]

• z = cos(x) sin(y), [−2π, 2π], [−2π, 2π]

3.3. PLOTTING FUNCTIONS OF TWO VARIABLES 95

• z = ln (x2 + y2) , [−2, 2], [−2, 2]

• z = arctan (x2 + y2) , [−2, 2], [−2, 2]

• z = exp (−x2 − y2) , [−2, 2], [−2, 2]

• z = exp (−1/ (x2 + y2)) , [−2, 2], [−2, 2]

References


3.3.1 Project

Use Matlab or some other software to plot all of the surfaces listed in thissection, identifying as many as you can. Experiment with different domains.Also, rotate the surfaces to achieve different perspectives.


Chapter 4

Line Integrals and SurfaceIntegrals

4.1 Line Integrals

In this section we discuss what it means to integrate a vector field along apath.

Let R = R(t) be a smooth curve whose domain is the closed, boundedinterval [a, b]. Let C be the image (viewed as a set of points in space) of R.Finally, let F be a vector field whose domain contains C.

The construction here is very similar to that of the Riemann-Stieltjesintegral. First we choose a partition of [a, b]:

a = t0 < t1 < · · · < tn−1 < tn = b.

Then we choose a selection of points:

τi ∈ [ti−1, ti], i = 1, 2, . . . , n.

Let

R(τi) = 〈ξi, ηi, ζi〉 and ∆Ri = R(ti)−R(ti−1), i = 1, 2, . . . , n.

See Figure 4.1.Then, we define∫

C

F · dR = limn∑i=1

F(ξi, ηi, ζi) ·∆Ri,

97

98 CHAPTER 4. LINE INTEGRALS AND SURFACE INTEGRALS

O

R

R

i-1

(ξ,,η,ζ)

i

Figure 4.1: The line integral

where the limit is taken as the number of subintervals [ti−1, ti] tends to infin-ity and the length of each subinterval tends to 0, provided this limit existsindependently of the choices made above. The integral is called a line inte-gral.

We state, without proof, the following basic theorem.

Theorem 4.1.1. If F is a continuous vector field and R is a smooth curve,then

∫C

F · dR exists and has the same value for all smooth parametrizationsof C with the same orientation.

Recall that if C is an oriented smooth path, then the same smooth pathwith the opposite orientation (or reverse direction) is written −C. It is easyto see that ∫

−CF · dR = −

∫C

F · dR.

Recall that in sketches we indicate the orientation of a smooth path by arrowsalong the path.

We also would like to consider paths that are not necessarily smooth, butconsist of finitely many smooth paths joined together.

Definition 4.1.1. A curve R whose domain is the closed, bounded interval

4.1. LINE INTEGRALS 99

[a, b] is said to be piecewise smooth if R is simple and there is a partition

a = t0 < t1 < · · · < tn−1 < tn = b

of [a, b] such that, for each i, Ri = R|[ti−1, ti] is a smooth curve.1

In this case we define∫C

F · dR =

∫C1

F · dR +

∫C2

F · dR + · · ·+∫Cn

F · dR,

where C is the image of R and Ci is the image of Ri, for each i.Notice that a smooth curve defined on a closed and bounded interval is,

according to this definition a piecewise smooth curve.We will call a set of points a piecewise smooth path if it is the image of

some piecewise smooth curve. A piecewise smooth path can be oriented inexactly the same way as a smooth path.

Whenever we consider line integrals in this text we will always assumethat any path involved is piecewise smooth and oriented, even if this isn’texplicitly stated.

If R = R(t) is a smooth curve and the line integral of F over C exists,then one can show, using the mean-value theorem, that∫

C

F · dR =

∫ b

a

F · dRdt

dt,

where F is evaluated at (x(t), y(t), z(t)).This can also be written as∫

C

F1 dx+ F2 dy + F3 dz =

∫ b

a

(F1dx

dt+ F2

dy

dt+ F3

dz

dt

)dt,

where F1, F2, and F3 are evaluated at (x(t), y(t), z(t)). Some authors actuallyuse this as the definition of the line integral.

For future reference, we pause here to note that the line segment fromP (p1, p2, p3) to Q(q1, q2, q3) can be parametrized by

x = p1 + (q1−p1)t, y = p2 + (q2−p2)t, z = p3 + (q3−p3)t, 0 ≤ t ≤ 1.

1If f is a function and S is a subset of the domain of f , then f |S is defined by

(f |S) (t) = f(t), t ∈ S.

f |S is called the restriction of f to S.


Example 4.1.1. Find∫C

F · dR if F(x, y, z) = xzi−xyj + yzk and C is thestraight line segment from (1, 0, 0) to (−1, 0, 4)

Solution. The straight line segment can be parametrized as

x = 1− 2t, y = 0, z = 4t, 0 ≤ t ≤ 1.

Thus, dx/dt = −2, dy/dt = 0, and dz/dt = 4. Therefore,∫C

F · dR =

∫C

xz dx− xy dy + yz dz

=

∫ 1

0

[(1− 2t)(4t)(−2) + 0 + 0] dt

=

∫ 1

0

(16t2 − 8t) dt =4

3.

If C is a closed path, then we will usually write the line integral as∮C

F · dR.

In this case the line integral is called the circulation of F around C.

Recall that a smooth path can always be parametrized by arc length. IfR = R(s) is a smooth curve parametrized by arc length we have∫

C

F · dR =

∫C

F · dRds

ds =

∫C

F ·Tds,

where T is the unit tangent vector to the path. Since T is a unit vector,F ·T is the component of F tangent to the path. Thus, if F is a force fieldand C is a non-closed path, then

∫C

F · dR is the work done by the forcefield moving an object from the initial point of C to the terminal point of C,along C. If C is a closed path, then

∮C

F · dR is the work done in moving anobject one time around C in the appropriate direction, regardless of whereon C the object starts out. (You should be able to figure out, intuitively,why this is the case.)

4.1. LINE INTEGRALS 101

Line Integrals in the Plane

All of these considerations apply to line integrals in the plane with the obviousmodifications. As usual, we leave it to the reader to fill in the details. We do,however, want to make a few additional remarks concerning the orientationof closed piecewise smooth paths in the plane.

The Jordan curve theorem states that each simple closed path in theplane divides the plane into three pairwise disjoint regions:

• a bounded open region called the inside of the path

• an unbounded open region called the outside of the path

• the path itself which is the boundary of both the inside of the path andthe outside of the path.

The Jordan curve theorem is the epitome of an “obvious” result that is verydifficult to prove in the general case.

For a closed piecewise smooth path the counterclockwise orientation canthen be defined as follows: If you traverse the path in the counterclockwisedirection, the inside of the path will be on your left.2 The opposite orientationis called clockwise.

Example 4.1.2. Find∮C

(3x2 + 5y) dx+ (2x+ 3y2) dy

around the circle x2 + y2 = 4 oriented counterclockwise.

Solution. The circle can be parametrized as

x = 2 cos t, y = 2 sin t, 0 ≤ t ≤ 2π.

Thus, dx/dt = −2 sin t and dy/dt = 2 cos t. Therefore,∮C

(3x2 + 5y) dx+ (2x+ 3y2) dy

=

∫ 2π

0

[(6 cos2 t+ 10 sin t)(−2 sin t) + (4 cos t+ 12 sin2 t)(2 cos t)

]dt

=

∫ 2π

0

(−12 cos t sin t− 20 sin2 t+ 8 cos2 t+ 24 sin2 t cos t) dt

= −12π.

2Omitting those points, if any, where the path is not smooth.


4.1.1 Exercises

For Exercises 1–4, let C be the portion of the helix parametrized by

x = cos t, y = sin t, z = t, 0 ≤ t ≤ 2π.

1. Find∫C

F · dR, where F(x, y, z) = xj.

2. Find the work done moving an object along C from (1, 0, 0) to (1, 0, 2π)by the force field F(x, y, z) = yi.

3. Find∫Cz dx.

4. Find∫Cz ds.

5. Find∫C

F · dR where F(x, y) = yi− xj and

(a) C is the straight line segment from (0, 0) to (1, 1)

(b) C is the segment of the curve y = x2 from (0, 0) to (1, 1)

6. Let C be the perimeter of the square with opposite vertices (0, 0) and(1, 1) oriented counterclockwise. Find

∮Cx dx.

7. Let C be the perimeter of the square with opposite vertices (0, 0) and(1, 1) oriented counterclockwise. Find

∮Cy dx.

8. Let C be the perimeter of the triangle with vertices (0, 0), (1, 0), and(0, 1) oriented counterclockwise. Find

∮Cx dx.

9. Let C be the perimeter of the triangle with vertices (0, 0), (1, 0), and(0, 1) oriented counterclockwise. Find

∮Cy dx.

10. Find∫C

F · dR, where F(x, y, z) = xi− yj + zk and C is parametrizedby

x = cos t, y = sin t, z = t3, 0 ≤ t ≤ 2π.

11. Find∫C

F · dR, where F(x, y, z) = −xi + yj + zk and

(a) C is parametrized by

R(t) = cos ti + sin tj + etk, 0 ≤ t ≤ π

4.2. CONSERVATIVE FIELDS 103

(b) C is the line segment from (1, 0, 0) to (−1, 0, eπ)

12. Find∮C

F · dR, where F(x, y, z) = yi + xj + sin(xyz)k and C is thecircle

x2 + y2 − 2y = 3, z = −1,

oriented counterclockwise as viewed from above.

13. Find∫C

F · dR, where F(x, y, z) = x2i + j + yzk and C is parametrizedby

x = t, y = 2t2, z = 3t, 0 ≤ t ≤ 1.

14. Find∫C

F · dR, where F(x, y, z) = 2xyi + (x2 + z) j + y3k and C is theline segment from (1, 0, 2) to (3, 4, 1).

15. Calculate∮C

F · dR, where F(x, y, z) = yk, over the perimeter of thetriangle with vertices P (a, 0, 0), Q(0, 2a, 0), and R(0, 0, 3a), where a isa positive constant and the path is oriented from P to Q to R and backto P .

16. Prove: If F is a continuous vector field and R is a smooth curve,then

∫C

F · dR has the same value for all smooth parametrizations ofC with the same orientation. (Hint: See the hint for Exercise 22 inSection 2.2.1. In this case h′ is positive.)

4.2 Conservative Fields

Domains and Simply Connected Regions

Before taking up the subject of conservative fields we need to briefly take upthe subject of simple connectedness.

A region R is said to be arcwise connected if given any two points P andQ in R there is a path lying entirely in R that joins P and Q. A region Ris said to be convex if given any two points P and Q in R the line segmentjoining P and Q lies entirely in R. A convex region is obviously arcwiseconnected.

An open region is arcwise connected if and only if it cannot be dividedinto two disjoint open regions. In this text, we will refer to an open, arcwiseconnected region as a domain. (This is not to be confused with the domain


of a function.) It is possible to show that any two points in a domain can bejoined by a smooth path that lies entirely in the domain.

A region is said to be simply connected if it is arcwise connected andevery closed path lying in the region can be continuously shrunk to a pointin the region without any part of the path passing outside the region. Fora more rigorous treatment of simple connectedness, see Singer and Thorpe[20].

Example 4.2.1. Consider the following regions.

• The region consisting of all points in space is a simply connected do-main. It is also convex.

• An open ball is a simply connected domain. It is also convex.

• Any convex region is simply connected.

• An open solid torus is a domain, but it is not simply connected. Anopen solid torus is like a doughnut (not including the surface). A closedpath surrounding the hole cannot be continuously shrunk to a point inthe solid torus without some part of the path passing outside the solidtorus. See Figure 4.2.

• The set {(x, y, z) : a < x2 + y2 < b}, where 0 ≤ a < b, is a domain, butit is not simply connected. (Here b can be replaced by ∞.)

• The set {(x, y, z) : a < x2 + y2 + z2 < b}, where 0 ≤ a < b, is a simplyconnected domain. (Here b can be replaced by ∞.)

• The set {(x, y) : a < x2 + y2 < b}, where 0 ≤ a < b, is a domain in theplane, but it is not simply connected. (Here b can be replaced by ∞.)

Conservative Fields and Potentials

Definition 4.2.1. Assume that F is a continuous vector field defined on adomain. F is said to be conservative if there exists a scalar field φ such thatF = grad φ.


Figure 4.2: A torus

The scalar field φ in this definition is called a potential function, or simplya potential, for F. Obviously, if φ is a potential for F, then so is φ + C forany constant function C.

In physics it is customary to call φ a potential for F if F = −gradφ.We’ve adopted the definition that we have for simplicity. Obviously, if φ is apotential for F according to one definition, then −φ is a potential accordingto the other. Therefore, the minus-sign makes no difference as to whether agiven vector field is conservative or not. See Exercise 17, Section 3.2.1.

Because of its resemblance to the fundamental theorem of calculus werefer to the following theorem as the fundamental theorem for line integrals.

Theorem 4.2.1. Let F be a continuous vector field defined on a domain D.F is conservative if and only if the line integral of F along every non-closedpiecewise smooth path in D depends only on the initial and terminal pointsof the path. In this case, one has∫

C

F · dR = φ(Q)− φ(P ),

where φ is any potential for F and P and Q are the initial and terminalpoints, respectively, of the path C.

If the line integral of F along every non-closed piecewise smooth path inD depends only on the initial and terminal points of the path, then we saythat the line integral is independent of path.


Proof. First, suppose∫C

F · dR is independent of path. Choose any point(x0, y0, z0) in D. Given any other point (x, y, z) ∈ D choose a smooth pathC1 from (x0, y0, z0) to (x, y, z) and define

φ(x, y, z) =

∫C1

F · dR.

This is well-defined since the line integral is independent of path.We seek to show that F = gradφ. By definition,

∂φ

∂x= lim

∆x→0

φ(x+ ∆x, y, z)− φ(x, y, z)

∆x.

Choose ∆x 6= 0 small enough so that the line segment C2 from (x, y, z) to(x+ ∆x, y, z) lies in D. (This is always possible since D is a domain.) Thenwe have

φ(x+ ∆x, y, z) = φ(x, y, z) +

∫C2

F · dR.

It follows that

∂φ

∂x= lim

∆x→0

1

∆x

∫C2

F · dR

= lim∆x→0

1

∆x

∫ x+∆x

x

F1(t, y, z) dt = F1(x, y, z).

The last equality follows from the mean-value theorem for integrals and thecontinuity of F1.

The proof that F2 = ∂φ/∂y and F3 = ∂φ/∂z are similar. This shows thatF is conservative.

Now, let C be a piecewise smooth path from P to Q, and C1 be a smoothpath from (x0, y0, z0) to P . Then we have

φ(Q) =

∫C1

F · dR +

∫C

F · dR

= φ(P ) +

∫C

F · dR.

Therefore, ∫C

F · dR = φ(Q)− φ(P ).


This completes one direction of the proof.For the other direction, assume that F = grad φ. Then for any smooth

path C from P to Q choose a parametrization with domain [t0, t1]. It followsthat ∫

C

F · dR =

∫C

∂φ

∂xdx+

∂φ

∂ydy +

∂φ

∂zdz

=

∫ t1

t0

(∂φ

∂x

dx

dt+∂φ

∂y

dy

dt+∂φ

∂z

dz

dt

)dt

=

∫ t1

t0

dφ

dtdt = φ(Q)− φ(P ).

The same equation now follows easily for piecewise smooth curves.

The proof of the following corollary we leave as an exercise.

Corollary 4.2.1. Let F be a continuous vector field defined on a domainD. F is conservative if and only if the line integral of F along every closedpiecewise smooth path in D is 0, that is,∮

C

F · dR = 0,

for every closed piecewise smooth path C in D.

It turns out that there is a simple way to determine when a vector fielddefined on a simply connected domain is conservative.

Theorem 4.2.2. Let F be a continuously differentiable vector field definedon a simply connected domain D. F is conservative on D if and only ifcurl F = 0 at every point in D.

For simplicity we will give a proof of this theorem where D is all of space.Afterwards we will discuss how the proof can be applied to somewhat moregeneral domains.

Proof. If F is conservative and F = gradφ, then by assumption φ is twicecontinuously differentiable. It follows that

curl F = curl gradφ = 0,

by Exercise 15(a) in Section 3.2.1.


As stated above, we will give the proof in the other direction for the casewhere the domain is all of space.

Assume curl F = 0 and define a piecewise smooth path C from (0, 0, 0)to (x, y, z) made up of the following straight line segments:

• C1 is the straight line segment from (0, 0, 0) to (x, 0, 0)

• C2 is the straight line segment from (x, 0, 0) to (x, y, 0)

• C3 is the straight line segment from (x, y, 0) to (x, y, z)

We then define

φ(x, y, z) =

∫C

F ·dR =

∫ x

0

F1(t, 0, 0) dt+

∫ y

0

F2(x, t, 0) dt+

∫ z

0

F3(x, y, t) dt.

First, by the fundamental theorem of calculus,

∂φ

∂z= F3(x, y, z).

(Note that the first two terms in the expression for φ don’t depend on z.)Next, we have

∂φ

∂y= F2(x, y, 0) +

∫ z

0

∂

∂yF3(x, y, t) dt.

Here the first term on the right is obtained from the fundamental theorem ofcalculus; while the second term follows from Leibnitz’s rule which allows usto interchange the integral and the partial derivative, since F3 is continuouslydifferentiable (see [21]).

Now, since we are assuming that curl F = 0, we have

∂

∂yF3(x, y, t) =

∂

∂tF2(x, y, t).

Substituting this into the equation above, we obtain

∂φ

∂y= F2(x, y, 0) +

∫ z

0

∂

∂tF2(x, y, t) dt

= F2(x, y, 0) + F2(x, y, z)− F2(x, y, 0) = F2(x, y, z).


We will leave the similar proof that

∂φ

∂x= F1(x, y, z)

to the reader.This shows that gradφ = F, and completes the proof.

Recall that if curl F = 0 on a domain D, then F is said to be irrotationalin D.

A few remarks may be in order here.

• The proof we’ve given shows that “conservative implies irrotational” forany domain D; it’s the converse that requires simple connectedness.

• Our proof of “irrotational implies conservative” is valid as long as thepath C defined therein lies in the domain D for every (x, y, z) ∈ D.

• If we modify our proof by replacing (0, 0, 0) by any other fixed point(x0, y0, z0) and the resulting path C lies in the domain D for every(x, y, z) ∈ D, then this method of proof will also be valid. This obvi-ously includes (open) balls and the interiors of rectangular solids whoseedges are parallel to the coordinate axes, as well as many other com-monly encountered domains.

The proof of Theorem 4.2.2 for a general simply connected domain is wellbeyond the scope of this text. One can however give an elementary prooffor domains that are “star-shaped.” A region is star-shaped if there exists apoint P in the region such that for every other point Q in the region the linesegment joining P and Q lies entirely within the region. We won’t give thisproof here, but rather refer the reader to [4]. A convex region is obviouslystar-shaped and a star-shaped region is simply connected.

We note that the method of proof of Theorem 4.2.2 can be used as amethod for computing a potential on certain domains; however, we preferthe following method.

Example 4.2.2. Show that

F(x, y, z) = (y + z)i + (x+ z)j + (x+ y + z)k

is conservative and find a potential.


Solution. One can easily show that curl F = 0 everywhere, so F is conser-vative.

If F = gradφ, then φ must satisfy, simultaneously,

∂φ

∂x= y + z,

∂φ

∂y= x+ z,

∂φ

∂z= x+ y + z.

Integrating the first equation with respect to x, we obtain

φ(x, y, z) = xy + xz + g(y, z),

where g(y, z) may depend on y and z, but not on x.Taking the partial derivative of this with respect to y and comparing this

with the second equation above we obtain

x+∂g(y, z)

∂y= x+ z,

so ∂g(y, z)/∂y = z. This implies that g(y, z) = yz + h(z), where h(z) maydepend on z, but not on x or on y.

Substituting this into the expression for φ we obtain

φ(x, y, z) = xy + xz + yz + h(z).

Next, taking the partial derivative of this with respect to z, and comparingit with the third equation above we obtain

x+ y + h′(z) = x+ y + z,

so h′(z) = z. Thus, h(z) = 12z2 + C, where C is an arbitrary constant.

Again substituting into the expression for φ we finally obtain

φ(x, y, z) = xy + xz + yz +1

2z2 + C.

The reader should verify that F = gradφ.

We conclude this section with a few more examples.

Example 4.2.3. Find a potential for

F(x, y, z) = (y + z)i + (x+ z)j + (x+ y + z)k

that satisfies φ(1,−1, 2) = 0.


Solution. In the last example, we found that

φ(x, y, z) = xy + xz + yz +1

2z2 + C.

Substituting, we find

φ(1,−1, 2) = 1 + C = 0,

which gives C = −1.Therefore,

φ(x, y, z) = xy + xz + yz +1

2z2 − 1

is a potential that satisfies the given condition.


F(x, y, z) = (y + z cosxz)i + xj + x cosxzk


Solution. One can easily show that curl F = 0 everywhere, so F is conser-vative.

If F = gradφ, then φ must satisfy, simultaneously,

∂φ

∂x= y + z cosxz,

∂φ

∂y= x,

∂φ

∂z= x cosxz.


φ(x, y, z) = xy + sinxz + g(y, z),

where g(y, z) may depend on y and z, but not on x.Taking the partial derivative of this with respect to y and comparing this


x+∂g(y, z)

∂y= x,

so ∂g(y, z)/∂y = 0. This implies that g(y, z) = h(z), where h(z) may dependon z, but not on x or on y.


Substituting this into the expression for φ, taking the partial derivativewith respect to z, and comparing with the third equation above we obtain

x cosxz + h′(z) = x cosxz,

so h′(z) = 0. Thus, h(z) = C, where C is an arbitrary constant.Substituting this into the expression for φ we obtain

φ(x, y, z) = xy + sinxz + C.


Example 4.2.5. Find∫C

F · dR, where

F(x, y, z) = (y + z cosxz)i + xj + x cosxzk

and R is given by

x = t, y = t2, z = t3, 2 ≤ t ≤ 3.

Solution. In the previous example, we showed that φ(x, y, z) = xy + sinxzis a potential for F (take C = 0). Therefore,∫

C

F · dR = φ(3, 9, 27)− φ(2, 4, 8) = 19 + sin 81− sin 16,

since the path begins at (2, 4, 8) and ends at (3, 9, 27).

An alternative method for computing∫C

F · dR if F is conservative isto use the fact that the line integral is independent of path. The idea is toreplace the path C by a piecewise smooth path, with the same initial andterminal points, that makes the integral easier to compute.

Example 4.2.6. Compute the line integral in Example 4.2.5 by replacing Cwith a different path.

Solution. Let

C1 : x = t, y = 4, z = 8, 2 ≤ t ≤ 3

C2 : x = 3, y = t, z = 8, 4 ≤ t ≤ 9

C3 : x = 3, y = 9, z = t, 8 ≤ t ≤ 27.


The line segments C1, C2, and C3 are parallel to the coordinate axes, jointhe endpoints of C, and are oriented the correct way.

Since we’ve already shown that F is conservative, it follows that∫C

F · dR =

∫C1

F · dR +

∫C2

F · dR +

∫C3

F · dR.

The integrals on the right-hand side of this equation are easy to calculate:∫C1

F · dR =

∫ 3

2

(4 + 8 cos 8t) dt = 4 + sin 24− sin 16∫C2

F · dR =

∫ 9

4

3 dt = 15∫C3

F · dR =

∫ 27

8

3 cos 3t dt = sin 81− sin 24.

Adding these together we obtain∫C

F · dR = 19 + sin 81− sin 16.

Conservative Fields in the Plane

As per our custom in this text, we have in this section focused on conservativevector fields in space. However, we want to mention here that Definition 4.2.1,Theorem 4.2.1, and Corollary 4.2.1 make sense and are valid in the plane.Theorem 4.2.2 is also valid if curl F = 0 is replaced by

∂F2

∂x=∂F1

∂y.

As usual, we leave it to the reader to consider the details. However, considerthe following example.


F(x, y) = (y sinx+ xy cosx)i + (x sinx+ 1)j



Solution. Here F1(x, y) = y sinx+ xy cosx and F2(x, y) = x sinx+ 1. Since,

∂F2

∂x=∂F1

∂y= sinx+ x cosx,

for all (x, y) in the plane, F is conservative.If F = gradφ, then φ must satisfy, simultaneously,

∂φ

∂x= y sinx+ xy cosx,

∂φ

∂y= x sinx+ 1.


φ(x, y) = xy sinx+ g(y),

where g(y) may depend on y, but not on x.Taking the partial derivative of this with respect to y and comparing this


x sinx+ g′(y) = x sinx+ 1,

so g′(y) = 1. Integrating gives g(y) = y+C, where C is an arbitrary constant.Substituting this into the expression for φ we obtain

φ(x, y) = xy sinx+ y + C.


Application

The methods used to solve Example 4.2.7 should seem very familiar to anyonewho has studied exact differential equations. The differential equation

F1(x, y) + F2(x, y)dy

dx= 0,

which is usually written

F1(x, y) dx+ F2(x, y) dy = 0,

is exact if and only if F = F1i + F2j is conservative. If φ is a potential forF, then φ(x, y) = C (C a constant) is an implicit solution of the differentialequation. In this case, the expression

F1(x, y) dx+ F2(x, y) dy

is called an exact differential (or an exact differential form).


Example 4.2.8. Show that the differential equation

(y sinx+ xy cosx) dx+ (x sinx+ 1) dy = 0

is exact and find an implicit solution.

Solution. In Example 4.2.7 we showed that

F(x, y) = (y sinx+ xy cosx)i + (x sinx+ 1)j

is conservative, and that

φ(x, y) = xy sinx+ y

is a potential for F. Therefore, the differential equation is exact, and

xy sinx+ y = C

is an implicit solution.

4.2.1 Exercises

In Exercises 1–3, for the given region R, determine if R is (a) a domain and(b) simply connected.

1. R = B1(P ) ∪B1(Q) for P (0, 0, 0) and Q(2, 0, 0)

2. R = B1(P ) ∪B1(Q) ∪ {(1, 0, 0)} for P (0, 0, 0) and Q(2, 0, 0)

3. R = {(x, y, z) : 1 < x2 + y2 < 4, 0 < z < 1}

4. Determine whether the vector field F(x, y, z) = yi + xj + zk is conser-vative. If F is conservative, find a potential for F.

5. Determine whether the vector field F(x, y, z) = yi− xj + zk is conser-vative. If F is conservative, find a potential for F.

6. Determine whether the vector field F(x, y, z) = 2xyi + (x2 + z) j + ykis conservative. If F is conservative, find a potential for F.

7. Assuming F is continuous, determine whether the vector field

F(x, y, z) = g(x)i + h(y)j + l(z)k

is conservative. If F is conservative, find a potential for F.


8. Let F(x, y, z) = z2i + 2yj + 2xzk.

(a) Show F is conservative, and then find a potential function for F.

(b) Find∫C

F · dR for the curve

x = exp(sin t), y = cos t, z =t

2π, 0 ≤ t ≤ 2π,

using the equation∫C

F · dR = φ(Q)− φ(P ).

(c) Find the line integral in (b) by replacing C with the line segmentfrom the initial point of C to the terminal point of C.

9. Determine the value of the line integral∫C

F · dR, where

F(x, y, z) =(e−y − ze−x

)i +(e−z − xe−y

)j +(e−x − ye−z

)k

and R is given by

x =1

ln 2ln(1 + t), y = sin

(πt

2

), z =

1− et

1− e, 0 ≤ t ≤ 1.

10. Determine whether the vector field F(x, y) = (x + y)i + (x + 2y)j isconservative. If F is conservative, find a potential for F.

11. Determine whether the vector field F(x, y) = (x + y)i − (x − y)j isconservative. If F is conservative, find a potential for F.

12. Show that the vector field F(x, y) = (x + sin y)i + (x cos y − 2y)j isconservative and find the potential φ such that φ(1,−1) = 0.

13. Determine whether the differential equation (x+y) dx+(x+2y) dy = 0is exact. If it is exact, find an implicit solution. (Hint: See Exercise 10.)

14. Determine whether the differential equation (x+ y) dx− (x− y) dy = 0is exact. If it is exact, find an implicit solution. (Hint: See Exercise 11.)

15. Show that the differential equation (x+sin y) dx+(x cos y−2y) dy = 0is exact, and find an implicit solution such that y(1) = −1. (Hint: SeeExercise 12.)

16. Let F(x, y) = (x2 + y) i + (x+ ey) j.

4.3. AREA INTEGRALS AND VOLUME INTEGRALS 117

(a) Show F is conservative, and then find a potential function for F.

(b) Find∫C

F · dR for the spiral

x = t cos t, y = t sin t, 0 ≤ t ≤ 2π,

using the equation∫C

F · dR = φ(Q)− φ(P ).

(c) Find the line integral in (b) by replacing C with the line segmentfrom the initial point of C to the terminal point of C.

17. Determine the value of the line integral∫C

F · dR, where

F(x, y) =(2xy2 − 3

)i +(2x2y + 4

)j

and R is given by

x =1

ln 2ln(1 + t), y = cos(πt), 0 ≤ t ≤ 1.

18. Let F(x, y) =−yi + xj

x2 + y2.

(a) Show that∂F2

∂x=∂F1

∂y.

(b) Compute∮C

F · dR, where C is the unit circle centered at theorigin and oriented counterclockwise.

(c) In light of the answers to (a) and (b), is F conservative? Explain.

19. Prove Corollary 4.2.1.


4.3 Area Integrals and Volume Integrals

We would like to quickly review area integrals and volume integrals, sincethey will play a role in the upcoming material. Since we will not go into greatdetail concerning these topics, the reader may be well served to spend sometime reviewing them in his or her favorite elementary calculus text. See, forexample, [22].


Preliminaries

Suppose f is a scalar field defined on a bounded domain D in space and onthe boundary of D. In this context, the term “rectangle” means “rectangularsolid.” Choose a rectangle containing D and partition this rectangle intosubrectangles Ri (ignoring those subrectangles that have no points in D).Let ∆Vi be the volume of Ri and choose points (xi, yi, zi) in Ri ∩ D. Wedefine ∫∫∫

D

f(x, y, z) dV = lim∑

f(xi, yi, zi)∆Vi,

where the limit is taken as the diagonals of all of the subrectangles approach0 and is independent of all of the choices made above. We will refer to thisintegral as a volume integral. dV is often referred to as the element of volumeor the volume element.

If f is continuous and the boundary of D is piecewise smooth, then thislimit exists.

The area integral over a planar region D is defined similarly. However, inthis case the “rectangles” are actual rectangles in the plane; two-dimensionalvolume is called area, so ∆Vi becomes ∆Ai, dV becomes dA, and the integralis written ∫∫

D

f(x, y) dA.

We leave it to the reader to fill in the details. dA is often referred to theelement of area or the area element.

Area Integrals

In practice, one usually uses Fubini’s theorem to evaluate an area integralas an iterated integral. This is certainly valid if f is continuous and theboundary of D is piecewise smooth.

Recall that if ρ = the area density of D, then∫∫D

ρ dA = mass of D.

If ρ = 1 on D, then we obtain∫∫D

dA = area of D.


x

y

D

y=x

(1,1)

1


Example 4.3.1. Let D be the region in the plane bounded by the trianglewith vertices (0, 0), (1, 0), and (1, 1). Find∫∫

D

xy dA.

Solution. Referring to Figure 4.3, we have

∫∫D

xy dA =

∫ 1

0

∫ x

0

xy dydx

=

∫ 1

0

xy2

2

∣∣∣x0dx

=

∫ 1

0

x3

2dx =

1

8.

Polar Coordinates

Many area integrals can be more easily evaluated by using polar coordinates.


r

x

y

(x,y)

θ

Figure 4.4: Polar coordinates


The transformations for polar coordinates are:

x = r cos θ, y = r sin θ,

where r ≥ 0. For polar coordinates the area element is given by

dA = r drdθ.

It often is helpful to remember the identity r2 = x2 + y2.

Example 4.3.2. Find the area inside the circle of radius a.

Solution. We can assume the circle is centered at the origin. Using polarcoordinates, we have ∫∫

D

dA =

∫ 2π

0

∫ a

0

r drdθ = πa2.

So, the area is πa2.

Volume Integrals

As with area integrals, one usually uses Fubini’s theorem to evaluate a volumeintegral as an iterated integral. This is certainly valid if f is continuous andthe boundary of D is piecewise smooth.

Recall that if ρ = the mass density of D, then∫∫∫D

ρ dV = mass of D.

If ρ = 1 on D, then we obtain∫∫∫D

dV = volume of D.

Example 4.3.3. A solid tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0),and (0, 0, 1) has density δ = x. Find its mass.


Solution. The plane through (1, 0, 0), (0, 1, 0), and (0, 0, 1) has the equationx+ y + z = 1 or z = 1− x− y. Therefore, the mass is given by∫∫∫

D

x dV =

∫ 1

0

∫ 1−x

0

∫ 1−x−y

0

x dz dy dx

=

∫ 1

0

∫ 1−x

0

x(1− x− y) dy dx

=

∫ 1

0

[x

(y − xy − y2

2

)]1−x

y=0

dx

=

∫ 1

0

x

[1− x− x(1− x)− (1− x)2

2

]dx

=

∫ 1

0

(x3

2− x2 +

x

2

)dx

=1

24.

Cylindrical and Spherical Coordinates

Many volume integrals can be more easily evaluated by using either cylindri-cal or spherical coordinates.

The transformations for cylindrical coordinates are:

x = r cos θ, y = r sin θ, z = z,

where r ≥ 0.3 For cylindrical coordinates the volume element is given by

dV = r drdθdz.

In cylindrical coordinates we also have r2 = x2 + y2.The transformations for spherical coordinates are:

x = ρ cos θ sinφ, y = ρ sin θ sinφ, z = ρ cosφ,

3We should note that some authors use different notation than we do for cylindricalcoordinates and for spherical coordinates. We feel that our notation is the most commonlyused at present. Also, in our notation r and θ have the same meaning as they do in polarcoordinates.


x

y

z

(x,y,0)

z

(x,y,z)

r

θ

Figure 4.5: Cylindrical coordinates


x

y

z

(x,y,0)

(x,y,z)

θ

ρ

φ

Figure 4.6: Spherical coordinates


where ρ ≥ 0. For spherical coordinates the volume element is given by

dV = ρ2 sinφ dρdφdθ.

It is often helpful to remember that ρ2 = x2 + y2 + z2.

Example 4.3.4. Find the volume inside the sphere of radius a.

Solution. We can assume that the sphere is centered at the origin. Usingspherical coordinates, we have∫∫∫

D

dV =

∫ 2π

0

∫ π

0

∫ a

0

ρ2 sinφ dρdφdθ

=

∫ 2π

0

∫ π

0

1

3a3 sinφ dφdθ

=

∫ 2π

0

1

3a3 · 2 dθ

=1

3a3 · 2 · 2π =

4

3πa3.

Thus, the volume is 43πa3.

Example 4.3.5. Find the volume of the region bounded by the surface

z = e−(x2+y2),

the cylinder x2 + y2 = 1, and the xy-plane.

Solution. Using cylindrical coordinates, we have∫∫∫D

dV =

∫ 2π

0

∫ 1

0

∫ e−r2

0

r dzdrdθ

=

∫ 2π

0

∫ 1

0

re−r2

drdθ

=

∫ 2π

0

1

2

(1− e−1

)dθ

= π(1− e−1

).

Thus, the volume is π(1− 1/e).


4.3.1 Exercises

1. If a point P has cylindrical coordinates (4, π/2, 2), find rectangular andspherical coordinates for P .

2. If the spherical coordinates of a point P are given by ρ = 12, θ = 3π/4,and φ = π/6, find rectangular and cylindrical coordinates for P .

3. Find and simplify equations in cylindrical coordinates and sphericalcoordinates for the equation 4y = x2 + y2. Identify the surface.

4. Write ρ sinφ = 2 as a cartesian equation. Identify this surface.

5. Find an equation in rectangular coordinates for the surface whosespherical coordinates satisfy ρ = 6 cosφ. Identify this surface.

6. Evaluate ∫ 2

1

∫ 3

0

(x+ y) dxdy.

7. Evaluate ∫ 4

1

∫ 2

0

(x+ y2) dxdy.

8. Evaluate ∫ 4

2

∫ 8−y

y

y dxdy.

9. Evaluate ∫ 1

−1

∫ −3y

y

(2x− y) dxdy.

10. Evaluate ∫ 1

0

∫ 3

3y

ex2

dxdy.

(Hint: Reverse the order of integration.)

11. Evaluate ∫ 3

0

∫ 9

y2ye−x

2

dxdy.

(Hint: Reverse the order of integration.)


12. Evaluate ∫ 1

0

∫ √1−x2

0

sin(x2 + y2) dydx.

13. Evaluate ∫ 1

0

∫ √1−x2

0

e−x2−y2 dydx.

14. Evaluate ∫ 1

−1

∫ √1−x2

0

1

1 + x2 + y2dydx.

15. Let D be the region in the plane bounded by the lines x+y = 4, y = 1,and the y-axis. Evaluate ∫∫

D

(xy + y2) dA.

16. Let D be the region in the plane bounded by y = x2, the x-axis, andx = 2. Evaluate ∫∫

D

xy dA.

17. Let D be the region in the plane bounded by the curves y =√x and

y = x. Evaluate ∫∫D

xy dA.

18. Let D be the region in the plane bounded by the triangle with vertices(3,−3), (6,−3), and (6, 3). Evaluate∫∫

D

(−3x− 2y) dA.

19. Let D be the region in the plane bounded by circle x2+y2 = 4. Evaluate∫∫D

(x2 + y2

)7/2dA.

20. Let D be the region in the plane bounded by circle x2+y2 = 9. Evaluate∫∫D

√x2 + y2 dA.


21. Evaluate ∫ 0

−1

∫ 3x

0

∫ 2x+3z

z

x dydzdx.

22. Evaluate ∫ 1

0

∫ x

0

∫ x−y

0

x dzdydx.

23. Evaluate ∫ 2

0

∫ z2

0

∫ z

x

(x+ z) dydxdz.

24. Compute the volume of the solid tetrahedron with vertices (0, 0, 0),(1, 0, 0), (0, 2, 0), and (0, 0, 3) by evaluating a volume integral. Checkyour answer by using geometric formulas and by using the triple scalarproduct.

25. For the solid tetrahedron of Exercise 24 suppose the density is givenby δ = z. Find its mass.

26. Using a volume integral, find the volume of the tetrahedron boundedby the coordinate planes and the plane

4x+ 3y + z = 12.

27. Using a volume integral, find the volume of the region bounded by theplane

3x+ 2y + z = 6

and the coordinate planes.

28. Find the volume of the region inside the paraboloid z = 9 − x2 − y2,outside the cylinder x2 + y2 = 4, and above the plane z = 0.

29. Find the volume of the region in the first octant bounded by the cylinderx2 + y2 = 4 and the plane y + z = 2.

30. Use a volume integral to compute the volume of the region bounded bythe cone z2 = x2 + y2 and the cylinder x2 + y2 = 9. Check your answerby using geometric formulas.

4.4. SURFACE INTEGRALS 129

31. Use cylindrical coordinates to find the volume of the region inside theparaboloid z = 25 − x2 − y2, outside the cylinder x2 + y2 = 16, andabove the plane z = 0.

32. Find the volume of the solid that lies above the cone φ = π/4 andbelow the sphere ρ = 5 cosφ.

33. The moment of inertia about the z-axis of a solid object occupying aregion R in space is given by

Iz =

∫∫∫R

(x2 + y2) δ dV,

where δ = δ(x, y, z) is the density.

Assume the density δ is constant and find Iz for the solid cylinder

R ={

(x, y, z) | x2 + y2 ≤ R2, 0 ≤ z ≤ h}.

(Hint: Use cylindrical coordinates.)

4.4 Surface Integrals

As one might expect, a careful study of surfaces is significantly more difficultthan a careful study of curves. Likewise, surface integrals are generally morecomplicated than line integrals.

To make up for this we will rely somewhat more on geometric intuitionin the case of surfaces than we did for curves. Also, we will discover that wewill be able to use known formulas for various surface areas from elementarygeometry to simplify many calculations involving surface integrals.

For a thorough treatment of the geometry of smooth surfaces we recom-mend Millman and Parker [16].

Before we move on to surfaces, we need to very briefly consider vector-valued functions of two variables.

Vector Functions of Two Variables

A vector (or vector-valued) function of two variables can be written:

F(u, v) = F1(u, v)i + F2(u, v)j + F3(u, v)k,


where F1, F2, and F3 are real-valued functions of two real variables. Unlessotherwise stated, the domain of F is the set of points in the plane for whichF1, F2, and F3 are all defined. The functions F1, F2, and F3 are called thecomponent functions of F.

Much of the material in Section 2.1, with obvious modifications, is appli-cable to vector functions of two variables, so we give a very brief treatment.

As with vector functions of one variable, the sum, dot product, and crossproduct of vector functions of two variables are all defined pointwise. Like-wise for the scalar product of a real-valued function of two variables and avector function of two variables. The scalar product of a constant and avector function of two variables is defined in the obvious way.

F is continuous on a subset of the plane if and only if F1, F2, and F3 are.We define

∂F

∂u=∂F1

∂ui +

∂F2

∂uj +

∂F3

∂uk,

wherever ∂Fi/∂u, i = 1, 2, 3, are all defined. We define ∂F/∂v and higherorder derivatives similarly.4

We say that F is continuously differentiable [twice continuously differen-tiable, etc.] on an open subset of the plane if and only if F1, F2, and F3

are.

Surfaces

As we did with curves, to avoid various types of pathologies, we will limitour consideration to certain “piecewise smooth” surfaces. We begin with thefollowing definition.

Definition 4.4.1. The image of a vector function R = R(u, v), whose do-main D is a closed bounded region in the plane for which the boundary is apiecewise smooth closed path, will be called a compact smooth surface if:

• R(u, v) 6= R(u1, v1) if (u, v) is in the interior of D, (u1, v2) ∈ D, and(u, v) 6= (u1, v1).

• R is continuously differentiable and

N(u, v) =∂R

∂u× ∂R

∂v4Many authors use subscripts to denote various partial derivatives. We will eschew

this notation here, so as to avoid any possibility of confounding partial derivatives withcomponent functions.


Figure 4.7: A compact smooth surface with boundary

is nonvanishing in the interior of D.

• lim(u,v)→(u0,v0)

N(u, v)

|N(u, v)|exists for each (u0, v0) in the boundary of D.

As with paths, we normally view a surface as a set of points in spacerather than as a set of directed line segments representing position vectors.

Roughly speaking, to create a compact smooth surface a closed boundedregion in the plane, whose boundary is a piecewise smooth path, may bestretched or contracted in any direction while being bent or warped, but nottorn or creased. The surface may not pass through itself; however, parts ofthe boundary path may be collapsed to a point or “glued” together as longas this doesn’t create any corners or points in the surface itself.

A compact smooth surface may have a boundary (in this case, it shouldperhaps more properly be called a surface-with-boundary), or it may enclosea region of space. The boundary of the surface, if it has one, will consist ofone or more piecewise smooth paths consisting of all or part of the image ofthe boundary path of the domain of R. Note that the boundary of a surfaceis not the same thing as the boundary of a region as described in Section 3.1.

A circular cylinder (not including the flat ends) is an example of a compactsmooth surface that has a boundary. The boundary being the two circles atthe ends of the cylinder. A sphere and a torus are both examples of compactsmooth surfaces without boundaries.


Frequently, we will write a surface in terms of its components:

x = x(u, v), y = y(u, v), z = z(u, v), (u, v) ∈ D.

Example 4.4.1. Letting ρ = a > 0 in the equations for spherical coordinates,we see that the sphere centered at the origin of radius a can be parametrizedby

x = a sinφ sin θ, y = a sinφ cos θ, z = a cosφ,

where 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π.We leave it as an exercise to show that this is a compact smooth surface.

Example 4.4.2. Using cylindrical coordinates we can parametrize the rightcircular cylinder symmetric about the z-axis of radius r > 0 and betweenz = a and z = b by

x = r cosu, y = r sinu, z = v, 0 ≤ u ≤ 2π, a ≤ v ≤ b.

We leave it as an exercise to show that this is a compact smooth surface.

One can show, using the implicit function theorem from advanced calcu-lus, that the level set corresponding to f(x, y, z) = 0 is a compact smoothsurface if it is bounded and arcwise connected, f is continuously differen-tiable, and grad f is nonvanishing on an open region containing the levelset.

Orientable Surfaces

Consider a compact smooth surface given by R = R(u, v). Since

∂R

∂uand

∂R

∂v

are both tangent to the surface,

N(u, v) =∂R

∂u× ∂R

∂v

is normal to the surface.Therefore,

n = n(u, v) =N(u, v)

|N(u, v)|


Figure 4.8: A Mobius strip

is a unit normal to the surface. From now on n will always denote a unitnormal vector. The surface is said to be orientable if n is continuous on theentire surface. In this case an orientation is the choice of either n or −n. Thisamounts to choosing one side of the surface. A surface that is not orientableis often called a one-sided surface.

The classic example of a compact smooth surface that is not orientableis the Mobius strip. See Figure 4.8. The reader can construct a model ofa Mobius strip by taking a strip of paper, giving it half a twist in the longdirection, and joining the ends together. If you then attempt to color oneside, you will end up coloring the entire surface.

It so happens that every compact smooth surface without boundary is,in fact, orientable. The proof of this, however, is well beyond the scope ofthis text.

If an oriented compact smooth surface is bounded by a piecewise smoothclosed path, then the orientation of the surface determines an orientation forthe boundary path as follows: The path is oriented so that when one followsthe path on the side of the chosen normal vector in the positive directionthe surface is on the left. See Figure 4.9. This orientation will be referred toas the induced orientation. If the surface lies in the xy-plane, or in a planeparallel to the xy-plane, and the chosen normal vector is k, then the inducedorientation on the boundary path is referred to as counterclockwise. In thiscase, the opposite orientation is referred to as clockwise.

If an oriented compact smooth surface bounds a region of space, then theusual or standard orientation is given by the outward unit normal vector.


Figure 4.9: The induced orientation

See Figure 4.10.

We also want to consider orientable compact surfaces that are made up offinitely many orientable compact smooth surfaces which are “glued” togetheralong portions of their boundaries in such a way that the resulting surface is“two-sided” and either encloses a region of space or has a boundary whichconsists of one or more piecewise smooth closed paths. Once an orientationis chosen, we will refer to such a surface as a compact oriented piecewisesmooth surface. Examples include the surfaces of tetrahedrons, cubes, andother polyhedra, which are not smooth due to the edges and the vertices.These all have an “inside” and an “outside” that can be used to orient theparticular surface. Figure 4.11 depicts a compact orientable piecewise smoothsurface that has a boundary.

On the other hand, although the surface depicted in Figure 4.12 can beobtained by glueing three rectangles together along portions of their bound-aries, it is not an oriented compact piecewise smooth surface, since it is not“two-sided” and its boundary is not a piecewise smooth path.

Since there are so many different ways to join parts of the boundaries ofsurfaces together, we won’t be able to give a more formal definition of anoriented compact piecewise smooth surface; however, in all of the examplesand exercises it will be clear how the surface can be constructed from a finitenumber of orientable compact smooth surfaces and how it can be oriented.

Henceforth, by “smooth surface” we will mean “oriented compact smoothsurface” and by “piecewise smooth surface” we will mean “oriented compactpiecewise smooth surface.”


Figure 4.10: A closed surface with several outward unit normal vectors

Figure 4.11: A piecewise smooth surface with boundary


Figure 4.12: A surface that is not piecewise smooth

Surface Area

Suppose R defines a compact smooth surface. Consider a rectangle that liesin the interior of the domain of R with vertices (u, v), (u+∆u, v), (u, v+∆v),and (u + ∆u, v + ∆v). R maps this rectangle into a little piece of surfacewhose area we can approximate by the area of the parallelogram with sidesR(u+∆u, v)−R(u, v) and R(u, v+∆v)−R(u, v). See Figure 4.13. Further,we approximate these sides by

R(u+ ∆u, v)−R(u, v) ≈ ∂R

∂u∆u

and

R(u, v + ∆v)−R(u, v) ≈ ∂R

∂v∆v,

where the partial derivatives are evaluated at (u, v). This leads to

∆S ≈∣∣∣∣∂R

∂u× ∂R

∂v

∣∣∣∣∆u∆v

as the approximate area of the little piece of surface.


(u,v) (u+Δu,v)

(u,v+Δv) (u+Δu,v+Δv)

R(u,v)R

RR

R

(u+Δu,v)

(u+Δu,v+Δv)(u,v+Δv)

Figure 4.13: Surface area

This motivates us to define the area of the entire surface by

S =

∫∫D

∣∣∣∣∂R

∂u× ∂R

∂v

∣∣∣∣ du dv,where D is the domain of R.

One can show that this is independent of the smooth parametrization ofthe surface.

If we write

dS =

(∂R

∂u× ∂R

∂v

)du dv

and

dS = |dS| =∣∣∣∣∂R

∂u× ∂R

∂v

∣∣∣∣ du dv,then we can write, for the above,

S =

∫∫S

dS =

∫∫S

|dS| =∫∫

S

n · dS,

where

n =∂R∂u× ∂R

∂v∣∣∂R∂u× ∂R

∂v

∣∣is a unit normal to the surface.5

dS is referred to as the element of surface area. Always remember thatdS depends on the particular surface.

5Perhaps unfortunately, it is customary to use S to denote both the surface and thearea of the surface.


Example 4.4.3. Find the surface area of a sphere of radius a.

Solution. The sphere of radius a (centered at the origin) can be parametrizedby

x = a sinφ sin θ, y = a sinφ cos θ, z = a cosφ,

where 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π. (See Example 4.4.1.) So, we can take

R(θ, φ) = a sinφ sin θi+a sinφ cos θj+a cosφk, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π.

A straightforward calculation (using cos2 α + sin2 α = 1 several times)yields ∣∣∣∣∂R

∂θ× ∂R

∂φ

∣∣∣∣ = a2 sinφ.

Thus, we find the surface area is∫ π

0

∫ 2π

0

a2 sinφ dφ dθ = 4πa2.

If a surface is given by z = f(x, y), (x, y) ∈ D, then it can be parametrizedby

x = u, y = v, z = f(u, v), (u, v) ∈ D.

So, we can take

R(u, v) = ui + vj + f(u, v)k, (u, v) ∈ D.

An easy computation then shows that∣∣∣∣∂R

∂u× ∂R

∂v

∣∣∣∣ =

√1 +

(∂f

∂u

)2

+

(∂f

∂v

)2

=

√1 +

(∂f

∂x

)2

+

(∂f

∂y

)2

.

So,

S =

∫∫D

√1 +

(∂f

∂x

)2

+

(∂f

∂y

)2

dx dy.

In this case, we can derive another useful expression for dS. If γ is theangle between

∂R

∂u× ∂R

∂vand k,


then it is easy to show that

| cos γ| = |n · k| =∣∣(∂R

∂u× ∂R

∂v

)· k∣∣∣∣∂R

∂u× ∂R

∂v

∣∣ =1√

1 +(∂f∂x

)2+(∂f∂y

)2.

Thus,

dS =1

| cos γ|dx dy.

The reader should be able to derive similar expressions for surfaces givenby x = f(y, z) and for surfaces given by y = f(x, z).

Example 4.4.4. Find the surface area of the paraboloid

z =1

2x2 +

1

2y2

for x2 + y2 ≤ R2, where R > 0 is a constant.

Solution. Since ∂z/∂x = x and ∂z/∂y = y, we have∫∫S

dS =

∫∫x2+y2≤R2

√1 + x2 + y2 dxdy

=

∫ 2π

0

∫ R

0

(1 + r2

)1/2r drdθ

= π

∫ R2+1

1

u1/2 du

=2

3π[(R2 + 1

)3/2 − 1].

So, the surface area is2

3π[(R2 + 1

)3/2 − 1].

Surface Integrals

Recall that by “smooth surface” we mean “oriented compact smooth surface”and by “piecewise smooth surface” we mean “oriented compact piecewisesmooth surface.”


The idea behind the definition of a surface integral is as follows. Supposef is defined on some domain containing the smooth surface S. Divide S inton “patches” with areas ∆S1,∆S2, . . . ,∆Sn. Choose a point (xi, yi, zi) in eachpatch. We then define the surface integral of f over S by∫∫

S

f dS = limn∑i=1

f(xi, yi, zi)∆Si,

where the limit is taken as the number of patches tends to infinity and eachpatch gets smaller and smaller in every direction, provided this limit existsindependently of the selections made above.

If f is continuous, then this integral does exist. Moreover, dS can becomputed using the formulas given in the previous section.

If S is piecewise smooth, then the surface integral over S is equal to thesum of the surface integrals over the smooth surfaces that constitute S.

As an example, if f is the area density of a surface (say in grams persquare centimeter) then

∫∫Sf dS is the mass of the surface (in grams). In

this text we are particularly interested in the case where f = F · n, where Fis a vector field and n is a unit normal to the surface.

The surface integral ∫∫S

F · n dS =

∫∫S

F · dS

is called the flux of F through S.Recall that earlier we interpreted div F(P ) as “the flux per unit volume

at P .” If we consider a small piecewise smooth closed surface S that enclosesa solid of volume V containing P , then

div F(P ) ≈ 1

V

∫∫S

F · n dS.

Taking the limit as the enclosing solid shrinks to P , we obtain

div F(P ) = lim1

V

∫∫S

F · n dS.

(Some authors use this as the definition of the divergence, and then derivethe definition we’ve given as a consequence.)


One form of Gauss’s law of electrostatics states that∫∫S

E · n dS = K∑

qi,

where S is a closed surface, E is the electric field, the sum is taken over allof the charges qi enclosed by S, and K is a constant depending on the unitsthat are used. Here n is the outward unit normal to the surface.

The following examples illustrate some of the methods that can be usedto evaluate surface integrals. The reader should study them all carefully.

Example 4.4.5. Compute∫∫

SF · n dS, where

S ={

(x, y, z) : x2 + y2 + z2 = 9}

and F(x, y, z) = xi+yj+zk. Assume here that n is the outward unit normalon S.

Solution. The surface S is the sphere centered at the origin of radius 3. So,the outward unit normal vector on S is

n =1

3(xi + yj + zk)

and

F · n =1

3

(x2 + y2 + z2

)= 3

on S. Therefore,∫∫S

F · n dS = 3

∫∫S

dS = 3(surface area of S) = 108π.


SF · n dS for F(x, y, z) = xi over the triangle

with vertices (1, 0, 0), (0, 2, 0), and (0, 0, 3), taking the normal on the sideaway from the origin.

Solution. The triangle lies in the plane that has

N = 〈−1, 2, 0〉× 〈−1, 0, 3〉 = 6i + 3j + 2k


as a normal vector. Since |N| = 7, we have

n =6

7i +

3

7j +

2

7k

which we note does point away from the origin.F · n = 6

7x and |n · k| = 2

7, so∫∫

S

F · n dS =

∫ 1

0

∫ 2−2x

0

6

7x · 7

2dy dx = 1.

(We leave it to the reader to perform the integration.)


SF·n dS, where S is the boundary of the region

bounded by the planes x = ±1, y = ±1, z = ±1, and F(x, y, z) = xi. Take nto be the outward unit normal to S.

Solution. For the faces given by

y = ±1, −1 ≤ x, z ≤ 1,

and

z = ±1, −1 ≤ x, y ≤ 1,

F · n = 0.For the face, say S1, given by

x = −1, −1 ≤ y, z ≤ 1,

n = −i, and for the face, say S2, given by

x = 1, −1 ≤ y, z ≤ 1

n = i. Thus, on both of these faces, F · n = 1.Therefore,∫∫

S

F · n dS =

∫∫S1

F · n dS +

∫∫S2

F · n dS = 4 + 4 = 8,

since the area of each face is 4


4.4.1 Exercises

In these Exercises, if S is a closed surface, then n denotes the outward unitnormal.

1. Find the surface area of the surface given by

x =1

2u2, y = uv, z = v2, 0 ≤ u ≤ 3, 0 ≤ v ≤ 2.

2. Determine the element of surface area dS = |dS| for the right circularcylinder parametrized by

x = r cosu, y = r sinu, z = v, 0 ≤ u ≤ 2π, 0 ≤ v ≤ h,

where r is the radius of the base and h is the height.

Use this to find the surface area of the cylinder (not including the topor the bottom).

3. Determine the element of surface area dS = |dS| for the right circularcone parametrized by

x = (r/h)v cosu, y = (r/h)v sinu, z = v,

0 ≤ u ≤ 2π, 0 ≤ v ≤ h,

where r is the radius of the base and h is the height.

Use this to find the surface area of the cone (not including the top).6

4. Find the surface area of z = 23x3/2 + 2

3y3/2 for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1.

5. Find the surface area of the paraboloid z = ax2 +ay2 for x2 + y2 ≤ R2.Here a is a nonzero constant and R is a positive constant.

6. Find the surface area of the hyperboloid z = ax2−ay2 for x2 +y2 ≤ R2.Here a is a nonzero constant and R is a positive constant. (Hint.Compare with Exercise 5.)

6A cone is not a smooth surface, but one can still use the formulas in this section tocalculate the area. If you cut off a little bit of the tip of the cone and take the appropriatelimit, you will get the same answer.


7. Find the surface area of the hyperboloid z = axy for x2+y2 ≤ R2. Herea is a nonzero constant and R is a positive constant. (Hint. Comparewith Exercise 6.)

8. Consider the torus (which looks like the surface of an inner tube) withmajor radius A and minor radius a. Derive the parametrization

x = (A+ a cos v) cosu, y = (A+ a cos v) sinu, z = a sin v,

0 ≤ u, v ≤ 2π.

Use this to compute the surface area of the torus.

9. Let F(x, y, z) = zk. Find ∫∫S

F · n dS

over the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1) and normalpointing away from the origin.

10. Let S be the surface of the cube whose faces are each parallel to one ofthe coordinate planes and with opposite vertices (0, 0, 0) and (2, 2, 2),and let F(x, y, z) = xi + yj + zk. Compute∫∫

S

F · n dS.

11. Let S be the surface of the cube whose faces are each parallel to one ofthe coordinate planes and with opposite vertices (0, 0, 0) and (2, 2, 2),and let F(x, y, z) = yi. Compute∫∫

S

F · n dS.

12. Let S be the surface of the cube whose faces are each parallel to one ofthe coordinate planes and with opposite vertices (0, 0, 0) and (2, 2, 2),and let F(x, y, z) = xyi. Compute∫∫

S

F · n dS.


13. Let S be the sphere of radius 2 centered at the origin, and let F = zk.Compute ∫∫

S

F · n dS.

(Hint. Use spherical coordinates. Here dS = 4 sinφ dφ dθ.)

14. Let S be the boundary of the region bounded by the planes z = 0,z = 3, and the cylinder x2 + y2 = 4, and let F(x, y, z) = xi + yj + zk.Compute ∫∫

S

F · n dS.

15. Let S be the surface in Exercise 14, and let F(x, y, z) = yi. Compute∫∫S

F · n dS.

16. Let S be the surface in Exercise 14, and let F(x, y, z) = xy2i. Compute∫∫S

F · n dS.

(Hint. Use cylindrical coordinates. Here dS = 2 dz dθ.)

17. Let F(x, y, z) = xi + yj + (z2 − 1)k. Find∫∫S

F · n dS

over the boundary of the region bounded by the planes z = 0, z = 1,and the cylinder x2 + y2 = a2.

18. Let


(x2 + y2 + z2)3/2.

Compute∫∫

SF · n dS, where S is the sphere x2 + y2 + z2 = a2. (Here

a is a positive constant.)

19. (a) Calculate∫∫

SF ·n dS for F(x, y, z) = e−xi + e−yj + e−zk over the

surface of a cube of side s centered at (x0, y0, z0) and whose facesare parallel to the coordinate planes.


(b) Divide the result above by the volume of the cube and calculatethe limit of the quotient as s → 0+. (Hint. If f is differentiableat x, then

f ′(x) = limh→0

f(x+ h/2)− f(x− h/2)

h.

)(c) Calculate the divergence of F at (x0, y0, z0), and compare the re-

sult to the answer to part (b).

20. Given the smooth surface R = R(u, v), let

E =

∣∣∣∣∂R

∂u

∣∣∣∣2 , F =∂R

∂u· ∂R

∂v, G =

∣∣∣∣∂R

∂v

∣∣∣∣2 .Show that

dS =√EG− F 2 dudv.

21. Use the formula in Exercise 20 to recalculate dS for the surface inExercise 1.

22. Use the formula in Exercise 20 to recalculate dS for the surface inExercise 2.

23. Prove that a sphere is an orientable compact smooth surface. (Hint.Verify the conditions in Definition 4.4.1 for the parametrization givenin Example 4.4.1. Then verify orientability.)

24. Prove that the cylinder of Example 4.4.2 is an orientable compactsmooth surface. (Hint. Verify the conditions in Definition 4.4.1 forthe parametrization given in Example 4.4.2. Then verify orientability.)

4.5 Plotting Parametric Surfaces

Matlab (with Symbolic Math Toolbox) makes it easy to plot surfaces thatare given parametrically. If you have access to Matlab try the following.You should pay careful attention to the syntax. Entering a command or amathematical expression incorrectly is a very common mistake. If you areusing some other software, then you will need to modify these commands.(We use “>” for the prompt. This may be different on your computer. Donot type the symbol “>”.)

4.5. PLOTTING PARAMETRIC SURFACES 147

> syms u v

> ezmesh(cos(u),sin(u),v,[0,2*pi],[-2,2])

You should be able to rotate the surface and view it from different per-spectives.

You can use “ezsurf” in place of “ezmesh” for a different looking plot.You may also want to use the optional command “axis equal.” For ex-

ample, try the following.

> syms u v

> ezmesh(sin(v)*cos(u),sin(v)*sin(u),cos(v),[0,2*pi],[0,pi]);

axis equal

(Type “axis equal” on the same line as the “ezmesh” command.) This is theunit sphere centered at (0, 0, 0). Now try it leaving out the command “axisequal.” This command also works with “ezsurf” and “ezplot.” For moreinformation about optional plotting commands for Matlab, see the on-linehelp.

We list below some parametric surfaces along with suggested domains.We encourage the reader to experiment by considering different domains.

Note that you only need to type “syms u v” once, at the beginning of,each session.

• x = cosu, y = sinu, z = v, [0, 2π], [−2, 2]

• x = v cosu, y = v sinu, z = v, [0, 2π], [−2, 2]

• x = sin v sinu, y = sin v cosu, z = cos v, [0, 2π], [0, π]

• x = cosh v cosu, y = cosh v sinu, z = sinh v,[0, 2π], [−2, 2]

• x = sinhu cos v, y = sinhu sin v, z =u

|u|coshu,

[−2, 2], [0, 2π]

• x = cos3 u sin3 v, y = sin3 u sin3 v, z = cos3 v,[0, 2π], [0, π]

• x = (2 + cos v) cosu, y = (2 + cos v) sinu, z = sin v,[0, 2π], [0, 2π]


• x = 2u2, y = 2uv, z = v2,[0, 2], [0, 3]

• x =(1 + 1

2v cos 1

2u)

cosu, y =(1 + 1

2v cos 1

2u)

sinu,z = 1

2v sin 1

2u, [0, 2π], [−1, 1]

References


4.5.1 Project

Use Matlab or some other software to plot all of the surfaces listed in thissection, identifying those that you can. Experiment with different domains.Also, rotate the surfaces to achieve different perspectives.

Chapter 5

Stokes-type Theorems

5.1 Green’s Theorem

We begin this section with the statement of Green’s theorem.

Theorem 5.1.1 (Green’s Theorem). Suppose D is a bounded domain inthe plane whose boundary is a piecewise smooth path C which is orientedcounterclockwise. Let P and Q be continuously differentiable scalar fieldsdefined on an open region containing D and C. Then

∮C

P dx+Qdy =

∫∫D

(∂Q

∂x− ∂P

∂y

)dA.

Note that for a surface in the xy-plane dS = dA = dx dy.

Example 5.1.1. Let C be the circle x2 +y2 = a2 and D be the disk x2 +y2 <a2. Verify the conclusion of Green’s theorem for

P (x, y) = −x2y, Q(x, y) = xy2.

Solution. We can parametrize C by

x = a cos t, y = a sin t, 0 ≤ t ≤ 2π.

149

150 CHAPTER 5. STOKES-TYPE THEOREMS

Then we have dx/dt = −a sin t and dy/dt = a cos t. So,∮C

P dx+Qdy =

∮C

−x2y dx+ xy2 dy

=

∫ 2π

0

(a4 cos2 t sin2 t+ a4 cos2 t sin2 t

)dt

= 2a4

∫ 2π

0

cos2 t sin2 t dt =πa4

2.

On the other hand, we have ∂Q/∂x = y2 and ∂P/∂y = −x2. Using polarcoordinates, we have∫∫

D

(∂Q

∂x− ∂P

∂y

)dA =

∫∫D

(y2 + x2

)dx dy

=

∫ 2π

0

∫ a

0

r2 · r dr dθ

=

∫ 2π

0

∫ a

0

r3 dr dθ =πa4

2.

This verifies the conclusion of Green’s theorem for this example.

For simplicity we will give a proof of Green’s Theorem for domains Dthat satisfy the following additional hypothesis:

Any line passing through an interior point of D and parallel to eithercoordinate axis intersects the boundary of D in exactly two points.

Proof. Consider Figure 5.1. Using the fundamental theorem of calculus, wehave

∫∫D

∂P

∂ydA =

∫ b

a

∫ g2(x)

g1(x)

∂P

∂ydy dx

=

∫ b

a

P (x, g2(x))− P (x, g1(x)) dx

= −∫C2

P dx−∫C1

P dx

= −∮C

P dx.

A similar argument gives the other term.

5.1. GREEN’S THEOREM 151

C1 : y = g1(x)

C2 : y=g2(x)

a b

y

x

Figure 5.1: Green’s theorem

Our proof of Green’s theorem can easily be extended to the case where Dcan be partitioned into a finite number of domains that satisfy the hypothesesof Green’s theorem as well as our additional hypothesis.1 This is illustratedin Figure 5.2. Note that the line integrals along the line that divides thedomain cancel.

It is also easy to extend Green’s theorem to domains that have finitelymany “holes.” For example, Green’s theorem can be applied to an annulus.We leave the details to the reader.

Corollary 5.1.1. Suppose D is a domain and C is a piecewise smooth paththat satisfy the hypotheses of Green’s theorem. Then∮

C

x dy = −∮C

y dx =1

2

∮C

x dy − y dx = area of D.

We leave the proof of this corollary as an exercise.

1That is, D = D1∪D2∪. . .∪Dk, where each Di is a domain that satisfies the hypothesesof Green’s theorem and our additional hypothesis, and boundary Di∩boundary Dj , i 6= j,is either empty or a piecewise smooth path.


Figure 5.2: A domain on which Green’s theorem holds

5.1. GREEN’S THEOREM 153

Example 5.1.2. Use Corollary 5.1.1 to find the area inside the ellipse

x2

a2+y2

b2= 1,

where a and b are positive constants.

Solution. We can parametrize the ellipse as follows:

x = a cos t, y = b sin t, 0 ≤ t ≤ 2π.

Thendy

dt= b cos t,

and ∫C

x dy =

∫ 2π

0

ab cos2 t dt = πab.

5.1.1 Exercises

1. Verify Green’s theorem where D is the annulus 1 < x2 + y2 < 4 andF(x, y) = −yi + xj.

2. Evaluate∮C

(x3 − x2y) dx+ xy2 dy, where C is the curve shown in Fig-ure 5.3 oriented counterclockwise. (C is made up of two semicirclesand two line segments.)

3. Use Green’s theorem to re-evaluate the line integral in Exercise 6 ofSection 4.1.1.





−4 −3 −2 −1 0 1 2 3 4−1

0

1

2

3

4

5

x

y

Figure 5.3: Exercise 1, Section 5.1.1

7. Use Corollary 5.1.1 to find the area inside the cardioid

x = (1 + sin θ) cos θ, y = (1 + sin θ) sin θ, 0 ≤ θ ≤ 2π.

8. Use Corollary 5.1.1 to find the area inside the hypocycloid with fourcusps (or astroid)

x = a cos3 t, y = a sin3 t, 0 ≤ t ≤ 2π.

(Here a is a positive constant.)2

9. Suppose that D is a domain in the plane whose boundary is a piecewisesmooth closed curve C.

Prove: If D is convex, then every line parallel to one of the coordinateaxes that intersects D intersects C in exactly two points.

Show, by example, that the converse is not true.

10. Extend our proof of Green’s theorem to the case where D can be parti-tioned into two domains that satisfy our additional hypothesis. (Hint:See Figure 5.2.)

2The careful reader may note that this is not a smooth parametrization; however, theconclusion of Green’s theorem is still valid.

5.2. THE DIVERGENCE THEOREM 155

11. Prove Corollary 5.1.1. (Hint: Choose P and Q appropriately.)

5.2 The Divergence Theorem

We begin this section with the statement of the divergence theorem. Thistheorem is also referred to as Gauss’s theorem.

Theorem 5.2.1 (The Divergence Theorem). Let D be a bounded domain inspace whose boundary is a closed piecewise smooth surface S with outwardunit normal n. Let F be a continuously differentiable vector field defined onan open region containing both D and S. Then∫∫∫

D

div F dV =

∫∫S

F · n dS.

Example 5.2.1. Verify the divergence theorem for the cube bounded by theplanes x = ±1, y = ±1, z = ±1, and

F(x, y, z) = xi + yj + zk.

Solution. It is easy to compute that div F = 3. Therefore,∫∫∫D

div F dV = 3 · volume of D = 3 · 23 = 24.

On the other hand, it is easy to show that F · n = 1, on each face of thecube. Therefore,∫∫

S

F · n dS = surface area of S = 6 · 22 = 24,

since there are six faces.

Thus, we’ve verified the divergence theorem for this example.

For simplicity we will give a proof of the divergence theorem for domainsD that satisfy the following additional hypothesis:

Each straight line through an interior point of D intersects the boundaryof D in exactly two points.


Proof. We can write

n = cosαi + cos βj + cos γk,

soF · n = F1 cosα + F2 cos β + F3 cos γ.

The statement of the divergence theorem then becomes∫∫∫D

(∂F1

∂x+∂F2

∂y+∂F3

∂z

)dV =

∫∫S

(F1 cosα + F2 cos β + F3 cos γ) dS.

We prove ∫∫∫D

∂F3

∂zdV =

∫∫S

F3 cos γ dS.

The two other equalities necessary to prove the theorem are proven in asimilar fashion. Consider Figure 5.4. In this figure S2 denotes the upperhalf of the surface, S1 denotes the lower half of the surface, and Rxy is theprojection of the surface in the xy-plane. (To avoid cluttering up the figure,we haven’t depicted the axes.)

For the upper surface

cos γ2 dS2 = dx dy, cos γ2 = n2 · k.Similarly, for the lower surface

cos γ1 dS1 = −dx dy, cos γ1 = n1 · k.It follows that∫∫

S

F3 cos γ dS =

∫∫S1

F3 cos γ1 dS1 +

∫∫S2

F3 cos γ2 dS2

= −∫∫

Rxy

F3 (x, y, g1(x, y)) dx dy

+

∫∫Rxy

F3 (x, y, g2(x, y)) dx dy

=

∫∫Rxy

(F3 (x, y, g2(x, y))− F3 (x, y, g1(x, y))) dx dy

=

∫∫Rxy

∫ g2(x,y)

g1(x,y)

∂F3

∂z(x, y, z) dz dx dy

=

∫∫∫D

∂F3

∂zdV.


S2 : z = g2(x,y)

S1 : z=g1(x,y)

Rxy

Figure 5.4: The divergence theorem


Figure 5.5: A domain on which the divergence theorem holds

Our proof of the divergence theorem can easily be extended to the casewhere D can be partitioned into a finite number of domains that satisfy thehypotheses of the divergence theorem as well as our additional hypothesis.3

This is illustrated in Figure 5.4. Note that the surface integrals over thesurface that divides the domain cancel.

It is also easy to extend the divergence theorem to domains that havefinitely many “holes.” For example, the divergence theorem can be appliedto concentric spheres. We leave the details to the reader.

An Application

The general form of Gauss’s law is∫∫S

E · n dS = K

∫∫∫D

ρ dV,

3That is, D = D1∪D2∪. . .∪Dk, where each Di is a domain that satisfies the hypothesesof the divergence theorem and our additional hypothesis, and boundary Di∩boundary Dj ,i 6= j, is either empty or a piecewise smooth surface.


where E is the electric field, ρ is the charge density, K > 0 is a constant thatdepends on the units used, and D is a region enclosed by S.

Assuming that the conclusion of the divergence theorem holds, we have∫∫∫D

div E dV = K

∫∫∫D

ρ dV.

Since this is true for any region D for which the divergence theorem holds,it follows that

divE = Kρ or ∇ · E = Kρ.

This is one of Maxwell’s equations.

5.2.1 Exercises

1. Verify the divergence theorem for the cube bounded by the coordinateplanes and the planes x = 1, y = 1, z = 1, and

F(x, y, z) = x2i + y2j + z2k.

2. Verify the divergence theorem for the domain given by

1 < x2 + y2 + z2 < 4, and

F(x, y, z) = xi + yj + zk.

3. Use the divergence theorem to re-evaluate the surface integral in Ex-ample 4.4.5 of Section 4.4.

4. Use the divergence theorem to re-evaluate the surface integral in Exer-cise 10 of Section 4.4.1.









12. Explain why the divergence theorem doesn’t apply to Exercise 18 ofSection 4.4.1.

13. Evaluate∫∫

SF · n dS, where

F(x, y, z) = (z2 − x)i− xyj + (y2 + 2z)k,

and S is the boundary of the region bounded by z = 4 − y2, x = 0,x = 3, and the xy-plane.

14. Given that the volume inside a sphere of radius a is (4/3)πa3, use thedivergence theorem to show that the surface area is 4πa2.

15. Show that our proof of the divergence theorem can be extended to thecase where the domain D can be partitioned into two domains thatsatisfy our additional hypothesis.

5.3 Stokes’ Theorem

We begin this section with the statement of Stokes’ theorem. Recall thatby “piecewise smooth surface” we mean “oriented compact piecewise smoothsurface,” and n is a unit normal vector which determines the orientation ofthe surface.

Theorem 5.3.1 (Stokes’ Theorem). Let S be a piecewise smooth surfacebounded by a piecewise smooth closed path C. Choose the orientation of C to

5.3. STOKES’ THEOREM 161

be consistent with S. Assume that F is a continuously differentiable vectorfield defined on some domain containing S and C. Then∮

C

F · dR =

∫∫S

curl F · n dS.

Example 5.3.1. Verify Stokes’ theorem for

F(x, y, z) = yi + zj + xk,

where S is the upper unit hemisphere z =√

1− x2 − y2 and C is the unitcircle in the xy-plane. Choose n so that it points away from the origin.

Solution. On the one hand, we can parametrize the unit circle in the xy-planeby

x = cos θ, y = sin θ, z = 0, 0 ≤ θ ≤ 2π.

(Observe that this is oriented the correct way.) Then we have

dx

dθ= − sin θ,

dy

dθ= cos θ,

dz

dθ= 0.

Thus, we find that ∮C

F · dR =

∫ 2π

0

− sin2 θ dθ = −π.

On the other hand, an easy computation shows that

curl F = −i− j− k.

Also, it is easy to see that

n = xi + yj + zk,

and socurl F · n = −x− y − z.

In light of Examples 4.4.1 and 4.4.3, we note that we can parametrize theupper unit hemisphere by

x = sinφ sin θ, y = sinφ cos θ, z = cosφ,


where 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/2, and that

dS = sinφ dθdφ.

Therefore,∫∫S

curl F · n dS =

∫ π/2

0

∫ 2π

0

(− sinφ sin θ − sinφ cos θ − cosφ) sinφ dθdφ

=

∫ π/2

0

∫ 2π

0

[− sin2 φ(sin θ + cos θ)− cosφ sinφ

]dθdφ

= −π.

This verifies Stokes’ theorem for this example.

The basic idea behind the proof of Stokes’ theorem is to “lift” Green’stheorem from the domain of the surface to the surface itself. To make thingsa little simpler, we will give a proof for the case where the surface is given byz = f(x, y), (x, y) ∈ D, where D satisfies the hypotheses of Green’s theorem,and f is twice continuously differentiable. Also, we will assume that S is asmooth surface, and that C and the boundary of D are smooth curves. Theproof of the more general result is similar.

Proof. Using results from Section 4.4, a straightforward calculation showsthat∫∫

S

curl F · n dS =

∫∫D

[(∂F3

∂y− ∂F2

∂z

)(−∂f∂x

)+

(∂F1

∂z− ∂F3

∂x

)(−∂f∂y

)+

(∂F2

∂x− ∂F1

∂y

)]dA.

On the other hand, let

x = x(t), y = y(t), t ∈ [a, b],

be a smooth parametrization of the boundary of D such that the orientationis counterclockwise. Then

x = x(t), y = y(t), z = f(x, y) t ∈ [a, b],

is a smooth parametrization of C. We leave it to the reader to check thatthis gives the correct orientation of C.

5.3. STOKES’ THEOREM 163

Then ∮C

F · dR =

∮C

F1 dx+ F2 dy + F3 dz

=

∫ b

a

(F1dx

dt+ F2

dy

dt+ F3

dz

dt

)dt.

By the chain rule,dz

dt=∂f

∂x

dx

dt+∂f

∂y

dy

dt.

Substituting this in the previous equation, doing a little algebra, and applyingGreen’s theorem, we obtain∮

C

F · dR =

∫ b

a

[(F1 + F3

∂f

∂x

)dx

dt+

(F1 + F3

∂f

∂y

)dy

dt

]dt

=

∮∂D

(F1 + F3

∂f

∂x

)dx+

(F1 + F3

∂f

∂y

)dy

=

∫∫D

[∂

∂x

(F2 + F3

∂f

∂y

)− ∂

∂y

(F1 + F2

∂f

∂x

)]dA.

Remember that on C we have F1 = F1(x, y, f(x, y)), etc. Therefore, usingthe chain rule, we have

∂

∂x

(F2 + F3

∂f

∂y

)=∂F2

∂x+∂F2

∂z

∂f

∂x+

(∂F3

∂x+∂F3

∂z

∂f

∂x

)∂f

∂y+ F3

∂2f

∂x∂y.

Similarly,

∂

∂y

(F1 + F3

∂f

∂x

)=∂F1

∂y+∂F1

∂z

∂f

∂y+

(∂F3

∂y+∂F3

∂z

∂f

∂y

)∂f

∂x+ F3

∂2f

∂x∂y.

Note that since we are assuming that f has continuous second partialderivatives,

F3∂2f

∂y∂x= F3

∂2f

∂x∂y.

So, again using a little algebra, we obtain∮C

F · dR =

∫∫D

[(∂F3

∂y− ∂F2

∂z

)(−∂f∂x

)+

(∂F1

∂z− ∂F3

∂x

)(−∂f∂y

)+

(∂F2

∂x− ∂F1

∂y

)]dA.

This completes the proof.


Stokes’ theorem can also be applied to any surface that can be partitionedinto finitely many surfaces which satisfy the hypotheses of Stokes’ theorem.4

For example a cylinder (not including the flat ends) can be partitioned intotwo such surfaces.

5.3.1 Exercises

1. Verify Stokes’ theorem for

F(x, y, z) = yi + zj + xk,

where S is the part of the sphere x2 + y2 + z2 = 1 that lies in the firstoctant x ≥ 0, y ≥ 0, z ≥ 0, and C is the boundary curve of S. Choosen so that it points away from the origin.

2. Let S be the surface of the cube generated by 〈1, 0, 0〉, 〈0, 1, 0〉, and〈0, 0, 1〉 except for the face with opposite vertices (0, 0, 0) and (1, 0, 1).Choose n pointing out of the cube. Verify Stokes’ theorem for thissurface with F given by

F(x, y, z) = yi + zj + xk.

3. Let S be the cylinder parametrized by

x = cosu, y = sinu, z = v, 0 ≤ u ≤ 1, 0 ≤ v ≤ 1.

Choose n pointing away from the origin. Verify Stokes’ theorem forthis surface with F given by

F(x, y, z) = yi + zj + xk.

4. Use Stokes’ theorem to calculate∮C

F · dR, where

F(x, y, z) = x2i + j + yzk,

over the perimeter of the triangle with vertices P (1, 0, 0), Q(0, 1, 0),and R(0, 0, 1) oriented from P to Q to R and back to P .

4That is, S = S1 ∪ S2 ∪ . . . ∪ Sk, where each Si is a piecewise smooth surface thatsatisfies the hypotheses of Stokes’ Theorem, and ∂Si ∩ ∂Sj , i 6= j, is either empty or apiecewise smooth path.

5.4. CYLINDRICAL AND SPHERICAL COORDINATES 165

5. Let F(x, y, z) = xk, and S be the triangle with vertices (1, 0, 0), (0, 1, 0),and (0, 0, 1) and normal vector pointing away from the origin.

(a) Calculate∫∫

SF · n dS, directly.

(b) Calculate∫∫

SF · n dS, using Stokes’ theorem. (Hint. Find, by

trial-and-error, a vector field G such that curl G = F.)

6. Use Stokes’ theorem to re-evaluate the line integral in Exercise 12 ofSection 4.1.1.

7. Use Stokes’ theorem to re-evaluate the line integral in Exercise 15 ofSection 4.1.1.

8. Assume S is a closed surface that can be partitioned into two sur-faces that satisfy the hypotheses of Stokes’ theorem. Show that if Fis a continuously differentiable vector field defined on an open regioncontaining S, then ∫∫

S

curl F · n dS = 0.

9. Let S be (a) a sphere, or (b) (the surface of) a torus. In each case,explain how S can be partitioned into two surfaces that satisfy thehypotheses of Stokes’ theorem. Conclude (see Exercise 8) that if Fis a continuously differentiable vector field defined on an open regioncontaining S, then, in each case,∫∫

S

curl F · n dS = 0.

10. Prove Green’s theorem assuming Stokes’ theorem.

(Hint: Start by assuming F = F1i + F2j and D is a domain in thexy-plane bounded by a simple closed path C such that the hypothesesof Stokes’ theorem hold. Apply Stokes’ theorem to this situation.)

5.4 Cylindrical and Spherical Coordinates

It is sometimes useful to be able to express the gradient, the divergence,and the curl in cylindrical coordinates or in spherical coordinates. One canderive these expressions in a rigorous fashion by using the equations relating


the different coordinate systems along with the chain rule. However, thesecalculations are messy, and not particularly instructive. Therefore, we havetaken a different, more heuristic, approach. The more ambitious reader isencouraged to check some or all of the formulas using the chain rule.

A Brief Remark on Notation

In this section and in the next we will occasionally use phraseology like: Iff(x, y, z) = x2 + y2 + z2, then in cylindrical coordinates f(r, θ, z) = r2 +z2. By this we mean that f maps the point with cylindrical coordinates(r, θ, z) to the real number r2 + z2. (A similar meaning applies to vector-valued functions.) This notation has the obvious drawback that an expressionsuch as f(1, π, 2) depends on whether (1, π, 2) is interpreted as the pointwith cartesian coordinates (1, π, 2) or the point with cylindrical coordinates(1, π, 2). This, however, will always be clear from context. We will use asimilar convention with polar and with spherical coordinates.

Also, when we write symbols such as ∂f/∂r, ∂f/∂θ, and ∂f/∂z we assumethat f is expressed as a function of r, θ, and z. Thus, for the example above,we have

∂f

∂r= 2r,

∂f

∂θ= 0, and

∂f

∂z= 2z,

where these partial derivatives are each evaluated at (r, θ, z) and interpretedas above. Again, we will use a similar convention with polar and with spher-ical coordinates.

Cylindrical Coordinates

Recall that the transformations for cylindrical coordinates are:

x = r cos θ, y = r sin θ, z = z,

where we assume here that r ≥ 0.We define orthogonal unit vectors analogous to i, j, and k as follows:

er = cos θi + sin θj

eθ = − sin θi + cos θj

ez = k.

Notice that these are not all constant vectors.


We can writeR = xi + yj + zk = rer + zez.

An easy computation shows that

dR = er dr + r der + ez dz + z dez

= er dr + reθ dθ + ez dz.

So,

ds = |dR| =(dr2 + r2dθ2 + dz2

)1/2

is the element of arc length in cylindrical coordinates. Also, for later use,recall that

dV = r dr dθ dz.

To obtain grad f in cylindrical coordinates we write

∇f = (er · ∇f) er + (eθ · ∇f) eθ + (ez · ∇f) ez.

Recall that if u is a unit vector, then u · ∇f is the directional derivativeof f in the direction of u. Thus,

er · ∇f =df

ds

∣∣θ,z constant

=∂f

∂r

eθ · ∇f =df

ds

∣∣r,z constant

=1

r

∂f

∂θ

ez · ∇f =df

ds

∣∣r,θ constant

=∂f

∂z.

Therefore,

∇f =∂f

∂rer +

1

r

∂f

∂θeθ +

∂f

∂zez.

Example 5.4.1. Transform to cylindrical coordinates and compute the gra-dient of

f(x, y, z) =(x2 + y2

)z

Solution. In cylindrical coordinates we have f(r, θ, z) = r2z. So,

∂f

∂r= 2rz,

∂f

∂θ= 0,

∂f

∂z= r2.

Therefore, in cylindrical coordinates

∇f(r, θ, z) = 2rzer + r2ez.


e

-e

I

II

r ∆θ

∆r

∆z

r

r

Figure 5.6: A cylindrical rectangular solid

To obtain an expression for the divergence in cylindrical coordinates werely on our interpretation of the divergence as flux per unit volume.

We first write

F = Frer + Fθeθ + Fzez,

and consider a small “cylindrical rectangular solid” centered at (r, θ, z) withedges of lengths approximately r∆θ, ∆r, and ∆z. Here we choose the facesso that the unit normal vectors are ±er, ±eθ, and ±ez. See Figure 5.6. Thisargument is similar to that of Section 3.2, so we will be somewhat briefer.

For the faces with sides of lengths approximately r∆θ and ∆z (i.e., thefaces labeled I and II in the figure) the flux is approximately

(r + ∆r/2)Fr(r + ∆r/2, θ, z)∆θ∆z − (r −∆r/2)Fr(r −∆r/2, θ, z)∆θ∆z

≈ ∂ (rFr)

∂r∆r∆θ∆z.

Similarly considering the other faces gives us

Fθ(r, θ + ∆θ/2, z)∆r∆z − Fθ(r, θ −∆θ/2, z)∆r∆z ≈ ∂Fθ∂θ

∆θ∆r∆z,


Δr

rΔθ

Figure 5.7: A cylindrical rectangle

and

rFz(r, θ, z + ∆z/2)∆θ∆r − rFz(r, θ, z −∆z/2)∆θ∆r ≈ r∂Fz∂z

∆z∆θ∆r.

Dividing by the approximate volume r∆r∆θ∆z, and letting the sides of theregion shrink to 0 around (x, y, z), we obtain

div F =1

r

∂

∂r(rFr) +

1

r

∂Fθ∂θ

+∂Fz∂z

.

To derive the expression for the curl of F in cylindrical coordinates weuse Stokes’ theorem. We consider small “cylindrical rectangles” centered at(r, θ, z) with “sides” of lengths approximately: (i) ∆r and r∆θ, (ii) ∆r and∆z, and (iii) r∆θ and ∆z, and such that the unit tangent vectors to thesevarious sides are ±er, ±eθ, and ±ez. One of these cylindrical rectangles isshown in Figure 5.7. Then, by Stokes’ theorem, we have

curl F · n ≈ 1

∆S

∮C

F · dR,

where ∆S is the surface area and C is the boundary of the small cylindricalrectangle with unit normal vector n and C is given the induced orientation.


In particular, if we choose n = ez, then, referring to Figure 5.7, we have∮C

F · dR ≈ (r + ∆r/2)Fθ(r + ∆r/2, θ, z)∆θ

− (r −∆r/2)Fθ(r −∆r/2, θ, z)∆θ

+ Fr(r, θ + ∆θ/2, z)∆r − Fr(r, θ −∆θ/2, z)∆r

≈ ∂

∂r(rFθ) ∆r∆θ − ∂Fr

∂θ∆θ∆r.

Dividing by the approximate surface area r∆r∆θ and letting all the sidesgo to 0 around (r, θ, z), we obtain

(curl F)z =1

r

∂

∂r(rFθ)−

1

r

∂Fr∂θ

.

In a similar fashion one can obtain

(curl F)r =1

r

∂Fz∂θ− ∂Fθ

∂z

and

(curl F)θ =∂Fr∂z− ∂Fz

∂r.

We leave it to the reader to show that this can be written as

curl F =

∣∣∣∣∣∣er reθ ez∂∂r

∂∂θ

∂∂z

Fr rFθ Fz

∣∣∣∣∣∣ ,where we expand the symbolic determinant out in the usual way.

Example 5.4.2. Transform each of the following vector fields to cylindricalcoordinates and compute the divergence and the curl.

F(x, y, z) =xi + yj

x2 + y2, G(x, y, z) =

−yi + xj

x2 + y2.

Solution. In cylindrical coordinates, F(r, θ, z) = 1rer, so

Fr(r, θ, z) =1

r, Fθ(r, θ, z) = 0, Fz(r, θ, z) = 0.


Therefore,

div F(r, θ, z) =1

r

∂

∂r(rFr) =

1

r,

and

curl F(r, θ, z) =1

r


∂∂θ

∂∂z

1/r 0 0

∣∣∣∣∣∣ =1

r· 0 = 0.

In cylindrical coordinates, G(r, θ, z) = 1reθ, so

Gr(r, θ, z) = 0, Gθ(r, θ, z) =1

r, Gz(r, θ, z) = 0.

Therefore,

div G(r, θ, z) =1

r

∂Gθ

∂θ= 0,

and

curl G(r, θ, z) =1

r


∂∂θ

∂∂z

0 1/r 0

∣∣∣∣∣∣ =1

r

∂

∂r

(1

r

)ez = − 1

r3ez.

Spherical Coordinates

Recall that the transformations for spherical coordinates are:

x = ρ cos θ sinφ, y = ρ sin θ sinφ, z = ρ cosφ,

where ρ ≥ 0. The volume element in spherical coordinates is

dV = ρ2 sinφ dρ dφ dθ.

We define the orthogonal unit vectors

eρ = sinφ cos θi + sinφ sin θj + cosφk

eθ = − sin θi + cos θj

eφ = cosφ cos θi + cosφ sin θj− sinφk.

Notice that R = ρeρ. So,

dR = eρ dρ+ eφρ dφ+ eθρ sinφ dθ,


andds = |dR| =

(dρ2 + ρ2 dφ2 + ρ2 sin2 φ dθ2

)1/2.

In the same manner as we did for cylindrical coordinates, we obtain thefollowing expression for the gradient in spherical coordinates:

grad f =∂f

∂ρeρ +

1

ρ

∂f

∂φeφ +

1

ρ sinφ

∂f

∂θeθ.

Example 5.4.3. Compute grad f in spherical coordinates for

f(x, y, z) =z

(x2 + y2 + z2)3/2.

Solution. In spherical coordinates f(ρ, θ, φ) = cosφ/ρ2. So,

∂f

∂ρ= −2 cosφ

ρ3,

∂f

∂φ= −sinφ

ρ2,

∂f

∂θ= 0.

Therefore,

grad f(ρ, θ, φ) = −2 cosφ

ρ3eρ −

sinφ

ρ3eφ.

If F is a vector field, then we can write

F = Fρeρ + Fφeφ + Fθeθ.

The divergence in spherical coordinates is given by

div F =1

ρ2

∂

∂ρ

(ρ2Fρ

)+

1

ρ sinφ

∂

∂φ(Fφ sinφ) +

1

ρ sinφ

∂Fθ∂θ

.

This can be derived in a similar fashion as the divergence in cylindricalcoordinates, so we leave it as exercise.

The curl in spherical coordinates is given by

curl F =1

ρ2 sinφ

∣∣∣∣∣∣eρ ρeφ (sinφ)eθ∂∂ρ

∂∂φ

∂∂θ

Fρ ρFφ (ρ sinφ)Fθ

∣∣∣∣∣∣ ,where again we expand this out in the usual way. This can be derived in asimilar fashion as the curl in cylindrical coordinates, so we leave this as anexercise, too.


Example 5.4.4. Compute the divergence and the curl for

F(ρ, θ, φ) = eρ + ρeφ + ρ cos θeθ.

Solution. Since,

Fρ(ρ, θ, φ) = 1, Fφ(ρ, θ, φ) = ρ, Fθ(ρ, θ, φ) = ρ cos θ,

we find that

div F(ρ, θ, φ) =2

ρ+ cotφ− sin θ

sinφ

and

curl F(ρ, θ, φ) = cotφ cos θeρ − 2 cos θeφ +2

ρeθ.

The Laplacian in Cylindrical and Spherical Coordinates

One can use the chain rule to derive expressions for the Laplacian in cylindri-cal and in spherical coordinated. The calculations are straightforward, butsomewhat messy. Another possibility is to use the expression

4f =∇2f =∇ ·∇f = div · grad f.

We will give these expressions here and leave their derivations as exercises.The Laplacian in cylindrical coordinates is given by

4f =1

r

∂

∂r

(r∂f

∂r

)+

1

r2

∂2f

∂θ2+∂2f

∂z2.

Note that this immediately implies the following expression for the Lapla-cian in polar coordinates:

4f =1

r

∂

∂r

(r∂f

∂r

)+

1

r2

∂2f

∂θ2.

The Laplacian in spherical coordinates is given by

4f =1

ρ2

∂

∂ρ

(ρ2∂f

∂ρ

)+

1

ρ2 sinφ

∂

∂φ

(sinφ

∂f

∂φ

)+

1

ρ2 sin2 φ

∂2f

∂θ2.


5.4.1 Exercises

1. Transform to cylindrical coordinates and compute the gradient of

f(x, y, z) = x2 + y2 + z2.

2. Transform to spherical coordinates and compute the gradient of

f(x, y, z) = x2 + y2 + z2.

3. Transform to cylindrical coordinates and compute (a) the divergenceand (b) the curl of


x2 + y2 + z2.

4. Transform to cylindrical coordinates and compute (a) the divergenceand (b) the curl of

F(x, y, z) =−yi + xj + zk

x2 + y2 + z2.

5. Transform to spherical coordinates and compute (a) the divergence and(b) the curl of


x2 + y2 + z2.

6. Use the divergence theorem to compute the flux∫∫

SF · n dS of

F(ρ, θ, φ) = ρneρ (n > −2)

through the surface bounded by the upper hemisphere ρ = 1,0 ≤ φ ≤ π/2, and the equatorial plane.

7. Find the Laplacian of f(r, θ, z) = rn in cylindrical coordinates.

8. Find the Laplacian of f(r, θ, z) = r2θ2 in cylindrical coordinates.

9. Find the Laplacian of f(ρ, θ, φ) = ρn in spherical coordinates.

10. Find the Laplacian of f(ρ, θ, φ) = ρ2φ in spherical coordinates.

5.5. APPLICATIONS 175

11. Find all the solutions to Laplace’s equation 4f = 0 for which f isindependent of ρ and θ, i.e., f(ρ, θ, φ) = f(φ).

12. A force field F is called a central force field if F(ρ, θ, φ) = F (ρ)eρ. Showthat a central force field is irrotational.

13. Complete the derivation of the curl in cylindrical coordinates.

14. Using arguments similar to those used in this section to derive thedivergence in cylindrical coordinates, derive the expression for the di-vergence in spherical coordinates. (Hint: Consider a “spherical rect-angular solid” with edges of lengths approximately ∆ρ, ρ sinφ∆θ, andρ∆φ.)

15. Using arguments similar to those used in this section to derive the curlin cylindrical coordinates, derive the expression for the curl in sphericalcoordinates.

16. Derive the expression given in this section for the Laplacian in cylin-drical coordinates. (Hint. Use 4f = div ˙grad f.)

17. Derive the expression given in this section for the Laplacian in sphericalcoordinates. (Hint. Use 4f = div ˙grad f.)

5.5 Applications

There are many applications of vector analysis. We have chosen to highlightMaxwell’s equations and electrostatics, because of the historical significanceof Maxwell’s equations, and because, as H. M. Shey states in his book Div,Grad, Curl, and all That [19], “much of vector [analysis] was invented forelectromagnetic theory and is ideally suited to it.”

Maxwell’s Equations: Introduction

In the October 2004 issue of Physics World, Robert P. Crease wrote aboutthe results of a reader survey of the greatest equations ever [3]. In the articleThe Greatest Equations Ever, the author reports that “most votes were givento Euler’s equation [eiπ + 1 = 0] and to Maxwell’s equations, which describehow an electromagnetic field varies in space and time.”


Although Maxwell’s equations are relatively simple, they daringlyreorganize our perceptions of nature unifying electricity and mag-netism and linking geometry, topology and physics. They are es-sential to understanding the surrounding world. And as the firstfield equations, they not only showed scientists a new way of ap-proaching physics but also took them on the first step toward aunification of the fundamental forces of nature.

In the survey, Maxwell’s equations well outpolled many famous equationssuch as E = mc2.

Also, according to Walter Isaacson’s recent biography of Albert Einstein,Einstein: His Life and Universe [12], Maxwell’s equations played a key rolein Einstein’s thinking leading up to the special theory of relativity. Isaacsonquotes Einstein as follows:

If I pursue a beam of light with the velocity c (velocity of light ina vacuum), I should observe such a beam of light as an electro-magnetic field at rest though spatially oscillating. There seemsto be no such thing, however, neither on the basis of experiencenor according to Maxwell’s equations.

Einstein also used Maxwell’s equations to derive the famous equationE = mc2. Again, quoting Isaacson quoting Einstein:

. . . the relativity principle, together with Maxwell’s equations, re-quires that mass be a direct measure of the energy contained ina body.

And, in his description of a speech Einstein gave at a scientific conferencein 1909, Isaacson comments:

. . . his [Einstein’s] love of Maxwell’s equations was well placed.They are among the few elements of theoretical physics to remainunchanged by both the relativity and quantum revolutions thatEinstein helped launch.

Maxwell’s Equations

The quantities occurring in Maxwell’s equation are the following:

• E — electric field


• B — magnetic field

• ρ — charge density

• J — current density

These may all vary in time as well as in space.For the remainder of this section, we will assume that all regions, surfaces,

curves, and functions are such that all of the physical laws stated below andany formal manipulations we perform (e.g., interchanging derivatives andintegrals) are, in fact, valid.

We have already derived the equation

∇ · E = K1ρ,

where K1 > 0 is a constant that depends on the units used.If S is a closed surface, then Gauss’s law for magnetic fields states that∫∫

S

B · n dS = 0.

This expresses the absence of magnetic charge (magnetic monopoles) in na-ture. Using the divergence theorem, this can be written

∇ ·B = 0.

Next we consider Faraday’s law:∮C

E · dR = −K2dΦ

dt,

where

Φ =

∫∫S

B · n dS

is the magnetic flux and K2 > 0 is a constant depending on the units used.Here we assume S, C, and n are as in the statement of Stokes’ theorem.

Since we are assuming that interchanging the derivative and the integralis valid, we have ∮

C

E · dR = −K2

∫∫S

∂B

∂t· n dS.


Applying Stokes’ theorem gives∫∫S

curl E · n dS = −K2

∫∫S

∂B

∂t· n dS.

Since this is true for all surfaces S that satisfy Stokes’ theorem we have

curl E = −K2∂B

∂tor ∇× E = −K2

∂B

∂t.

Finally, we have the Ampere-Maxwell law:∮C

B · dR = K3d

dt

∫∫S

E · n dS +K4I,

where K3 and K4 are positive constants that depend on the units used.Here we assume S is a surface with boundary curve C given the inducedorientation, and I is the current enclosed by C. One can show that

I =

∫∫S

J · n dS.

We leave it as an exercise to show that, using Stokes’ theorem, the Ampere-Maxwell law can be rewritten as

∇×B = K3∂E

∂t+K4J.

In summary, Maxwell’s equations can be written:

∇ · E = K1ρ

∇× E = −K2∂B

∂t∇ ·B = 0

∇×B = K3∂E

∂t+K4J,

where K1, K2, K3, and K4 are positive constants that depend on the unitsused.

Perhaps the most commonly used units in this setting are rationalizedMKSA units. In these units K1 = 1/ε0, K2 = 1, K3 = ε0µ0, and K4 = µ0,where ε0, the permittivity of free space, is approximately 8.854 × 10−12


coulombs2 per newton-meter2 and µ0, the permeability of free space, is ap-proximately 1.257× 10−6 newtons per ampere2.

We rewrite Maxwell’s equations with these constants:

∇ · E =1

ε0ρ

∇× E = −∂B

∂t∇ ·B = 0

∇×B = ε0µ0∂E

∂t+ µ0J.

Electromagnetic Waves

One can use Maxwell’s equations to show that in the absence of charges(ρ = 0) and currents (J = 0) both E and B satisfy the wave equation:

∇2u =1

c2

∂2u

∂t2,

where c is the speed of the wave. Notice that here ∇2 denotes the vectorLaplacian.

We use the vector identity:

∇× (∇× F) =∇(∇ · F)−∇2F.

In the absence of charges and currents, Maxwell’s equations become:

∇ · E = 0

∇× E = −∂B

∂t∇ ·B = 0

∇×B = ε0µ0∂E

∂t.

Using the vector identity above we have:

ε0µ0∂2E

∂t2=∇× ∂B

∂t= −∇× (∇× E)

= −∇(∇ · E) +∇2E.


Since ∇ · E = 0, we have

∇2E = ε0µ0∂2E

∂t2.

A similar argument shows that B satisfies the same equation. The speed cis given by

c = 1/√ε0µ0.

This value is approximately 3 × 108 meters per second — a value that isdoubtless familiar to the reader.

Electrostatics

For a static (non-time-varying) electric field E, the magnetic field B mustalso be static, i.e., ∂B/∂t = 0. Thus, by the second of Maxwell’s equations,

∇× E = 0,

so E is conservative (at least on simply connected regions). Therefore,

E = −∇Φ,

for some scalar field Φ which is determined up to an additive constant. (Notethe sign.)

Now, using the first of Maxwell’s equations we obtain:

div E = −div (∇Φ) =1

ε0ρ,

or

∇2Φ = − 1

ε0ρ.

The reader may recall that this is called Poisson’s equation.

If the region has no charges, then we have

∇2Φ = 0.

The reader may recall that this is called Laplace’s equation.


A Spherical Capacitor

We are now in the position to find the electric field for a spherical capacitorgiven certain boundary conditions. A spherical capacitor consists of twoconcentric spheres. Let R1 denote the radius of the inner sphere and R2

denote the radius of the outer sphere. Suppose that the inner sphere ismaintained at a potential of V0 and the outer one at a potential of 0. Clearly,by symmetry, Φ is independent of φ and θ:

Φ(ρ, φ, θ) = Φ(ρ).

Therefore, in this case, Laplace’s equation becomes

1

ρ2

d

dρ

(ρ2dΦ

dρ

)= 0,

with boundary conditions Φ(R1) = V0, Φ(R2) = 0.The solution to the differential equation is

Φ = −c1

ρ+ c2,

where c1, c2 are arbitrary constants. (Integrate twice.)Plugging into the boundary conditions and solving for c1 and c2 we obtain

c1 =V0R1R2

R1 −R2

, c2 =V0R1

R1 −R2

.

So,

Φ =V0R1

R1 −R2

(1− R2

ρ

).

Therefore,

E = −grad Φ = −∂Φ

∂ρeρ =

V0R1R2

R2 −R1

(1

ρ2

)eρ.

References

Since this is not a textbook in electricity and magnetism, we have barelyskimmed the surface of this material. Every elementary physics text in elec-tricity and magnetism should contain at least some discussion of this ma-terial. See, for example, [9]. For an elementary book dealing solely withMaxwell’s equations, see [8]. A more advanced text on Maxwell’s equationsis [11]. And for an elementary treatment of electrostatics via vector analysisthat also mentions Maxwell’s equations, we recommend [19].


5.5.1 Exercises

In these Exercises, assume that all regions, surfaces, curves, and functionsare such that all of the physical laws stated in this section and any necessaryformal manipulations are, in fact, valid.

1. Can F(x, y, z) = xzi− xyj + yzk represent an electric field in an openregion of space with no net charge? Why or why not?

2. For what values of n can F(ρ, θ, φ) = Kρneρ (K constant) represent anelectric field?

3. An (infinite) coaxial capacitor is made up of two infinite coaxial cylin-drical conductors with radii R1 and R2 (R1 < R2). Suppose the innerconductor (radius R1) is held at a constant potential V0 and the outer(radius R2) at 0 potential. Use Laplace’s equation to find the potentialΦ between the cylinders. Then use this to find the electric field. (Hint:Use cylindrical coordinates. See Exercise 7 in Section 5.4.1.)

4. Show that the electric field due to a point charge at the origin is pro-portional to eρ/ρ

2. (Hint: Assume E(ρ, θ, φ) = E(ρ).)

5. Prove the vector identity:

∇× (∇× F) =∇(∇ · F)−∇2F.

6. Prove the vector identity:

∇ · (F×G) = G · ∇× F− F · ∇×G.

7. Prove Poynting’s theorem:

∂u

∂t+∇ · S = −J · E,

where u = 12

(ε0|E|2 + |B|2/µ0) is the electromagnetic energy densityand S = E×B/µ0 is called the Poynting vector. (Hint. Use Exercise 6.)

8. Derive the equation

∇×B = K3∂E

∂t+K4J

from the Ampere-Maxwell law.


9. Use Maxwell’s equations to show that the magnetic field B satisfies thewave equation:

∇2B = ε0µ0∂2B

∂t2.

10. Suppose that φ and ψ are twice-continuously differentiable functions ofone variable. Show that

u(x, t) = φ(x− ct) + ψ(x+ ct)

is a solution to the one-dimensional wave equation:

∂2u

∂x2=

1

c2

∂2u

∂t2.

(This is called d’Alembert’s solution.)

11. Use the result of Exercises 10 to derive the solution

u(x, t) =1

2[f(x− ct) + f(x+ ct)]

to the initial-value problem

∂2u

∂x2=

1

c2

∂2u

∂t2, u(x, 0) = f(x),

∂u

∂t(x, 0) = 0,

where f is twice-continuously differentiable.

12. Use the result of Exercises 10 to derive the solution

u(x, t) =1

2c

∫ x+ct

x−ctg(s) ds

to the initial-value problem

∂2u

∂x2=

1

c2

∂2u

∂t2, u(x, 0) = 0,

∂u

∂t(x, 0) = g(x),

where g is continuously differentiable.

13. Using the results of Exercises 11 and 12, show that

u(x, t) =1

2[f(x− ct) + f(x+ ct)] +

1

2c

∫ x+ct

x−ctg(s) ds


is a solution to the initial-value problem

∂2u

∂x2=

1

c2

∂2u

∂t2, u(x, 0) = f(x),

∂u

∂t(x, 0) = g(x),

where f is twice-continuously differentiable and g is continuously dif-ferentiable.

14. Suppose that f is a twice-continuously differentiable function of onevariable. Let k1, k2, k3 be constants, and let u0 be a constant vector.Show that

u(x, y, z, t) = u0f(k1x+ k2y + k3z − ωt),

where ω = ±c√k2

1 + k22 + k2

3, is a solution to the wave equation:

∇2u =1

c2

∂2u

∂t2,

15. Use Maxwell’s equations to derive the continuity equation

∇ · J +dρ

dt= 0.

Appendix A

The Modern Stokes’ Theorem

The careful reader may have noted that Green’s theorem, the divergencetheorem, and Stokes’ theorem are similar in the following respect: Roughlyspeaking, the integral of some function over a particular region is equal tothe integral of a related function over the boundary of the region (subjectto orientation). This might lead one to speculate on the possibility thatthese theorems could somehow be viewed as special cases of a more generaltheorem. This is, in fact, true. In this appendix we will state a theorem (themodern version of Stokes’ theorem) that applies in a very general setting, andsubsequently derive Green’s theorem, the divergence theorem, and Stokes’theorem, as well as the fundamental theorem for line integrals, as relativelysimple corollaries.

The modern version of Stokes’ theorem applies to the higher dimensionalanalogues of curves and surfaces (these are called manifolds); however, wewill consider only one, two, and three dimensional objects.

Remark. There are several more-or-less elementary sources for the materialin this appendix. See, for example, [2], [5], [6], [14], [17], [18], and [21].We’ve chosen to adapt the approach followed in Spivak [21], but without theprofusion of difficult definitions. Instead, we simply state throughout thebasic formulas that are used in calculations, and limit our consideration totwo and three dimensional euclidean space. We hope that our approach willprove amenable to those meeting this material for the first time.

185

186 APPENDIX A. THE MODERN STOKES’ THEOREM

Notation

In this appendix we will adopt some notational changes that we hope willnot be too confusing.

First, we will identify the vector 〈x, y, z〉 with the point (x, y, z), essen-tially replacing angle brackets with parentheses. In conformity with this wewill write

(x1, y1, z1) + (x2, y2, z2) = (x1 + x2, y1 + y2, z1 + z2) ,

(x1, y1, z1) · (x2, y2, z2) = x1x2 + y1y2 + z1z2,

and similarly for other operations we’ve defined involving vectors.We will also drop the practice of using boldfaced type for vectors. So, for

example, the vector function

F(x) = 〈F1(x), F2(x), F3(x)〉

will be writtenF (x) = (F1(x), F2(x), F3(x)) ,

and similarly for vector functions of two variables. Recalling that gradf is avector function, this means we will write

gradf =

(∂f

∂x,∂f

∂y,∂f

∂z

),

and similarly for the curl and any other vector-valued operators.The same notational changes will be applied to vectors in the plane.Moreover, we will always use the letters “x, y, z” to stand for the coor-

dinates in space. Thus, we will always write a curve, say R, as

R(x) = (f(x), g(x), h(x))

instead of, say,R(t) = (f(t), g(t), h(t)) .

Likewise, we will always write a surface, again say R, as

R(x, y) = (f(x, y), g(x, y), h(x, y))

instead of, say,R(u, v) = (f(u, v), g(u, v), h(u, v)) .

Again, similar considerations will apply to planar curves and surfaces.

A.1. DIFFERENTIAL FORMS 187

A.1 Differential Forms

Developing differential forms from scratch requires quite a bit of work, notto mention linear algebra. Therefore, we will not define differential forms.The important thing for us is that they can all be expressed in terms ofbasic 1-forms (dx, dy, dz) and an operation between them, called the wedgeproduct, in a relatively simple way. Once differential forms are expressed inthis way they can be formally manipulated according to fairly simple rules.

As mentioned before, we will consider differential forms only in the planeand in space. (In Section A.4 we will ask the reader to dabble in four dimen-sional space.)

• A 0-form is a function. In this case, either f(x, y) or f(x, y, z).

• A 1-form in the plane can be written

ω = f(x, y) dx+ g(x, y) dy.

• A 1-form in space can be written

ω = f(x, y, z) dx+ g(x, y, z) dy + h(x, y, z) dz.

• A 2-form in the plane can be written

ω = f(x, y) dx ∧ dy.


ω = f(x, y, z) dx ∧ dy + g(x, y, z) dy ∧ dz + h(x, y, z) dz ∧ dx.


ω = f(x, y, z) dx ∧ dy ∧ dz.

These are all of the nontrivial differential forms that exist in the plane andin space.

The symbol “∧” is read “wedge.” The expression dx ∧ dy, for example,is the wedge product of the 1-forms dx and dy. We will have more to sayabout the wedge product a little later on.


A differential form is continuous [differentiable, etc.] on an open regionA, if the relevant functions f , g, h above are continuous [differentiable, etc.]on A. In what follows we will assume that all differential forms are at leastcontinuously differentiable.

We can add forms of the same type in the obvious way. For example,

f dx ∧ dy + g dx ∧ dy = (f + g) dx ∧ dy.

Scalar multiplication is defined so that the usual algebraic properties hold.So, for example, if

ω = f dx ∧ dy + g dy ∧ dz,

then

c ω = cf dx ∧ dy + cg dy ∧ dz.

With these definitions, the set of n-forms defined in a particular openregion in euclidean space is a vector space over the real numbers. The additiveidentity will be denoted simply by 0. So, for every differential form ω, wehave 0ω = 0 and 1ω = ω.

If f is a function and ω is any differential form, then fω is defined point-wise.

Given any two differential forms defined on an open region of either theplane or of space, say ω and η, one can define the wedge product, ω ∧ η. Iff is a function, then f ∧ ω is defined to be fω. For other differential formsthe definition is quite complicated, but the following properties will sufficefor our purposes.

Suppose ω, ω1, ω2 are l-forms, η, η1, η2 are k-forms, and θ is an m-formdefined on an open region of the plane or space. Then we have the following:

• (ω ∧ η) ∧ θ = ω ∧ (η ∧ θ)

• (ω1 + ω2) ∧ η = ω1 ∧ η + ω2 ∧ η

• ω ∧ (η1 + η2) = ω ∧ η1 + ω ∧ η2

• (fω) ∧ η = ω ∧ (fη) = f(ω ∧ η)

• dx ∧ dy = −dy ∧ dx, dy ∧ dz = −dz ∧ dy, and dx ∧ dz = −dz ∧ dx

• dx ∧ dx = dy ∧ dy = dz ∧ dz = 0

A.1. DIFFERENTIAL FORMS 189

Since the wedge product is associative, both (ω ∧ η) ∧ θ and ω ∧ (η ∧ θ)are written ω∧η∧θ. (This is why we were able to write dx∧dy∧dz withoutparentheses.) Note, however, that the wedge product is not commutative.

Using these properties we can calculate the wedge product of two differ-ential forms. For example,

(f1 dx+ g1 dy) ∧ (f2 dx+ g2 dy)

= f1f2 dx ∧ dx+ f1g2 dx ∧ dy + f2g1 dy ∧ dx+ g1g2 dy ∧ dy= (f1g2 − f2g1) dx ∧ dy.

The Differential

Given a differential form ω, we now define the differential of ω, dω.Recall that if f = f(x, y), then

df =∂f

∂xdx+

∂f

∂ydy,

and if f = f(x, y, z), then

df =∂f

∂xdx+

∂f

∂ydy +

∂f

∂zdz.

We define the differential of the other types of differential forms as follows:

• For ω = f dx+ g dy,

dω = df ∧ dx+ dg ∧ dy.

• For ω = f dx+ g dy + h dz,

dω = df ∧ dx+ dg ∧ dy + dh ∧ dz.

• For ω = f dx ∧ dy,dω = df ∧ dx ∧ dy.

• For ω = f dx ∧ dy + g dy ∧ dz + h dz ∧ dx,

dω = df ∧ dx ∧ dy + dg ∧ dy ∧ dz + dh ∧ dz ∧ dx.


• For ω = f dx ∧ dy ∧ dz,

dω = df ∧ dx ∧ dy ∧ dz = 0.

The pattern here should be reasonably clear.These expressions can be expanded using the properties of the wedge

product. For example, if ω = f(x, y) dx+ g(x, y) dy, then

dω = df ∧ dx+ dg ∧ dy

=

(∂f

∂xdx+

∂f

∂ydy

)∧ dx+

(∂g

∂xdx+

∂g

∂ydy

)∧ dy

=

(∂g

∂x− ∂f

∂y

)dx ∧ dy.

A.1.1 Exercises

1. For ω = f1 dx+g1 dy+h1 dz and η = f2 dx∧dy+g2 dy∧dz+h2 dz∧dx,find ω ∧ η.

2. For each of the following compute df or dω.

(a) f = x2y

(b) f = xy + yz

(c) ω = (xy + yz) dx

(d) ω = x2y dy ∧ dx− xz dx ∧ dy

3. For each type of differential form defined in Section A.1, prove that

d2ω = d(dω) = 0.

Assume all of the functions involved are twice continuously differen-tiable.

A.2 The Modern Stokes’ Theorem

Before proceeding any further, we would like to remind the reader of thedefinition of the composition of two functions. If F is defined at G(P ), thenthe composition of F with G, F ◦G, at P is given by

F ◦G(P ) = F (G(P )).

A.2. THE MODERN STOKES’ THEOREM 191

So, for example, if R(x) = (f(x), g(x), h(x)) is a curve in space and F =F (x, y, z) is defined at R(x), then

F ◦R(x) = F (f(x), g(x), h(x)) .

If R is a curve or a surface either in the plane or in space, we wantto consider an operator R∗. Again, the explicit definition is complicated;however, the following properties can be used for calculations.

• R∗(ω1 + ω2) = R∗(ω1) +R∗(ω2)

• R∗(Fω) = (F ◦R)R∗(ω)

• R∗(ω ∧ η) = R∗(ω) ∧R∗(η)

• If R(x) = (f(x), g(x)) is a smooth curve in the plane or R(x) =(f(x), g(x), h(x)) is a smooth curve in space, then

R∗(dx) =df

dxdx

R∗(dy) =dg

dxdx

R∗(dz) =dh

dxdx.

• If R(x, y) = (f(x, y), g(x, y), h(x, y)) is a smooth surface in space, then

R∗(dx) =∂f

∂xdx+

∂f

∂ydy

R∗(dy) =∂g

∂xdx+

∂g

∂ydy

R∗(dz) =∂h

∂xdx+

∂h

∂ydy.

Example A.2.1. Let R(x) = (f(x), g(x), h(x)) and ω = F dx+Gdy+H dz.Compute R∗(ω).


Solution.

R∗(ω) = R∗(F dx+Gdy +H dz)

= R∗(F dx) +R∗(Gdy) +R∗(H dz)

= (F ◦R)R∗(dx) + (G ◦R)R∗(dy) + (H ◦R)R∗(dz)

= (F ◦R)df

dxdx+ (G ◦R)

dg

dxdx+ (H ◦R)

dh

dxdx

=

[(F ◦R)

df

dx+ (G ◦R)

dg

dx+ (H ◦R)

dh

dx

]dx

Here F ◦ R(x) = F (f(x), g(x), h(x)), G ◦ R(x) = G (f(x), g(x), h(x)), andH ◦R(x) = H (f(x), g(x), h(x)).

Example A.2.2. Let R(x) = (f(x, y), g(x, y), h(x, y)) and ω = F dx ∧ dy.Compute R∗(ω).

Solution.

R∗(ω) = R∗(F dx ∧ dy)

= (F ◦R)R∗(dx ∧ dy)

= (F ◦R) (R∗ dx ∧R∗ dy)

= (F ◦R)

[(∂f

∂xdx+

∂f

∂ydy

)∧(∂g

∂xdx+

∂g

∂ydy

)]= (F ◦R)

(∂f

∂x

∂g

∂y− ∂f

∂y

∂g

∂x

)dx ∧ dy

Here F ◦R(x, y) = F (f(x, y), g(x, y), h(x, y)).

The Integral of an n-Form over an n-Chain

We are now in a position to define the integral of an n-form over an n-chain.In this text, a 1-chain is a piecewise smooth curve, a 2-chain is a piecewisesmooth surface, and a 3-chain is a compact region whose boundary is apiecewise smooth surface. In what follows, we will assume that all functionsare at least continuously differentiable.

We define the integral of an n-form over an n-chain for n = 1, 2, 3 asfollows.


• If I = [a, b], then ∫I

F dx =

∫ b

a

F dx.

• If D is a compact region in the plane whose boundary is a piecewisesmooth curve, then ∫

D

F dx ∧ dy =

∫∫D

F dA.

• If V is a compact region in space whose boundary is a piecewise smoothsurface, then ∫

V

F dx ∧ dy ∧ dz =

∫∫∫V

F dV.

• If C is a smooth path and R is a smooth parametrization of C whosedomain is I = [a, b], then ∫

C

ω =

∫I

R∗ω.

Again, if C is a closed path, this is usually written as∮C

ω.

• If S is a smooth surface and R is a smooth parametrization of S whosedomain is D, then ∫

S

ω =

∫D

R∗ω.

It can be shown that this integral is independent of parametrization aslong as the parametrizations have the same orientation. Also, this integralis linear, since the integrals on the right above are linear.

The integral of a differential form over a piecewise smooth path or a piece-wise smooth surface is defined as the sum of the integrals over the smoothpaths or smooth surfaces that constitute the piecewise smooth path or piece-wise smooth surface.

Example A.2.1 shows that the integral of the 1-form F1 dx+F2 dy+F3 dzover the 1-chain C is equal to the line integral∫

C



as defined in Chapter 4. Similarly, for the 1-form F1 dx+F2 dy in the plane.We will consider surface integrals a little later on.

Example A.2.3. Compute ∫S

z dx ∧ dy,

where the surface S is parametrized by

R(x, y) = (x+ y, x− y, xy), 0 ≤ x, y ≤ 1.

Solution. In the notation we’ve been using, F = z, f = x + y, g = x − y,h = xy, and D = {(x, y) : 0 ≤ x, y ≤ 1}. Using the result of the last example,we have ∫

S

z dx ∧ dy =

∫D

R∗(z dx ∧ dy)

=

∫D

xy

(∂f

∂x

∂g

∂y− ∂f

∂y

∂g

∂x

)dx ∧ dy

= −2

∫∫D

xy dA

= −2

∫ 1

0

∫ 1

0

xy dxdy = −1

2.

The Modern Version of Stokes’ Theorem

In the modern version of Stokes’ theorem we will assume the following.

• If M is a compact region in space whose boundary ∂M is a piece-wise smooth surface, then the orientation of ∂M is determined by theoutward unit normal.

• If M is a piecewise smooth surface whose boundary ∂M is a piecewisesmooth curve, then ∂M is given the induced orientation. Recall thismeans that the curve is oriented so that when one follows the pathon the side of the chosen normal vector in the positive direction thesurface is on the left. If M is such a surface in the plane, then we’llassume its orientation is given by n = k. The induced orientation onC in this case is referred to as “counterclockwise.”


We also need to make the following definition. If C is a piecewise smoothpath with initial point P and terminal point Q, then∫

∂C

f = f(Q)− f(P ).

Obviously, this depends on the orientation of C.

Theorem A.2.1 (The Modern Stokes’ Theorem). Suppose M is an (n+ 1)-chain whose boundary is an n-chain. If ω is an n-form defined on an openregion containing M , then ∫

M

dω =

∫∂M

ω.

If M is an (n+ 1)-chain without boundary, then∫M

dω = 0.

The proof of this theorem is beyond the scope of this text. However,in the next section we will show how the major classical theorems of vectoranalysis (Green’s theorem, the divergence theorem, and Stokes’ theorem) canbe derived as relatively simple corollaries of the modern version of Stokes’theorem.

Example A.2.4. Verify the modern Stokes’ theorem for the following.

M = B ={

(x, y, z) : x2 + y2 + z2 ≤ 1},

∂M = S ={

(x, y, z) : x2 + y2 + z2 = 1}, and

ω = z dx ∧ dy.

Solution. On the one hand,

dω = dz ∧ dx ∧ dy = dx ∧ dy ∧ dz,

so ∫B

dω =

∫B

dx ∧ dy ∧ dz =

∫∫∫B

dV =4

3π.

On the other hand, we can parametrize S by

R(x, y) = (sin x cos y, sinx sin y, cosx), 0 ≤ x ≤ π, 0 ≤ y ≤ 2π.


So, we take D = {(x, y) : 0 ≤ x ≤ π, 0 ≤ y ≤ 2π}. We leave it to the readerto show that

∂R

∂x× ∂R

∂y

points outward on S. Using Example A.2.2, a straightforward calculationshows that

R∗(ω) = z cosx sinx dx ∧ dy = cos2 x sinx dx ∧ dy.

Therefore,∫S

ω =

∫D

R∗(ω) =

∫D

cos2 x sinx dx ∧ dy

=

∫ 2π

0

∫ π

0

cos2 x sinx dx dy =4

3π.

A.2.1 Exercises

1. Compute∫Sω, where

ω = xy dy ∧ dz + x dz ∧ dx+ 3zx dx ∧ dy,

and S is parametrized by

R(x, y) = (x+ y, x− y, xy), 0 ≤ x, y ≤ 1.

2. Verify the modern Stokes’ theorem for

ω = x dx+ xy dy,

where M is the region inside the square with opposite vertices (0, 0)and (1, 1).

3. Verify the modern Stokes’ theorem for ω = x dz, where M is the surfaceparametrized by

R(x, y) =(xy, x+ y, x2 + y2

), 0 ≤ y ≤ x, 0 ≤ x ≤ 1.

A.3. STOKES-TYPE THEOREMS, REVISITED 197

4. Verify the Modern Stokes’ theorem for the following:

M ={

(x, y, z) : x2 + y2 + z2 ≤ 1}, and

ω = z3 dx ∧ dy.

5. A differential form ω is exact if there exists a differential form η suchthat ω = dη. Show that if ω is an exact n-form and M is an (n + 1)-chain as described in Section A.2, then∫

∂M

ω = 0.

Assume things are as differentiable as you like. (Hint. See Exercise 3in Section A.1.1.)

6. In Example A.2.4, show that

∂R

∂x× ∂R

∂y

points outward on S.

A.3 Stokes-Type Theorems, Revisited

We begin with the fundamental theorem for line integrals.

Theorem A.3.1. If C is a non-closed piecewise smooth path with initialpoint P and terminal point Q, either in the plane or in space, and F is acontinuously differentiable scalar field defined on an open region containingC, then ∫

C

gradF · dR = F (Q)− F (P ).

If C is a piecewise smooth closed path, and F is a continuously differentiablescalar field defined on an open region containing C, then∫

C

gradF · dR = 0.


Proof. We will assume C is a space curve; the proof for a plane curve issimilar. We will also assume, for simplicity, that C is smooth. We will justprove the first part of the theorem; the second part is similar.

In this case n = 0, so we will take ω = F and M = C. Let R(x) =(f(x), g(x), h(x)), a ≤ x ≤ b. Using the modern version of Stokes’ theorem,we have

∫C

gradF · dR =

∫ b

a

(∂F

∂x

df

dx+∂F

∂y

dg

dx+∂F

∂z

dh

dx

)dx

=

∫[a,b]

R∗(dF ) =

∫C

dF =

∫∂C

F

= F (Q)− F (P ).

Next, we use the modern version of Stokes’ theorem to prove Green’stheorem.

Theorem A.3.2 (Green’s Theorem). Suppose D is a compact region in theplane whose boundary is a piecewise smooth closed curve C oriented counter-clockwise. Also, assume P and Q are continuously differentiable on an openregion containing D. Then∮

C

P dx+Qdy =

∫∫D

(∂Q

∂x− ∂P

∂y

)dA.

Proof. Again for simplicity we will assume that C is a smooth curve. In thiscase n = 1, M = D, and ∂M = C. We take ω = P dx+Qdy. At the end ofSection A.1.1, we calculated that

dω =

(∂Q

∂x− ∂P

∂y

)dx ∧ dy.

Recall that our new definition of∮CP dx + Qdy is equivalent to our old


definition. So, using the modern version of Stokes’ theorem we have∮C

P dx+Qdy =

∫D

d(P dx+Qdy)

=

∫D

(∂Q

∂x− ∂P

∂y

)dx ∧ dy

=

∫∫D

(∂Q

∂x− ∂P

∂y

)dA.

Before considering the divergence theorem and Stokes’ theorem, we needa few more preliminaries. In the sequel, for technical reasons, we will alwaysassume that a unit normal vector to a smooth surface can be extended toa continuously differentiable function on an open region of space containingthe surface.

Theorem A.3.3. If S is a smooth surface oriented by the unit normal vectorn = (n1, n2, n3) and F is continuous on an open region containing S, then∫

S

F (n1 dy ∧ dz + n2 dz ∧ dx+ n3 dx ∧ dy) =

∫∫S

F dS.

Here∫∫

SF dS is the surface integral of F over S, as previously defined in

Section 4.4.

Proof. Let

R(x, y) = (f(x, y), g(x, y), h(x, y)) , (x, y) ∈ D,

be a smooth parametrization of S such that the orientation is as describedin the theorem. We need to calculate

R∗ (F (n1 dy ∧ dz + n2 dz ∧ dx+ n3 dx ∧ dy))

= F (n1R∗(dy ∧ dz) + n2R

∗(dz ∧ dx) + n3R∗(dx ∧ dy)) ,

where F , n1, n2, and n3 are all evaluated at R(x, y).


We know from Section 4.4 that

n1 =

(∂g

∂x

∂h

∂y− ∂h

∂x

∂g

∂y

)/∣∣∣∣∂R∂x × ∂R

∂y

∣∣∣∣n2 =

(∂h

∂x

∂f

∂y− ∂f

∂x

∂h

∂y

)/∣∣∣∣∂R∂x × ∂R

∂y

∣∣∣∣n3 =

(∂f

∂x

∂g

∂y− ∂g

∂x

∂f

∂y

)/∣∣∣∣∂R∂x × ∂R

∂y

∣∣∣∣ .In Example A.2.2, we calculated

R∗(dx ∧ dy) =

(∂f

∂x

∂g

∂y− ∂g

∂x

∂f

∂y

)dx ∧ dy.

Similarly,

R∗(dz ∧ dx) =

(∂h

∂x

∂f

∂y− ∂f

∂x

∂h

∂y

)dx ∧ dy

R∗(dy ∧ dz) =

(∂g

∂x

∂h

∂y− ∂h

∂x

∂g

∂y

)dx ∧ dy.

It follows that

R∗ (F (n1 dy ∧ dz + n2 dz ∧ dx+ n3 dx ∧ dy)) = F

∣∣∣∣∂R∂x × ∂R

∂y

∣∣∣∣ dx ∧ dy.Therefore,∫S

F (n1 dy ∧ dz + n2 dz ∧ dx+ n3 dx ∧ dy)

=

∫D

F

∣∣∣∣∂R∂x × ∂R

∂y

∣∣∣∣ dx ∧ dy =

∫∫D

F

∣∣∣∣∂R∂x × ∂R

∂y

∣∣∣∣ dA =

∫∫S

F dS.

In light of Theorem A.3.3, we write

dS = n1 dy ∧ dz + n2 dz ∧ dx+ n3 dx ∧ dy.

dS is a 2-form in space defined on an open region containing S; although, itis not necessarily the differential of a 1-form.

Thus, the integral of the differential form F dS over S is equal to thesurface integral

∫∫SF dS, as previously defined.

To prove the divergence theorem and Stokes’ theorem we use the followinglemma.


Lemma A.3.1. If S is a smooth surface, then∫∫S

n3 dS =

∫S

dx ∧ dy∫∫S

n2 dS =

∫S

dz ∧ dx∫∫S

n1 dS =

∫S

dy ∧ dz.

Proof. We show the first equality. The others are proved similarly. Let Dbe the domain of R and write R(x, y) = (f(x, y), g(x, y), h(x, y)), (x, y) ∈ D.Then, using Example A.2.2, we have∫

S

dx ∧ dy =

∫∫D

R∗(dx ∧ dy)

=

∫∫D

(∂f

∂x

∂g

∂y− ∂f

∂y

∂g

∂x

)dA.

On the other hand,∫∫S

n3 dS =

∫∫D

[(∂f

∂x

∂g

∂y− ∂g

∂x

∂f

∂y

)/∣∣∣∣∂R∂x × ∂R

∂y

∣∣∣∣] ∣∣∣∣∂R∂x × ∂R

∂y

∣∣∣∣ dA=

∫∫D

(∂f

∂x

∂g

∂y− ∂f

∂y

∂g

∂x

)dA.

Now, we can use the modern version of Stokes’ theorem to give a proofof the divergence theorem and Stokes’ theorem.

Theorem A.3.4 (The Divergence Theorem). Let D be a compact region inspace whose boundary is a piecewise smooth closed surface S with outwardunit normal n = (n1, n2, n3). Let F be a continuously differentiable vectorfield defined on an open region containing D. Then∫∫∫

D

divF dV =

∫∫S

F · n dS.

Proof. Again, for simplicity we will assume that S is a smooth surface. Inthis case n = 2, M = D, and ∂M = S. We take

ω = F1 dy ∧ dz + F2 dz ∧ dx+ F3 dx ∧ dy.


By Lemma A.3.1, we have∫∫S

F · n dS =

∫∫S

(F1n1 + F2n2 + F3n3) dS

=

∫S

F1 dy ∧ dz + F2 dz ∧ dx+ F3 dx ∧ dy

=

∫S

ω.

On the other hand, it is easy to show that

dω = divF dx ∧ dy ∧ dz.

Therefore, by the modern version of Stokes’ theorem, we have∫∫∫D

divF dV =

∫D

divF dx ∧ dy ∧ dz

=

∫D

dω =

∫S

ω =

∫∫S

F · n dS.

Finally, we have the following.

Theorem A.3.5 (Stokes’ Theorem). Let S be a compact piecewise smoothsurface whose boundary is a piecewise smooth closed curve C with the inducedorientation. Let F be a continuously differentiable vector field defined on anopen region containing S. Then∮

C

F · dR =

∫∫S

curlF · n dS.

Proof. Again, for simplicity we will assume that S is a smooth surface andC is a smooth curve. In this case n = 1, M = S, and ∂M = C. We take

ω = F1 dx+ F2 dy + F3 dz.

A straightforward calculation shows that

dω =

(∂F3

∂y− ∂F2

∂z

)dy∧dz+

(∂F1

∂z− ∂F3

∂x

)dz∧dx+

(∂F2

∂x− ∂F1

∂y

)dx∧dy.


Again using Lemma A.3.1, we see that∫S

dω =

∫∫S

curlF · n dS.

Therefore, by the modern version of Stokes’ theorem we have∮C

F · dR =

∮C


=

∮C

ω =

∫S

dω =

∫∫S

curlF · n dS.

A.3.1 Exercises

1. Show that if

ω = F1 dy ∧ dz + F2 dz ∧ dx+ F3 dx ∧ dy,

then

dω = divF dx ∧ dy ∧ dz.

2. Show that if

ω = F1 dx+ F2 dy + F3 dz,

then

dω =

(∂F3

∂y− ∂F2

∂z

)dy ∧ dz +

(∂F1

∂z− ∂F3

∂x

)dz ∧ dx

+

(∂F2

∂x− ∂F1

∂y

)dx ∧ dy.

3. Complete the proof of Lemma A.3.1.


A.4 Project

In this project you will be asked to extend part of the treatment given inthis appendix to four dimensional euclidean space, hereafter know as 4-space.You will then be asked to use these results to calculate the volume of a balland the volume of a sphere in 4-space.1

Let x, y, z, and t be the coordinates in 4-space.

• Write out, in standard form, all of the nontrivial differential forms in4-space.

• To the properties of the wedge product add the following:

dx ∧ dt = −dt ∧ dx, dy ∧ dt = −dt ∧ dy, dz ∧ dt = −dt ∧ dz, and

dt ∧ dt = 0.

Let

ω = f1 dx+ g1 dy + h1 dz

η = f2 dx+ g2 dy + h2 dz

θ = f3 dx+ g3 dy + h3 dz,

and calculate ω ∧ η ∧ θ.

• Extend the definition of the differential of a differential form to 4-space.

• Extend the properties of the operator R∗ to 4-space.

• Without explicitly defining chains in 4-space, extend the definition ofthe integral of an n-form over an n-chain to 4-space.

• As mentioned before, the modern version of Stokes’ theorem is valid inhigher dimensions. Use the modern version of Stokes’ theorem to showthat the volume of the ball of radius a in 4-space is π2a4/2. (Hint: Thesphere of radius a, centered at the origin, can be parametrized by

R(x, y, z) = (a cosx sin y sin z, a sinx sin y sin z, a cos y sin z, a cos z),

where 0 ≤ x ≤ 2π, 0 ≤ y ≤ π, and 0 ≤ z ≤ π. Apply the modernversion of Stokes’ theorem with ω = −t dx ∧ dy ∧ dz.)

1What we call the volume of a sphere in 4-space other authors call the area or surfacearea.

A.4. PROJECT 205

• If F = (F1, F2, F3, F4) is a vector field in 4-space, then we define thedivergence of F by

divF =∂F1

∂x+∂F2

∂y+∂F3

∂z+∂F4

∂t.

The divergence theorem applies to “nice” regions in 4-space:∫∫∫∫M

divF dV =

∫∫∫∂M

F · n dV,

where dV is the volume element of the region being integrated over.

Use the divergence theorem and the previous result to show that thevolume of the sphere in 4-space, or hypersphere, of radius a is 2π2a3.(Hint: Choose F appropriately.)


Appendix B

Planar Motion in PolarCoordinates: Kepler’s Laws

In this appendix we consider motion in a plane using polar coordinates withan eye to deriving Kepler’s laws from Newton’s (second) law of motion andNewton’s law of universal gravitation. For simplicity, we will assume that allfunctions appearing in this appendix are twice continuously differentiable.

B.1 Planar Motion in Polar Coordinates

Velocity and Acceleration

Recall that the transformations for polar coordinates are:

x = r cos θ, y = r sin θ,

where we will assume r ≥ 0. It often is helpful to remember the identityr2 = x2 + y2.

The most appropriate orthogonal unit vectors to consider here are:

er = cos θi + sin θj

eθ = − sin θi + cos θj.

Notice that these are not constant vectors.Note that

derdθ

= eθ anddeθdθ

= −er.

207

208 APPENDIX B. PLANAR MOTION IN POLAR COORDINATES

In polar coordinates the motion of an object in the plane can be written

R(t) = r(t) cos θ(t)i + r(t) sin θ(t)j = r(t)er(t).

The velocity can then be written

v =dR

dt=dr

dter + r

derdt

=dr

dter + r

derdθ

dθ

dt

=dr

dter + r

dθ

dteθ.

A straightforward calculation (which we leave to the reader) gives theacceleration:

a =dv

dt=

[d2r

dt2− r

(dθ

dt

)2]

er +

[rd2θ

dt2+ 2

dr

dt

dθ

dt

]eθ.

Motion in a Plane

Suppose F is the force acting on an object whose motion lies in a plane,which we can take to be the xy-plane. Then, by Newton’s (second) law ofmotion,

F = Frer + Fθeθ = ma,

where m is the (constant) mass of the object.Using the previous expression for the acceleration, we have

Fr = md2r

dt2−mr

(dθ

dt

)2

Fθ = mrd2θ

dt2+ 2m

dr

dt

dθ

dt.

Multiplying the last equation by r, we obtain

rFθ =d

dt

(mr2dθ

dt

)or Fθ =

1

r

d

dt

(mr2dθ

dt

).

B.1.1 Exercise

1. Derive the expression for the acceleration in polar coordinates from thevelocity.

B.2. KEPLER’S LAWS 209

B.2 Kepler’s Laws

Kepler’s laws of planetary motion were formulated by Johannes Kepler c.1605based on voluminous data obtained by Tycho Brahe. Newton later “proved”Kepler’s laws using his (second) law of motion and law of universal gravita-tion.

Kepler’s laws of planetary motion can be stated as follows:

1. The orbit of each planet is an ellipse with the Sun at one of the foci.

2. A line from the Sun to a planet sweeps out equal areas in equal timeintervals.

3. If T is the period of the orbit and a is the length of the semi-major axis,then for every planet in the Solar System T 2/a3 has the same value.

Kepler’s laws are valid for any “planetary system” that satisfies the fol-lowing assumptions.

• The central body is much more massive than any orbiting body. (OurSun is more that a 1000 times more massive than Jupiter, which is byfar the most massive planet.)

• The gravitational force on any orbiting body due to all other objects inthe universe is much less than the gravitational force due to the centralbody.

In this case, the force on a orbiting body can be modeled by a central,inverse-square force field with the central body at the origin, i.e.,

F =

(−GMm

|R|3

)R =

(−GMm

r2

)er,

where G is the universal gravitation constant, M is the mass of the centralbody, and m is the mass of the orbiting body.

Kepler’s Second Law

We will demonstrate the second law first, since it depends only on the forcefield being a central force field.


Let A(t) denote the area swept out by the position vector R = R(t)during the time interval [t0, t] for some fixed t0. Let

∆A = A(t0 + ∆t)− A(t0) = A(t0 + ∆t).

Then, by the intermediate value theorem,

∆A =1

2r2∆θ,

where r is evaluated at some point in [t0, t0 +∆t] and ∆θ = θ(t0 +∆t)−θ(t0).Therefore,

∆A

∆t=

1

2r2 ∆θ

∆t.

Letting ∆t −→ 0+, we obtain

dA

dt=

1

2r2dθ

dt,

where r is evaluated at t0. (This follows from the continuity of r.)If Fθ = 0, then

d

dt

(mr2dθ

dt

)= 0,

from which it follows that dA/dt is constant.That is, for a central force field the rate at which the position vector R

sweeps out area is a constant.

This demonstrates Kepler’s second law.

Kepler’s First Law

Since the demonstration of Kepler’s first law is somewhat involved, we willproceed in steps.1

Step 1. Using Newton’s law F = ma, the expression for the accelerationin polar coordinates, and equating components, we have

d2r

dt2− r

(dθ

dt

)2

=−GMr2

r2dθ

dt= K,

1This approach is adapted from a series of Exercises in [7].


where K is a constant.

Step 2. Substituting for dθ/dt in the first equation and multiplying bydr/dt, we obtain

d2r

dt2dr

dt− K2

r3

dr

dt=−GMr2

dr

dt.

Since,

d

dt

[1

2

(dr

dt

)2]

=dr

dt

d2r

dt2,

the integral ofdr

dt

d2r

dt2,

with respect to t, is

1

2

(dr

dt

)2

.

Integrating the other two terms with respect to t and multiplying by 2,we obtain (

dr

dt

)2

+K2

r2=

2GM

r+ C,

where C is a constant.

Step 3. Letting p = 1/r or r = 1/p, we easily obtain

1

p4

(dp

dt

)2

+K2p2 = 2GMp+ C.

Step 4. Assuming dθ/dt is never 0, that is, the planet doesn’t stop or“backup,” we can (at least in theory) solve θ = θ(t) for t and write t = t(θ)and p = p(t(θ)). Then, using the equation for dθ/dt in Step 1, we have(

dp

dθ

)2

=

(dp

dt

)2(dt

dθ

)2

=

(dp

dt

)2r4

K2=

(dp

dt

)21

K2p4

or

K2

(dp

dθ

)2

=1

p4

(dp

dt

)2

.


Then, using the result from Step 3, we obtain

K2

[(dp

dθ

)2

+ p2

]= 2GMp+ C.

Step 5. Differentiating the last equation with respect to θ results in

2dp

dθ

d2p

dθ2+ 2p

dp

dθ=

2GM

K2

dp

dθ

ord2p

dθ2+ p =

GM

K2.

Step 6. In elementary differential equations one learns (or, at least,should learn) that the (general) solution to this differential equation can bewritten

p = A cos(θ − δ) +GM

K2,

where A is a nonnegative constant and δ is a constant.By rotating the axes, if necessary, we can take δ = 0. Therefore, by

choosing our axes appropriately, we have

p = A cos θ +GM

K2.

Step 7. This last equation can be written

r =K2/(GM)

1− e cos θ,

where e is a nonnegative constant.This is known to be the equation of a conic section where the origin is

one of the foci. The number e is called the eccentricity of the conic section.If e = 0, then the orbit is a circle, which we consider to be a special typeof ellipse for which the foci coincide. If 0 < e < 1, then the orbit is anellipse. If e = 1, then the orbit is a parabola. If e > 1, then the orbit is ahyperbola. (See, for example, [22].) Since the orbit of a planet is obviouslynot a parabola or a hyperbola, it must be an ellipse.

This demonstrates Kepler’s first law.


Kepler’s Third Law

We now proceed to Kepler’s third law. We will use the same notation weused in our derivation of the first and second laws. Again, we will proceedin steps.

Step 1. Suppose that 0 ≤ e < 1. The polar coordinate equation

r =K2/(GM)

1− e cos θ

can be written, in cartesian coordinates, as

(x− h)2

a2+y2

b2= 1,

where

a =K2/(GM)

1− e2, b =

K2/(GM)√1− e2

, and h = ae.

We leave it to the reader to verify this.

Step 2.Recall that the area swept out by the position vector R satisfies

dA

dt=

1

2K.

If we start to measure the area from time t = 0, then we have

A =1

2Kt.

Since the area of the ellipse swept out by the position vector R during oneperiod is πab (see Example 5.1.2), it follows that

πab =1

2KT.

Step 3. Using the results of Step 1 and Step 2, it follows that

T 2 =4π2

GMa3.

Again, we leave the verification of this to the reader.

This demonstrates Kepler’s third law.


Remark B.2.1. The average distance between the Sun and the Earth isdefined to be 1 astronomical unit (AU).2 Since (fortunately for us) the Earth’sorbit is very nearly circular, the length of the semi-major axis of the Earth’sorbit is very close to 1 AU. Therefore, in our Solar System, if we measuredistance in AU and time in years, Kepler’s third law can be written T 2 = a3.

B.2.1 Exercises

1. For the orbit of the planet Uranus the length of the semi-major axis isaround 19.2 AU. Find the period of the orbit.

2. Look up the orbital properties of the largest five moons of Uranus onWikipedia or obtain them from some other source. Plot the periodT versus the semi-major axis a on log-log graph paper by hand or usecomputer software to obtain a log-log plot. Use a for the horizontal axisand T for the vertical axis. (Alternatively, you can plot log T versuslog a on regular graph paper.) The points should all lie on a straightline. What is the slope of this line? (Hint. If you do this by hand,choose units that make the numbers reasonable, for example, T in tensof thousands of kilometers and a in days.)

3. Verify Step 3 in the derivation of Kepler’s first law.

4. Verify, by direct substitution, that

p = A cos(θ − δ) +GM

K2,

where A is a nonnegative constant and δ is a constant, is a solution of

d2p

dθ2+ p =

GM

K2.

5. Verify the equation in Step 7 in the derivation of Kepler’s first law.

6. Verify Step 1 in the derivation of Kepler’s third law.

7. Verify Step 3 in the derivation of Kepler’s third law.

21 AU ≈ 93, 000, 000 mi ≈ 150, 000, 000 km

Appendix C

Solutions to Selected Exercises

1.1.1

1. 〈0,−2, 1〉

2. 7i

3. (a) 4i− 7j− 3k

(b)√

14

4. (a) 〈18, 14, 2〉(b) 2

√131

5. (a) 2j + 3k

(b)√

13

6.1√6〈1, 2,−1〉

7.2√41

(−4i + 5j)

8.2√41

(4i− 5j)

9. − 1√5

(6i + 3j)

215

216 APPENDIX C. SOLUTIONS TO SELECTED EXERCISES

10.4√6

(i− j + 2k)

1.2.1

1. 6

2. 14

3. −3

4. cos−1

(6√574

)

5. cos−1

(14

15

)

6. cos−1

(− 3

2√

21

)7.

1

5√

2

8.5

3√

6

9. − 4

3√

26

10.1√10

11. cos−1(√

5/6)

12. − 9√41

13.22√

2

5

14.27√17

217

15.

⟨0,−24

41,30

41

⟩+

⟨3,

65

41,52

41

⟩

16.

(28

9i +

28

9j− 14

9k

)+

(−1

9i +

8

9j +

14

9k

)17. 〈−2/3,−1/3, 2/3〉+ 〈14/3,−14/3, 7/3〉

18. 1200√

3 ft-lb ≈ 2078.5 ft-lb

19. 6 QPR ≈ 46.25◦, 6 PQR ≈ 74.54◦, 6 PRQ ≈ 59.21◦

1.3.1

1. 〈13,−15,−12〉

2. −5i− 5j + 5k

3. 4i− 3j + 2k

4.√

538

5. 5√

3

6.√

29

7. 8i− 56j− 73k

8. ± 1√70〈6, 3, 5〉

9. (i× i)× j 6= i× (i× j)

10.√

790/2

11.√

2909/2

12. 9√

10/2

13.√

641/2

14.√

21/2


15. (3,−2, 1), 9

(−1, 0, 1), 9

16. 4

17. 1/6

18. 2

1.4.1

1. x = −1 + t, y = 2 + 2t, z = 3− 4t

x+ 1 = (y − 2)/2 = (3− z)/4

2. x = 1 + 2t, y = 1 + t, z = 1− 5t

(x− 1)/2 = y − 1 = (1− z)/5

3. x = −2 + 8t, y = 1 + t, z = 5− 8t

4. x = 2− 3t, y = −1 + 3t, z = 2t

2− x3

=y + 1

3=z

2

5. They do not intersect.

6. (−2, 4, 0)

7. (−1,−5, 0)

8. cos−1

(√6

11

)

9.

√43

15

10. 9x− 5y − z = 10

11. 26x− 27y − 20z = 123

12. 21x− 6y + 7z = −38

219

13. 4x− 7y − 24z = −37

14. x− 2y − 2z = 1

15. x+ 11y + 3z = 20

16. −3x+ 2y + 9z = 11

17. 2x+ 5y − 9z = 22

18.5

3√

3

19.10√

6

20.10√107

21. x− 2y + z = 0

22. 3x+ 2y − z = 4

2.1.1

1. cos(2t)i + πeπtj + 15t4k

2. [3, 5)

F′(t) =4

t− 5i +

6√t− 5

j + 12t3k

F′′(t) = − 4

(t− 5)2i− 3

(t− 5)3/2j + 36t2k

3. [−1,∞)

F′(t) =3

2 + ti +

6√t+ 1

j + 12t5k

F′′(t) = − 3

(2 + t)2i− 3

(t+ 1)3/2j + 60t4k

4. 120i− 260

3j


5.3325

3i− 38j + 65k

6. (−t3 − 7t+ 5) i + (−t4 − 3t+ 7) j + (3t2 + 2t+ 4) k

2.2.1

1. R(1) = i + j

v(1) = 2i− 2j

2. −32i + j + 8k

33

3. (a)1

2t2 + 1(2ti + 2t2j + k)

(b)1

3(2i + 2j + k)

4.1√

9t2 + 25[(3 cos t− 3t sin t)i + (3 sin t+ 3t cos t)j + 4k]

5. −yai +

x

aj

6.1− x

2=y + 1

3= 2− z

7.1− x√

3=y − 3

√3

3= z − π

3

8. x = −9− t, y = −2 + 2t, z = −3 + 2t

9. x = 8 + 4t, y = 6 + 32t, z = 1− t

10. x = 1 + t, y = 1− 12t, z = 2− 3t

11. (a) v(t) = −2 sin ti− j + 2 cos tk

(b) 2π√

5

12. y = − 2

πx+

π

2

221

13. (a) π√

5

(b) x =π

2+ t, y = −1, z = −2t

14. 68

15. 14/3

16. π√

4π2 + 1 +1

2ln(2π +

√4π2 + 1

)17. 6a

18. x =

√2s

3cos

(1

2ln

(2s

3

)), y =

√2s

3sin

(1

2ln

(2s

3

)), z =

√2s

3

19.1

|t|√

5(t sin ti + t cos tj + 2tk)

20. 3√

5π2/2

21. ln(√

2 + 1)

2.3.1

1. 9t/ (9t2 + 25)1/2(

9t2 + 36− 81t2

9t2 + 25

)1/2

2.2

(1 + 4x2)3/2

3. 1/a

(−1/a)(xi + yj)

k

0

4. 1

0


−(3/5) sin si + cos sj− (4/5) sin sk

−(3/5) cos si− sin sj− (4/5) cos sk

−(4/5)i + (3/5)k

5.1

5|t|

6. R(t) = (3 + 2 sin t)i +√

2 cos tj +√

2 cos tk, 0 ≤ t ≤ 2π

3.1.1

1. (a) {(x, y, z) : x2 + y2 + z2 < 1}(b) {(x, y, z) : x2 + y2 + z2 = 1}(c) {(x, y, z) : x2 + y2 + z2 > 1}(d) open

2. (a) {(x, y, z) : x2 + y2 + z2 < 1}(b) {(x, y, z) : x2 + y2 + z2 = 1}(c) {(x, y, z) : x2 + y2 + z2 > 1}(d) closed

3. (a) the empty set

(b) {(x, y, z) : x2 + y2 + z2 = 1}(c) {(x, y, z) : x2 + y2 + z2 6= 1}(d) closed


(b) {(x, y, z) : x2 + y2 ≤ 1, z = 0}(c) {(x, y, z) : x2 + y2 > 1, z = 0} ∪ {(x, y, z) : z 6= 0}(d) neither

5. (a) {(x, y) : x2 + y2 < 1}(b) {(x, y) : x2 + y2 6= 1}(c) {(x, y) : x2 + y2 > 1}

223

(d) open


(b) all of space

(c) the empty set

(d) neither

7. 〈−4 cos(−1), 4 sin(−1), 4 cos(−1)〉

8.3

577

(√3− 2

)9. −69/5

10. −9

11.24√

5

12. i− 2j

6√

5

13.54√13

〈−1,−3〉6√

10

14. (a) −1

(b) 2

(c) 〈0,−1〉

15.8√11

16. −4/3

17. 3√

5

2i + k


18. If k < 0, then the level set is the empty set. If k = 0, then the level setis the origin. If k > 0, then the level set is the surface of the cube withvertices (±k,±k,±k).

19. 2x− 6y + z = 7

20. 4x− 2y − z = 1

21. −16x+ 4y − 7z = 14

22. 9x+ 8y − 9z = 34

23. 3x− 2y + z = 18

24. 4x+ 8y − 2z = 42

25. (a) 2x+ 4y + 6z = 12

(b) x = 1 + 2t, y = 1 + 4t, z = 1 + 6t

26. (a) 6√

3(x− y) + 24z = π√

3 + 6

(b) x =π

3−√

3

4t, y =

π

3+

√3

4t, z =

1

4− t

27. 2x+ 4y − z = 5

28. (x0, 2x0 + 3, 9) , x0 arbitrary

29. (2,−2, 1) and (−2, 2, 1)

3.2.1

1. x2 + y2 = 25

2. y = −12/x, x > 0

3. z = 4x = −3y

4. ex−1 = y/2 = z/3

5. (a) z − x+ y

(b) zi + xj− yk

225

6. (a) 2x+ y

(b) zi

7. (a) 2xz + 2yx+ 2

(b) i + x2j + y2k

8. (a) 2y

(b) (3y2 − 1) i

9. (a) −y sin(xy) + x cos(xy)

(b) (y cos(xy) + x sin(xy)) k

10. (a) exyz(yz + xz + xy)

(b) exyz [(xz − xy)i + (xy − yz)j + (yz − xz)k]

11. (a) 4y

(b) zi− xk

(c) 2j

12. (a) e−xy [−y sin yzi + (−x sin yz + z cos yz)j + y cos yzk]

(b) e−xy [(x2 − z2) sin yz − 2xz cos yz]

13. − (x2 + y2 + z2)−1/2

+ C

14. ∼ 393 kg/s, ∼ 6230 gal./min.

4.1.1

1. π

2. −π

3. 2π

4. 2π2√

2

5. (a) 0

(b) −1/3


6. 0

7. −1

8. 0

9. −1/2

10. 32π6

11. (a)1

2(e2π − 1)

(b)1

2(e2π − 1)

12. 0

13. 41/6

14. 26

15. 3a2

4.2.1

1. (a) No

(b) No

2. (a) No

(b) Yes

3. (a) Yes

(b) No

4. conservative; xy +1

2z2 + C

5. not conservative

6. conservative; x2y + yz + C

227

7. conservative; G(x) + H(y) + L(z) + C, where G, H, L are any an-tiderivatives of g, h, l, resp.

8. (a) xz2 + y2 + C

(b) 1

(c) 1

9. 3/e

10. conservative;1

2x2 + xy + y2 + C

11. not conservative

12.1

2x2 + x sin y − y2 + sin 1 +

1

2

13. exact;1

2x2 + xy + y2 = C

14. not exact

15.1

2x2 + x sin y − y2 = sin 1 +

1

2

16. (a)1

3x3 + xy + ey + C

(b) 8π3/3

(c) 8π3/3

17. −10

18. (a) both equaly2 − x2

(x2 + y2)2

(b) 2π

(c) No. Corollary 4.2.1 implies that F is not conservative. Note thatthe domain of F is not simply connected.


4.3.1

1. (0, 4, 2)(2√

5, π/2, cos−1(1/√

5))

2.(−3√

3, 3√

3, 6√

3)(

6, 3π/4, 6√

3)

3. r = 4 sin θ

ρ sinφ = 4 sin θ

circular cylinder of radius 2 with axis x = 0, y = 2

4. x2 + y2 = 4

circular cylinder of radius 2 with axis the z-axis

5. x2 + y2 + z2 = 6z or x2 + y2 + (z − 3)2 = 9

sphere radius 3 centered at (0, 0, 3)

6. 9

7. 48

8. 32/3

9. 8

10.1

6(e9 − 1)

11.1

4(1− e−81)

12.π

4(1− cos 1)

13.π

4(1− e−1)

14.π

2ln 2

15. 225/8

229

16. 16/3

17. 1/24

18. −117

19. 1024π/9

20. 18π

21. −15/4

22. 1/8

23. 32/105

24. 1

25. 3/4

26. 24

27. 6

28. 25π/2

29. 2π − 8/3

30. 36π

31. 81π/2

32. 125π/8

33. δπhR4/2 = MR2/2, (M = mass)

4.4.1

1. 34

2. 2πrh

3. πr√r2 + h2


4.4

15

(35/2 − 25/2 + 1

)5.

π

6a2

[(1 + 4a2R2)

3/2 − 1]

6.π

6a2

[(1 + 4a2R2)

3/2 − 1]

7.2π

3a2

[(1 + a2R2)

3/2 − 1]

8. 4π2aA

9. 1/6

10. 24

11. 0

12. 8

13. 32π/3

14. 36π

15. 0

16. 12π

17. 3πa2

18. 4π

19. (a) s2(−e−x0+s/2 + e−x0−s/2

− e−y0+s/2 + e−y0−s/2 − e−z0+s/2 + e−z0−s/2)

(b) −e−x0 − e−y0 − e−z0

(c) −e−x0 − e−y0 − e−z0

231

5.1.1

1. 6π

2. 60π

3. 0

4. −1

5. 0

6. −1/2

7. 3π/2

8. 3πa2/8

5.2.1

1. 3

2. 28π

3. 108π

4. 24

5. 0

6. 8

7. 32π/3

8. 36π

9. 0

10. 12π

11. 3πa2

12. F is not continuous (tends to infinity) at the origin.

13. −16


5.3.1

1. −3π/4

2. −1

3. −2

4. (a) 1/6

(b) 1/6

5. 1/6

6. 0

7. 3a2

5.4.1

1. 2 (rer + zez)

2. 2ρeρ

3.rer + zezr2 + z2

(a)1

r2 + z2

(b) 0

4.reθ + zezr2 + z2

(a)r2 − z2

(r2 + z2)2

(b)2rz

(r2 + z2)2 (rer + reθ + zez)

5.eρρ

(a) 1/ρ2

233

(b) 0

6. 2π

7. n2rn−2

8. 4θ2 + 2

9. n(n+ 1)ρn−2

10. 6φ+ cotφ

11. c1 ln | cscx+ cotx|+ c2

5.5.1

1. No. ∇ · F 6= 0

2. n = −2

3.V0 ln (r/R2)

ln (R1/R2),

−V0

r ln (R1/R2)er

A.1.1

1. (f1g2 + g1h2 + h1f2) dx ∧ dy ∧ dz

2. (a) 2xy dx+ x2 dy

(b) y dx+ (x+ z) dy + y dz

(c) −(x+ z) dx ∧ dy + y dz ∧ dx(d) (2xy + x) dx ∧ dy ∧ dz

A.2.1

1. −25/12

2. 1/2

3. −1/3

4. 2π/3


B.2.1

1. around 84 years

2. very close to 3/2

Bibliography

[1] Tom M. Apostol. Mathematical Analysis. Addison-Wesley PublishingCompany, Reading, Massachusetts, second edition, 1975.

[2] R. Creighton Buck. Advanced Calculus. Waveland Press, Long Grove,Illinois, third edition, 1978.

[3] Robert P. Crease. The greatest equations ever. Physics World,17(10):14–15, October 2004.

[4] Harry F. Davis and Arthur David Snider. Introduction to Vector Anal-ysis. Hawkes Publishing, seventh edition, 2000.

[5] Harold M. Edwards. Advanced Calculus: A Differential Forms Ap-proach. Birkhauser Boston, Cambridge, Massachusetts., 1994.

[6] Harley Flanders. Differential Forms with Applications to the PhysicalSciences. Dover, New York, 1989.

[7] Harley Flanders, Robert Korfhage, and Justin Price. Calculus. Aca-demic Press, New York, 1970.

[8] Daniel Fleisch. A Student’s Guide to Maxwell’s Equations. CambridgeUniversity Press, New York, 2009.

[9] David Halliday, Robert Resnick, and Kenneth S. Krane. Physics, vol-ume 2. Wiley, Hoboken, NJ, fifth edition, 2001.

[10] Brian R. Hunt, Ronald L. Lipsman, and Jonathan M. Rosenberg. AGuide to MATLAB: for Beginners and Experienced Users. CambridgeUniversity Press, Cambridge, U.K., second edition, 2006.

[11] Paul G. Huray. Maxwell’s Equations. Wiley, Hoboken, NJ, 2009.

235

236 BIBLIOGRAPHY

[12] Walter Isaacson. Einstein: His Life and Universe. Simon & Schuster,New York, 2007.

[13] Beniot B. Mandelbrot. The Fractal Geometry of Nature. W. H. Freemanand Company, New York, 1983.

[14] Jerrold E. Marsden and Anthony J. Tromba. Vector Calculus. W. H.Freeman and Company, New York, third edition, 1988.

[15] P. C. Matthews. Vector Calculus. Springer-Verlag, London, 1998.

[16] Richard S. Millman and George D. Parker. Elements of DifferentialGeometry. Prentice-Hall, Inc., Englewood Cliffs, New Jersey, 1977.

[17] James R. Munkres. Analysis on Manifolds. Westview Press, Boulder,Colorado, 1991.

[18] Walter Rudin. Principles of Mathematical Analysis. McGraw-Hill, NewYork, third edition, 1976.

[19] H. M. Shey. Div, Grad, Curl, and All That. W. W. Norton & Company,New York, fourth edition, 2005.

[20] I. M. Singer and J. A. Thorpe. Lecture Notes on Elementary Topologyand Geometry. Springer-Verlag, New York, 1967.

[21] Michael Spivak. Calculus on Manifolds. Addison-Wesley PublishingCompany, Reading, Massachusetts, 1965.

[22] James Stewart. Calculus. Thomson Brooks/Cole, Belmont, California,sixth edition, 2007.

Index

acceleration, 59in polar coordinates, 208normal component of, 59tangential component of, 59

Ampere-Maxwell law, 178analytic, 93antiderivative, 43arc length, 53

parametrization by, 53astroid, 57, 154

ballopen, 69, 104volume in 4-space, 204volume in space, 125

binormal vector, 61

c, 180capacitor

coaxial, 182spherical, 181

cardioid, 154Cauchy-Riemann equations, 93Cauchy-Schwarz inequality, 14chain, 192chain rule, 41charge density, 177circle, 46, 71circulation, 100compba, 17composition (of functions), 190

conservation of energy, 93continuity equation, 184cross product, 20

basic properties of, 21curl, 87–89

in cylindrical coordinates, 169in spherical coordinates, 172

current density, 177curvature, 58, 60

radius of, 65curve, 45, see also path

closed, 50fractal, 50nowhere-differentiable, 49parametric equations of, 47piecewise smooth, 98simple, 50smooth, 50space-filling, 50tangent to, 48unit speed, 55

cylindrical coordinates, 122

d’Alembert’s solution, 183del (operator), 90differential equation, 114

exact, 114differential form, 187–189

differential of, 189exact, 197integral of, 192

237

238 INDEX

directed line segment, 6directional derivative, 72disk

area of, 121open, 71

distancebetween geometric objects, 32

divergence, 83–87, 140in cylindrical coordinates, 168in spherical coordinates, 172

divergence theorem, 155–158, 201in 4-space, 205

domain, 103dot product, 13

basic properties of, 13

Einstein, Albertand Maxwell’s equations, 176

electric field, 176electromagnetic energy density, 182element of arc length, 55


element of area, 118in polar coordinates, 121

element of surface area, 137, 138element of volume, 118


Faraday’s law, 177flow curve, 81flow line, see flow curveflux, 84, 140force field, 92

conservative, 92Frenet equations, 62Fubini’s theorem, 118

fundamental theoremfor line integrals, 105, 197of calculus, 42of space curves, 62

Gauss’s law, 141, 158Gauss’s law for magnetic fields, 177Gauss’s theorem, 155gradient, 72


Green’s theorem, 149–151, 198

helix, 47, 63hypersphere, 205hypocycloid with four cusps, 57, 154

i, 5, 10inner product, see dot productintegral

area, 118definite, 42indefinite, 43line, see line integralof a differential form, 192surface, see surface integralvolume, 118

j, 5, 10Jordan curve theorem, 101

k, 5Kepler’s laws, 208–214

Lagrange multipliers, 80Laplace’s equation, 90, 180Laplacian, 90

in cylindrical coordinates, 173in polar coordinates, 173in spherical coordinates, 173

INDEX 239

vector, 90Leibnitz’s rule, 108level set, 74level surface, 74line, 27

parametric equations of, 27symmetric equations of, 28vector equation of, 27

line integral, 98–101, 193independent of path, 105

magnetic field, 177magnetic flux, 177magnetic monopoles, 177manifold, 185Matlab, 65, 94, 146Maxwell’s equations, 159, 175–180Mobius strip, 133

n-chain, see chainn-form, see differential formNewton’s law, 93, 209, 210Newton’s law of gravity, 209normal vector, 75

orthogonal unit vectors, 166, 171osculating circle, 65osculating plane, 64

parallelepiped, 24volume of, 24

parallelogram, 11, 23area of, 23

parallelogram law, 20path, 50, see also curve

oriented, 51clockwise, 101, 133counterclockwise, 101, 133

piecewise smooth, 99

smooth, 50permeability of free space, 179permittivity of free space, 178plane, 30

vector equation of, 30Poisson’s equation, 90, 180polar coordinates, 119potential function, 105Poynting vector, 182Poynting’s theorem, 182principle unit normal vector, 59projba, 17Pythagorean theorem, 14

region, 69arcwise connected, 103boundary of, 69bounded, 71closed, 69closure of, 80compact, 71convex, 103exterior of, 69interior of, 69open, 69simply connected, 104

Riemann-Stieltjes integral, 97right-hand rule, 3, 21

scalar, 1scalar field, 71scalar multiplication, 4, 8scalar product, see dot productspeed, 48sphere, 71

surface area of, 138volume in 4-space, 205

spherical coordinates, 122

240 INDEX

Stokes’ theorem, 160–164, 202modern (version), 194–195

surface, 130–134boundary of, 131compact smooth, 130orientable, 133orientation, 133

usual or standard, 133piecewise smooth, 134unit normal to, 133

surface area, 137surface integral, 140–142

tangent plane, 75torsion, 61torus, 131, 144

solid, 104triangle inequality, 14triple scalar product, 24

unit tangent vector, 48

vector field, 80conservative, 104–111irrotational, 89solenoidal, 87

vector product, see cross productvector space, 5vector(-valued) function, 37

component functions of, 37continuous, 38differentiable, 39integration of, 42of two variables, 129

vector(s), 1addition of, 4, 7angle between, 14components of, 4decomposition of, 17

displacement, 7equality of, 5magnitude of, 4, 7orthogonal, 15parallel, 4, 15perpendicular, 15plane, 10position, 7unit, 4

velocity, 48in polar coordinates, 208

wave equation, 179wedge product, 188

properties of, 188work, 18, 100

xy-axes, 2xyz-axes, 2

Math 440

Documents

x2 y2 z2lt

x2 y2 z2gt

x2 y2 z2

symbolic math

circlex2 y2

a2 b2

a3 b3

b3 c3