Mathematics 261: Calculus III Guided Notes · 2018-10-29 · 4 CHAPTER 12. VECTORS AND THE GEOMETRY OF SPACE 12.2 Vectors De nition 1. A vector is A vector is represented by The book

Mathematics 261: Calculus III

Guided Notes

Dr.s Eric and Erin Bancroft, Grove City College

Fall 2018

Contents

Contents i

12 Vectors and the Geometry of Space 1

12.1 3 Dimensional Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

12.2 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

12.3 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

12.4 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

12.5 Equations of Lines and Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

12.6 Cylinders and Quadric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

13 Vector Functions 21

13.1 Vector Functions and Space Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

13.2 Derivatives and Integrals of Vector Functions . . . . . . . . . . . . . . . . . . . . . . . . 24

13.3 Arc Length And Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

13.4 Motion in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

14 Partial Derivatives 31

14.1 Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

14.2 Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

14.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

14.4 Tangent Planes and Linear Approximation . . . . . . . . . . . . . . . . . . . . . . . . . 41

14.5 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

14.6 Directional Derivatives and the Gradient Vector . . . . . . . . . . . . . . . . . . . . . . 45

14.7 Maximum and Minimum Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

14.8 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

10 Polar Coordinates and Area 55

i

ii CONTENTS

10.3 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

10.4 Areas and Lengths in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . 58

15 Multiple Integrals 61

15.1 Volumes and Double Integrals over Rectangles . . . . . . . . . . . . . . . . . . . . . . . 61

15.2 Double Integrals over General Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

15.3 Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

15.6 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

15.7 Triple Integrals in Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 78

15.8 Triple Integrals in Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . 80

16 Vector Calculus 83

16.1 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

16.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

16.3 The Fundamental Theorem of Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . 88

16.4 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

16.6 Parametric Surfaces and Their Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

16.7 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

Chapter 12

Vectors and the Geometry of Space

12.1 3 Dimensional Coordinate Systems

We move from a 2D to a 3D coordinate system by

Example 1. Graph the points (2, 1, 3), (2,−1, 3), (2, 1,−3), and (−2, 1,−3)

We call 3D space

Example 2. Graph the following in R2 and R3:(a) x = 2 (b) y = 1 (c) x2 + y2 = 4

1

2 CHAPTER 12. VECTORS AND THE GEOMETRY OF SPACE

Recall: The distance between points P = (x1, y1) and Q = (x2, y2) in R2 was

|PQ| =

which we can prove using

Formula 12.1.1. (Distance Formula) For P = (x1, y1, z1) and Q = (x2, y2, z2) in R3

|PQ| =

Example 3. Find the distance from P = (6,−1, 1) to Q = (−2, 3, 0).

Formula 12.1.2. (Equation of a Sphere) For a sphere with center (h, k, l) and radius r we have

12.1. 3 DIMENSIONAL COORDINATE SYSTEMS 3

Example 4. Find an equation of the sphere which passes through the point (0, 1,−1) and is centeredat (3,−1, 0).

Example 5. What is the center and radius of the sphere x2 + y2 + z2 + 4x− 2y + 9z = 2?

Example 6. For each of the following draw a picture and describe in words.

1. z < 5

2. (x− 1)2 + y2 + (z − 1)2 ≤ 2

3. (x− 1)2 + y2 + (z − 1)2 > 2

4. x2 + z2 = 1


12.2 Vectors

Definition 1. A vector is

A vector is represented by

The book uses a bold, lower case letter to denote a vector (u), we’ll use an arrow over a lowercase letter (~u).

Definition 2. ~u and ~v are equivalent or equal if they have

Note: The zero-vector ~0 has length and direction .

Definition 3. (Adding Vectors) Position ~v so its tail is at the tip of ~u, ~u+ ~v =

Definition 4. A scalar is a real number. The book uses a lower case letter. I will typically use

Definition 5. If α is a scalar and ~v is a vector, then the scalar multiple α~v is

Definition 6. ~u− ~v =

12.2. VECTORS 5

Definition 7. The components of ~v are found by placing the tail of ~v at and looking

at the x, y, z coordinates of the .

Given points A(x1, y1, z1) and B(x2, y2, z2) the vector ~AB is ~AB =

Definition 8. The magnitude or length of a vector ~a = 〈a1, a2, a3〉 is

Example 7. Find the vector representation of the directed line segment with initial point A(4,−1, 2)and terminal point B(0, 2, 1). What is its magnitude?

Vector Operations: For ~a = 〈a1, a2, a3〉,~b = 〈b1, b2, b3〉, α ∈ R

Addition: ~a+~b =

Subtraction: ~a−~b =

Scalar Multiplication: α~b =

Example 8. If ~a = 〈−6, 1, 8〉 and ~b = 〈4,−5, 3〉 find

1. ~a+~b

2. 2~b− 3~a

3. ~a− 4~b


Properties of Vectors: If ~a,~b, and ~c are vectors and α and β are scalars, then

1. ~a+~b = ~b+ ~a

2. ~a+ (~b+ ~c) = (~a+~b) + ~c

3. ~a+~0 = ~a

4. ~a+ (−~a) = ~0

5. α(~a+~b) = α~a+ α~b

6. (α + β)~a = α~a+ β ~a

7. (αβ)~a = α (β~a)

8. 1~a = ~a

Definition 9. The standard basis vectors are

Example 9. Suppose ~a = 2ı + 4k,~b = −5ı + 2 + 7k. What is 2~a+~b?

Definition 10. A unit vector is a vector . Given a vector ~a, the

unit vector ~u which is ~u =

12.3. THE DOT PRODUCT 7

Example 10. Find the unit vector in the direction of ~a = 〈7, 2, 1〉.

12.3 The Dot Product

Definition 11. The dot product of two vectors ~a = 〈a1, a2, a3〉 and ~b = 〈b1, b2, b3〉 is

1. ~a q~b =

2. ~a q~b =

Example 11. Find ~a q~b if

1. |~a| = 3, |~b| = 5, and the angle between them is π/4.

2. ~a = 〈2, 3,−1〉,~b = 〈0, 1,−2〉

3. ~a = −4ı + − 2k, ~b = 3ı− + k


Special Cases of the Dot Product:

•

•

•

Note: We can think of ~a q~b as measuring

Example 12. Are ~a = ı + 3 + 2k and ~b = 2ı + − 12k orthogonal?

Properties of the Dot Product: If ~a,~b, and ~c are vectors and α is a scalar then

1. ~a q~a =

2. ~a q~b ~b q~a3. ~a q (~b+ ~c) =

4. (α~a) q~b = α(~a q~b) = ~a q (α~b)5. ~0 q~a =

Example 13. Find the angle between ~a = 〈−1, 2,−2〉 and ~b = 〈1, 2, 3〉.

12.3. THE DOT PRODUCT 9

Projections:

Definition 12. The scalar projection of ~b onto ~a is

comp~a ~b =

The vector projection of ~b onto ~a is

proj~a ~b =

Example 14. Let ~a = 〈−1, 2,−2〉, ~b = 〈1, 2, 3〉.

1. Find comp~a ~b and proj~a ~b.

2. Find comp~b ~a and proj~b ~a.


12.4 The Cross Product

Definition 13. If ~a and ~b are vectors in R3, then the cross product of ~a and ~b is

Note:

1.

2.

Example 15. 1. ı× =

2. × ı =

12.4. THE CROSS PRODUCT 11

Definition 14. For ~a = 〈a1, a2, a3〉 and ~b = 〈b1, b2, b3〉, the cross product of ~a and ~b is

~a×~b = 〈 a2b3 − a3b2, a3b1 − a1b3, a1b2 − a2b1 〉

The determinant of a 2× 2 matrix

(a bc d

)is

The determinant of a 3× 3 matrix

a1 a2 a3

b1 b2 b3

c1 c2 c3

is

Example 16. 1.

∣∣∣∣1 52 −3

∣∣∣∣2.

∣∣∣∣∣∣2 0 1−1 1 31 4 −2

∣∣∣∣∣∣


Example 17. Find ~a×~b

1. ~a = 〈1,−1, 2〉, ~b = 〈3, 1,−3〉

2. ~a = 〈1,−1, 5〉, ~b = 〈−2, 3,−4〉

Properties of the Cross Product: If ~a,~b and ~c are vectors and α is a scalar then

1. ~a×~b =

2. (α~a)×~b = α(~a×~b) = ~a× (α~b)

3. ~a× (~b+ ~c) =

4. (~a+~b)× ~c = ~a× ~c+~b× ~c

5. ~a q (~b× ~c) =

6. ~a× (~b× ~c) = (~a q~c)~b− (~a q~b)~cNote: We can also use cross products to compute

12.4. THE CROSS PRODUCT 13

Example 18. 1. Find the area of the parallelogram with vertices A(2, 1, 0), B(3, 3, 1), D(4, 1, 0) andC(5, 3, 1).

2. Find the volume of the parallelepiped determined by ~a = ı + 3 − k,~b = −2ı + + 3k, and~c = −ı + 3 + k

3. Do the points A(0, 1, 1), B(−2, 0,−1), C(−3, 1, 0) and D(−4, 2, 1) lie on the same plane?


12.5 Equations of Lines and Planes

In R2 we can find an equation of a line with

In R3 we can find an equation of a line with

Writing this in component form we get

parametric equations of a line

If neither a, b nor c are 0 we can

symmetric equations of a line

Definition 15. We call a, b and c the of the line.

12.5. EQUATIONS OF LINES AND PLANES 15

Example 19. 1. Find symmetric equations of the line through the points A(1,−3, 5) and B(.2, 4, 0).

2. Find a vector equation for the line parallel to the line −x = y − 4 = 2z + 5 through the point(2, 4,−6).

Note: The equations for a line are . Any of ~v

will work. So for our previous example we could have used ~v = to get

or ~v = to get . We could also pick ,

e.g. ~r0 = gives .

What if we want to describe a line segment?


Example 20. Determine whether the lines are parallel, skew or intersecting:

x =2y + 1

6=z + 5

2− 3x+ 3 = −6y = z + 8

Planes: A plane is determined by a and a

(called the and denoted by ).

Vector Equation of a plane

12.5. EQUATIONS OF LINES AND PLANES 17

Scalar Equation of a plane

Example 21. Find the equation of the plane. . .

1. . . . through the point (1,−2, 3) with normal vector 〈−1,−7, 5〉.

2. . . . through the point (−2, 0, 1) that contains vectors equivalent to both ~a = 〈1,−1, 2〉 and~b = 〈3, 1,−3〉.

3. . . . through the point (0, 2, 6) that contains the line x = 2t, y = 1− 3t, z = 5.

How planes interact in R3:

Two planes are either or they .

Example 22. Consider the planes x+ 3− 4z = −10 and −3x− 9y + 12z = 5:


If two planes are not parallel, then we can find the equation of the line of intersection and the anglebetween them.

Example 23. Finding the equation of the line of intersection of two planes:

−x− 3y + 4z = −7 x+ 2y − 4z = −10.

First we need a point on both planes.

Angle between two planes

Distance from a point P (x1, y1, z1) to a plane ax+ by + cz + d = 0 D =|ax1 + by1 + cz1 + d|√

a2 + b2 + c2

12.6. CYLINDERS AND QUADRIC SURFACES 19

12.6 Cylinders and Quadric Surfaces

To sketch the graph of a 3D surface we’ll determine the curves of intersection of the surface with planesparallel to the coordinate planes by holding one variable constant. We call these curves traces orcross-sections of the surface.

Example 24. Sketch the traces of z = x2 + 1 and then sketch the graph.

-

6

x = k traces

y

z6

-

y = k traces

x

z6

-

z = k traces

x

y


Chapter 13

Vector Functions

13.1 Vector Functions and Space Curves

Definition 16. A vector valued function (abbr. v.v.f.) is a function whose domain is

and whose range is a set of . The variable is typically t and we can think

of it as a .

Example 25. Find the domain of

1. ~r(t) =⟨√

t, t2, 3t

⟩

2. ~r(t) =⟨

sin t, ln t,√

1− t⟩

Limits of Vector Functions

If ~r(t) = 〈 f(t), g(t), h(t) 〉 then limt→a

~r(t) =

provided that the limit of each of the component functions .

Note: All of the Calc. I and II limit rules apply!

21

22 CHAPTER 13. VECTOR FUNCTIONS

Example 26. 1. limt→2

⟨t,t2 − 4

t2 − 2t,1

t

⟩=

2. limt→0

t2 ı− 5

t + et k =

3. limt→∞

⟨5t3 + 7

3t3 − 6t, e−t,

sin t

t

⟩=

Definition 17. A vector function ~r(t) is continuous at a if

Definition 18. The set C of all points (x, y, z) in space where x = f(t), y = g(t), z = h(t) and t ∈ Iand where f, g, and h are continuous real valued functions is called a .

Example 27. Sketch the curve with given vector equation and indicate the direction of increasing t.

1. ~r(t) = 〈 t2, t2, t 〉

2. ~r(t) = t ı + cos t + sin t k

13.1. VECTOR FUNCTIONS AND SPACE CURVES 23

Example 28. Find a vector function that represents the curve of intersection of the two surfaces.

1. z = x3, y = x3 + z2 + 2

2. z = x+ y2, z = 2x+ y

3. y2 + z2 = 9, x = 5y

We will need to find parametric equations that are defined by both surfaces.

Tips:

•

•


13.2 Derivatives and Integrals of Vector Functions

Definition 19. If ~r(t) = 〈 f(t), g(t), h(t) 〉 then the derivative of ~r(t) is ~r ′(t) =(As long as f, g and h are all differentiable. The usual differentiation rules apply.)

Evaluating ~r ′(t) at a specific t value gives the to the curve at thepoint as long as ~r ′(t) exists and is not nonzero.

Definition 20. The tangent line to the curve C at a point P is the line through P

the tangent vector ~r ′(t). The unit tangent vector is ~T (t) =

Example 29. 1. Find the unit tangent vector for ~r(t) = 〈t, ln t, t2〉 when t = 2.

2. Find a vector equation for the tangent line to the curve from the previous question at the point(1, 0, 1).

Differentiation Rules

1. ddt

[~u(t) + ~v(t)] = ~u ′(t) + ~v ′(t)

2. ddt

[α~u(t)] = α~u ′(t)

3. ddt

[f(t) ~u(t)] = f ′(t) ~u(t) + f(t) ~u ′(t)

4. ddt

[~u(t) q~v(t)] = ~u ′(t) q~v(t) + ~u(t) q~v ′(t)5. d

dt[~u(t)× ~v(t)] = ~u ′(t)× ~v(t) + ~u(t)× ~v ′(t)

6. ddt

[~u(f(t))] = f ′(t) ~u ′(f(t))

Example 30. For ~u(t) = 〈 t, t2, t3 〉 , ~v(t) = 〈 e2t, t, 6 〉 , f(t) = 1t

find

1. ddt

[f(t)~u(t)]

2. ddt

[~u q~v]

13.3. ARC LENGTH AND CURVATURE 25

Integrals of Vector Functions

Definition 21. For ~r(t) = 〈f(t), g(t), h(t)〉,∫~r(t) dt =

and∫ ba~r(t) dt =

Example 31. 1.∫〈 t, e2t, cos t 〉 dt =

2.∫ ⟨

ln t, 2t,√t+ 1

⟩dt =

3.∫ 2

0〈 t, t2, t3 〉 dt =

13.3 Arc Length And Curvature

Definition 22. The arc length of a curve ~r(t) = 〈 f(t), g(t), h(t) 〉 traversed exactly once between t = aand t = b is

L =

∫ b

a

√(f ′(t))2 + (g′(t))2 + (h′(t))2 dt =

∫ b

a

√(dx

dt

)2

+

(dy

dt

)2

+

(dz

dt

)2

dt =

∫ b

a

|~r ′(t)| dt

Example 32. Find the arc length of ~r(t) = 〈t, t2, 23t3〉 between (0, 0, 0) and (2, 4, 16

3).

A curve can be represented by more than one vector function depending on

and we can switch from one to another. For example if ~r(t) = 〈t, 1t, 4t2〉

2 ≤ t ≤ 5 is the vector function for the curve C, we can let t = to get another vector function

for C: ~r(t) = .


Note: Arc length is , in other words no matter

which parametrization we use, we’ll get the same for the same section of

curve.

Definition 23. The arc length function s(t) is s(t) =

∫ t

a

|~r ′(u)| du where s(t) is the length of C

between ~r(a) and ~r(t).

If we take the derivative of both sides of the arc length function we get

It is useful to parametrize a curve with respect to because arc length is

independent of parameter so it won’t affect the outcome if we change to a different coordinate system.

To do this we’ll .

Example 33. Reparametrize the curve ~r(t) = (5 − t) ı + t + (3t + 7) k with respect to arc lengthmeasured from the point t = 0 in the direction of increasing t.

Definition 24. ~r(t) is smooth on an interval I if .

Definition 25. Given a smooth vector function ~r(t) with unit tangent vector ~T and parametrized with

respect to arc length the curvature of ~r(t) is κ =

∣∣∣∣∣d~Tds∣∣∣∣∣.

13.3. ARC LENGTH AND CURVATURE 27

Since reparametrizing in terms of arc length is hard, we’d like a formula where the parameter is t. . .

Thus we have κ(t) =

Another more convenient formula is (see proof in the book on pg. 880):

Computational Curvature Formula: κ(t) =|~r ′(t)× ~r ′′(t)||~r ′(t)|3

For a plane curve y = f(x) we have

Plane Curve Curvature Formula: κ(x) =|f ′′(x)|

[1 + (f ′(x))2]3/2

Example 34. 1. Find κ(t) and κ(1) for ~r(t) =⟨t2, 2t,−1

3t3⟩

2. Find κ(x) and κ(0) for f(x) = x− 14x2.


The unit tangent vector points .

The principal unit normal vector points .

Unit Normal Vector: ~N(t) =~T ′(t)

|~T ′(t)|

Binormal Vector: ~B(t) = ~T (t)× ~N(t)

Example 35. Find ~T (t), ~N(t) and ~B(t) for ~r(t) = 〈3 sin t, 4t− 7, 3 cos t〉.

13.4 Motion in Space

If the position of an object in space is given by the vector function ~r(t), then its velocity is ,

its speed is , and its acceleration is .

13.4. MOTION IN SPACE 29

We are often interested in resolving the acceleration vector into two components, one in the di-rection of the tangent vector and the other in the direction of the normal vector.

Note that: ~v(t) = |~v(t)| ~T (t). Using our derivative rules we get:

Then, substituting in using our equations for curvature and the normal vector we find:

~a(t) = |~v|′ ~T + κ |~v|2 ~N

For computational purposes it may be easier to use: aT =~r ′(t) � ~r ′′(t)|~r ′(t)|

and aN =|~r ′(t)× ~r ′′(t)||~r ′(t)|

Example: If the position function of an object is given by ~r(t) = 〈3 sin t, 4t − 7, 3 cos t〉, then find itsvelocity and acceleration vectors. Also find the tangential and normal components of the accelerationvector.


Chapter 14

Partial Derivatives

14.1 Functions of Several Variables

Definition 26. A looks like f(x, y) or f(x, y) = z where x

and y are the and z is the . The domain of

f(x, y) is .

Example 36. Find and sketch the domain of the following functions.

1. f(x, y) =√

4− x2 − y2

2. g(x, y) = ln(4− x− y)

3. h(x, y) = ex/y

4. j(x, y) =1

xy

31

32 CHAPTER 14. PARTIAL DERIVATIVES

Definition 27. If f is a function of two variables with domain D then the of f is theset of all points (x, y, z) in R3 such that z = f(x, y) and (x, y) is in D .

Definition 28. A function f(x, y) = ax+ by + c is called a .

Definition 29. The of a function f of two variables are the curves withequations f(x, y) = k where k is a constant in the range of f .

A function of 3 variables, f(x, y, z) lives in 4D space (which is hard to visualize) so we look at its, which are the surfaces with equations f(x, y, z) = k where k is a constant

in the range of f .

Example 37. Sketch 4 level curves of f(x, y) = 9 − x − 3y. Then sketch the graph of f in the firstoctant.

14.2. LIMITS AND CONTINUITY 33

Example 38. What do the level surfaces of f(x, y, z) = y2 + z2 look like?

14.2 Limits and Continuity

Definition 30. Given a function f(x, y) we say that the limit of f(x, y) as (x, y) approaches (a, b)

is L if the values of f(x, y) can be made to L for (x, y)

to (but not equal to) (a, b).

The formal ε, δ definition is:

Example 39. Prove that lim(x,y)→(0,0)

x2y2

x2 + y2= 0


Recall: In Calc I limx→a f(x) existed if and only if

Now in R3 we have to approach a point (a, b) since we are on a sur-

face. So, the limit must exist and be the same value no mater which path we take to get to (a, b).

To show a limit DNE we just need to

Theorem 14.2.1. If f(x, y)→ L1 as (x, y)→ (a, b) along a path C1 and f(x, y)→ L2 as (x, y)→ (a, b)

along a different path C2 and L1 6= L2 then

Typically we will use paths that make calculations easier such as:

Note: All limit properties from Calc I and II can be extended to functions of several variables.

To show that a limit does exist we have to prove that the distance from f(x, y) to the limit L is

using .

Example 40.

1. lim(x,y)→(1,2)

y

x2 + y2

14.2. LIMITS AND CONTINUITY 35

2. lim(x,y)→(0,0)

y

x2 + y2

3. lim(x,y)→(0,0)

x2y2

x2 + y2

4. lim(x,y)→(0,0)

2x− y2

2x2 + y

5. lim(x,y)→(0,0)

xy√x2 + y2

6. lim(x,y)→(0,0)

xy cos(y)

3x2 + y2

7. lim(x,y)→(0,0)

x2yey

x4 + 4y2


Definition 31. A function f(x, y) is continuous at a point (a, b) if and only if

Note: Multi-variable polynomials and multi-variable rational functions are continuous

.

Example 41. lim(x,y)→(1,2)

3x2y + 2xy + 5x+ y2

Example 42.

1. Where is g(x, y) =

{2x−y22x2+y2

if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0)continuous?

2. Where is h(x, y) =

{x2y2

x2+y2if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0)continuous?

14.3. PARTIAL DERIVATIVES 37

14.3 Partial Derivatives

Given a multivariable function, we can treat all of the variables except one as constant and then differ-

entiate with respect to that one variable. This is know as a of the function.

Notation: For a function of two variables z = f(x, y):

• The partial derivative with respect to x is written

fx, fx(x, y),∂f

∂x,∂

∂x(f(x, y)) ,

∂z

∂x, or Dxf

• The partial derivative with respect to y is written

fy, fy(x, y),∂f

∂y,∂

∂y(f(x, y)) ,

∂z

∂y, or Dyf

Graphically/Geometrically: Graphically, ∂f∂x

tells us the instantaneous rate of change of the function

if we hold y fixed and move parallel to the x-axis in the positive direction, while ∂f∂y

tells us theinstantaneous rate of change of the function if we hold x fixed and move parallel to the y-axis in thepositive direction. See Figure 14.1 below.

(a) As we move in the +x-direction from (−1, 1),∂f∂x (−1, 1) is positive, so f is increasing in that direc-tion.

(b) As we move in the +y-direction from(2, 3

2

),

∂f∂x

(2, 3

2

)is negative, so f is decreasing in that direc-

tion.

Figure 14.1


Limit Definition

As with derivatives in calculus I, there is a limit definition for partial derivatives:

∂f

∂x= lim

h→0

f(x+ h, y)− f(x, y

h∂f

∂y= lim

h→0

f(x, y + h)− f(x, y

h

We won’t be using the limit definition to find partial derivatives in this class, but we would need it ifwe wanted to go through a later example.

Example 43. Find the first partial derivatives:

1. f(x, y) = x2y2 + y2 + 2x3y

2. f(x, y) = ex2y

3. f(x, y) = xex2y

4. h(x, y, z) =yzex

x2 sin y

5. f(x, y, z) = 2x2y + eyz +√z ln(x)

6. f(x, y, z) = zex2+xy

7. f(x, y, z) =x

(xy − z)2

14.3. PARTIAL DERIVATIVES 39

Higher Order Derivatives Notation:

We can find second order derivatives by simply differentiating the first order partial derivativesagain. We can find third or higher order derivatives in a similar manner.

The notation for second order derivatives is:

(fx)x = fxx =∂

∂x

(∂f

∂x

)=∂2f

∂x2=∂2z

∂x2= Dxxf

(fy)y = fyy =∂

∂y

(∂f

∂y

)=∂2f

∂y2=∂2z

∂y2= Dyyf

(fx)y = fxy =∂

∂y

(∂f

∂x

)=

∂2f

∂y ∂x=

∂2z

∂y ∂x= Dxyf

(fy)x = fyx =∂

∂x

(∂f

∂y

)=

∂2f

∂x ∂y=

∂2z

∂x ∂y= Dyxf

For third order derivatives, we have

fxxx, fyyx, fyxy, etc.

∂3f

∂x3,

∂3f

∂x ∂y2,

∂3f

∂y ∂x ∂y, etc.

There are 23 = 8 possible third order partial derivatives. In general there are

(number of indep variables)n

nth-order partial derivatives.

Note: fyxx =∂3f

∂x∂x∂y

Example 44. Find all second partial derivatives:

1. f(x, y) = x2y2 + y2 + 2x3y (from previous example)


2. f(x, y) =√xy + ln y

x+ 2xy

Theorem 14.3.1. (Clairaut’s Theorem) Suppose f(x, y) is a function such that fxy and fyx are contin-

uous on an open disk R. Then for every point (a, b) ∈ R, .

One example of a function whose mixed partials are different (so, it does not satisfy the hypotheses of

Clairaut’s Theorem) is

f(x, y) =

xy(x2 − y2)

x2 + y2, if (x, y) 6= (0, 0)

0, if (x, y) 6= (0, 0)

We could use the limit definitions of the derivatives to show that fxy(0, 0) = 1 but fyx(0, 0) = −1.

Implicit Differentiation (Round 1): Suppose z is defined implicitly as a function of x and y by the

equation z2 − x2y + xz3 + yz = 3. Find ∂z∂x

and ∂z∂y

.

14.4. TANGENT PLANES AND LINEAR APPROXIMATION 41

14.4 Tangent Planes and Linear Approximation

In Calculus I we used the derivative to find the equation of the line tangent to the curve at a point.Now we’ll use partial derivatives to find the equation of the plane tangent to the surface at a point.Suppose f has continuous partial derivatives. Then our equation of the plane tangent to the surfacez = f(x, y) at the point (x0, y0, z0) is

z − z0 = fx(x0, y0) · (x− x0) + fy(x0, y0) · (y − y0)

orz = f(x0, y0) + fx(x0, y0) · (x− x0) + fy(x0, y0) · (y − y0)

This equation is also called the linearization of f at (a, b) = (x0, y0) and written as the function:

L(x, y) = f(a, b) + fx(a, b) (x− a) + fy(a, b) (y − b)

which we use to approximate f(x, y) when (x, y) is near (a, b), i.e., L(x, y) ≈ f(x, y) near (a, b).

Example 45. Find an equation of the plane tangent to the graph of f(x, y) = 4− x2− y2 at the point(1,−1, 2).

Example 46. Reformat the equation you found in the previous example as the linearization of f anduse it to approximate f(0.8,−1.1).

Example 47. Find the linearization L(x, y) at the point (3, 0, 2) for f(x, y) =√x+ e4y. Use it to

approximate f(3.01, 0.01).


Differentials: The differentials dx and dy are independent variables. Then the total differential dz is:

dz = fx(x, y) dx+ fy(x, y) dy

We’ll take dx = ∆x = x− a and dy = ∆y = y − b to get:

dz = fx(a, b) (x− a) + fy(a, b) (y − b)

So, using the notation of differentials, the linear approximation can be written as

f(x, y) ≈ f(a, b) + dz.

Note:

• dz represents change in

• ∆z represents change in

Example 48. Let z = 2x2 − 5xy.

1. Find dz.

2. Suppose (x, y) changes from (1, 2) to (1.1, 1.8). Compute the values of ∆z and dz.

14.5 Chain Rule

Theorem 14.5.1. (The Chain Rule) Suppose that u is a differentiable function of the n variablesx1, x2, . . . , xn and each xj is a differentiable function of the m variables t1, t2, . . . , tm. Then u is afunction of t1, t2, . . . , tm and

∂u

∂ti=

∂u

∂x1

∂x1

∂ti+

∂u

∂x2

∂x2

∂ti+ · · ·+ ∂u

∂xn

∂xn∂ti

for each i = 1, 2, . . . ,m.

Rather than try to memorize this and apply it to specific situations, we can use a “tree diagram”to find the correct form of the chain rule.

14.5. CHAIN RULE 43

Example 49. y = f(x), x = g(t).

Example 50. z = f(x, y), x = g(t), y = h(t).

z = 4− x2 + xy − y2, x = 3t2 + 1, y = sin(t)

Example 51. z = f(x, y), x = g(s, t), y = h(s, t).

z = 4− x2 + xy − y2, x = t2 + st− 1, y = est


Example 52. w = f(x, y, z, t), x(p, q, r), y(p, q, r), t(p, q, r)

Example 53. z = f(x, y), x(p, q, r, s), y(p, q, r, s)

Example 54. The radius of a right circular cone is increasing at a rate of 1.8 in/s while its height isdecreasing at a rate of 2.5 in/s. At what rate is the volume of the cone changing when the radius is 120in. and the height is 140 in?

14.6. DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 45

Implicit Differentiation (easier way): Suppose that the equation F (x, y) = 0 defines y implicitlyas a function of x. Then since both y and x are functions of x we get:

Suppose that the equation F (x, y, z) = 0 defines z implicitly as a function of x and y, then

∂z

∂x= −

∂F∂x∂F∂z

= −FxFz,

∂z

∂y= −Fy

Fz

Example 55. Find ∂z∂x

and ∂z∂y

for z2 − x2y + z3x+ zy = 3.

14.6 Directional Derivatives and the Gradient Vector

We are able to find the derivatives at some (x, y) in the positive x direction (fx) or the positive ydirection (fy), but what if we want a derivative that points in a different direction?

Definition 32. Given a differentiable function f(x, y) and unit vector ~u = 〈a, b〉, the directionalderivative of f in the direction of u is

Duf(x, y) =

For a function of three variables f(x, y, z) and unit vector ~u = 〈a, b, c〉, we have

Duf(x, y, z) =

(Similar formulae for higher dimensions.)

Example 56. f(x, y) = 9− x2 − y2. Find the directional derivative at the point (2, 2) in the directionof ~u1 = 〈−1, 0〉 and ~u0 = 〈−1√

2, −1√

2〉.


Note: If ~u makes an angle θ with the x-axis then we can write ~u = 〈cos θ, sin θ〉 to get

D~uf(x, y) = fx cos θ + fy sin θ

Example 57. Continuing the previous example, suppose ~u is the unit vector given by the angle π/3.

Definition 33. Given a function f(x, y), the gradient of f is

∇f(x, y) = 〈fx(x, y), fy(x, y)〉 =∂f

∂xı +

∂f

∂y

(〈fx, fy, fz〉 for 3 variables).

Note: The gradient and directional derivative related to each other as follows:

D~uf(x, y) = ∇f q~uTheorem 14.6.1. Given a function f of two or three variables and point x (in two or three dimensions),the maximum value of the directional derivative at that point, Duf(x), is |∇f(x)| and it occurs when ~uhas the same direction as the gradient vector ∇f(x).

Proof.

Example 58.

1. For f(x, y) = 9− x2 − y2 find the direction in which f has the maximal rate of change and whatthis maximal rate of change is if we start at (2, 2).

14.6. DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 47

2. The temperature at a point (x, y, z) is given by T (x, y, z) = 200e−x2−3y2−9z2 where T is measured

in ◦C and x, y, z in meters.

(a) Find the rate of change of temperature at the point P (2,−1, 2) in the direction toward thepoint (3,−3, 3)

(b) In which direction does the temperature increase fastest at P?

(c) Find the maximum rate of increase at P .

Another property of the gradient is: Given function f and a point (in two or three dimensions),the gradient vector at that point is perpendicular to the level curve/surface of f which passes throughthat point. This leads to the following:

Tangent Planes to Level Surfaces

Given a level surface S of a function F (x, y, z) of three variables (so S is of the form F (x, y, z) = k),we can use the property above to find the equation of the plane tangent to S at a point (x0, y0, z0). Weobtain the normal vector from the gradient:

~n = ∇F (x0, y0, z0)

And so an equation of the tangent plane is

Fx(x0, y0, z0)(x− x0) + Fy(x0, y0, z0)(y − y0) + Fz(x0, y0, z0)(z − z0) = 0


Normal Line

The normal line to the tangent plane is the line with direction vector ∇F (x0, y0, z0) that passes through(x0, y0, z0). Its vector equation is

~r(t) = 〈Fx(x0, y0, z0) · t+ x0, Fy(x0, y0, z0) · t+ y0, Fz(x0, y0, z0) · t+ z0 〉

and its symmetric equations are

x− x0

Fx(x0, y0, z0)=

y − y0

Fy(x0, y0, z0)=

z − z0

Fz(x0, y0, z0)

14.7 Maximum and Minimum Values

To find local maxima/minima of a function f(x, y) of two variables, we do the following:

1. Take both partial derivatives, fx and fy, and set them equal to zero. This gives us a system ofequations.

2. Solve the system of equations for x and y. This will give our critical point(s), i.e., points (a, b)such that fx(a, b) = 0 and fy(a, b) = 0. (The first derivative test tells us that these are possiblelocal maxima or minima of f .)

3. Use the Second Derivative Test for Functions of Two Variables to test these points:

Let f(x, y) be a function, (a, b) a point such that fx(a, b) = 0 and fy(a, b) = 0, and let

D(x, y) =

∣∣∣∣fxx fxyfyx fyy

∣∣∣∣= fxxfyy − fyxfxy= fxxfyy − (fyx)

2

= fxxfyy − (fxy)2

Then:

i If D(a, b) > 0 and fxx(a, b) > 0 (or fyy(a, b) > 0), then f(x, y) has a local minimum at (a, b).

ii If D(a, b) > 0 and fxx(a, b) < 0 (or fyy(a, b) < 0), then f(x, y) has a local maximum at (a, b).

iii If D(a, b) < 0, then f(x, y) has neither a relative minimum or minimum at (a, b). In this case,f(x, y) has a saddle point at (a, b).

iv If D(a, b) = 0, then this test is inconclusive.

14.7. MAXIMUM AND MINIMUM VALUES 49

Example 59. Find the relative maxima and minima for the following functions:

1. f(x, y) = 12x2 + 2xy + 3y2 − x+ 2y

2. f(x, y) = x3 − y2 − 3x+ 4y

3. f(x, y) = 2x2 + 3xy + 5y2


4. f(x, y) = (2x− x2)(2y − y2)

5. (opt) f(x, y) = xe−2x2−2y2

Absolute Maximum and Minimum of Functions of Two Variables

Recall from Calc 1: Finding the absolute maximum and minimum values of a continuous function f(x)on the closed interval [a, b].

Example 60. Find the absolute max/min of f(x) = x2 on I = [0, 1].

To find the absolute maximum and minimum of a continuous function on a closed, bounded set B:

1. Find the value of f at any critical points of f in B.

2. Find the absolute maximum and minimum of f along the boundary of B. This typically involvesfinding an equation for the boundary, substituting that equation into f for one of the variables,then finding the absolute max/min using techniques from calculus I. If the boundary of B isdefined by multiple functions, then find the absolute max/min for each piece of the boundary.

3. The largest value from 1. and 2. is the absolute maximum, the smallest is the absolute minimum.

14.7. MAXIMUM AND MINIMUM VALUES 51

Example 61. Find the absolute max/min of f(x, y) = x2 + xy on B = {(x, y) : |x| ≤ 2, |y| ≤ 1}.


14.8 Lagrange Multipliers

The method of Lagrange multipliers come from the fact that the gradient vector at a point is perpen-dicular to the level curve/surface of f which passes through that point. So if any two curves/surfacespass through a point, then the gradient vectors of the associated functions will be parallel at that point.

Suppose we want to optimize the function f(x, y, z) subject to a constraint g(x, y, z) = k. Wefirst solve the system of equations

∇f(x, y, z) = λ∇g(x, y, z)

g(x, y, z) = k

or, equivalently,

fx = λgx

fy = λgy

fz = λgz

g = k

for x, y, and z (we don’t necessarily need to know the value of λ). Then we evaluate f at all of thesolution points (x, y, z). The largest resulting value is the maximum of f , subject to the constraint,and the smallest is the minimum of f , subject to the constraint.

The method can also be used to optimize functions of two independent variables subject to aconstraint:

∇f(x, y) = λ∇g(x, y)

g(x, y) = k

and to optimize functions subject to multiple constraints:

∇f(x, y, z) = λ∇g(x, y, z) + µ∇h(x, y, z)

g(x, y, z) = k

h(x, y, z) = l

or

∇f(x, y) = λ∇g(x, y) + µ∇h(x, y)

g(x, y) = k

h(x, y) = l

14.8. LAGRANGE MULTIPLIERS 53

Example 62. Optimize f(x, y, z) = x4 + y4 + z4 subject to the constraint x2 + y2 + z2 = 1.

Example 63. Optimize f(x, y, z) = x2 + y2 + z2 subject to the constraint x+ y + z = 12.

Example 64. Optimize f(x, y, z) = 3x+ y subject to the constraint x2 + y2 = 10.


Chapter 10

Polar Coordinates and Area

10.3 Polar Coordinates

Instead of using rectangles to get to a point in the xy-plane, we could use an angle and a distance/radius.For example, to plot the point (3, 4) using Cartesian/rectangular coordinates, we start at the origin, go3 units in the positive x direction and then go 4 units in the positive y direction. We could also getto the same point by standing at the origin facing the positive x-axis, turning θ = 0.92radians in thecounterclockwise direction, and then moving 5 units we write this as (r, θ) = (5, 0.92).

These are known as .

The origin is called the and the pos-

itive x-axis is called the . Co-

ordinates look like (r, θ).

Example 65. Plot the following polar coordinates:

1. (3, π4)

2. (−3, π4)

3. (3, 5π4

)

55

56 CHAPTER 10. POLAR COORDINATES AND AREA

4. (5,−π6)

5. (5,−25π6

)

Converting from Polar to Rectangular:

x = y =

For example the polar coordinate point (5, −π6

) is

Converting from Rectangular to Polar:

r = θ =

Note: when you are changing from rectangular to polar coordinates you must be sure that θ puts youinto the correct quadrant!! Add to θ to get into the 2nd or 3rd quadrant.

Example 66. Convert the following Cartesian coordinates to polar coordinates:

1. (1, 1)

2. (1,−1)

3. (−1, 1)

4. (−1,−1)

10.3. POLAR COORDINATES 57

Sketching and Describing Polar Regions and Equations:

Example 67. Sketch the following polar regions/equations:

1. 1 ≤ r, −π4≤ θ ≤ π

2. −1 ≤ r ≤ 1, −π3≤ θ ≤ π

2

3. 1 ≤ r ≤ 2, 0 ≤ θ ≤ 3π2

4. 0 ≤ r < 4, π4≤ θ ≤ 3π

4

5. r = 3

6. θ = π/3


10.4 Areas and Lengths in Polar Coordinates

The area of a sector of a circle is A = .

Given a polar function r = f(θ), α ≤ θ ≤ β, where r is continuous and non-negative on this θ-interval,we want to find the area bounded by r between θ = α and θ = β.

We divide the interval [α, β] into n subintervals each with angle ∆θ = β−αn

and let θi be the midpoint of

the ith subinterval. Then the area of the ith subinterval is

and the total area is . Taking the limit as n→∞ gives

10.4. AREAS AND LENGTHS IN POLAR COORDINATES 59

Example 68. Find the area of the inner loop of r = 1 + 2 cos(θ).

Example 69. Set up (but do not evaluate) the in-tegral to find the region bounded by both r1(θ) =−6 cos θ and r2(θ) = 2− 2 cos θ (the region wherethe dot is in the picture).

-6 -5 -4 -3 -2 -1

-3

-2

-1

1

2

3


Chapter 15

Multiple Integrals

15.1 Volumes and Double Integrals over Rectangles

Riemann Sum Approximations of Volumes over Rectangles

In Calc I we used a Riemann sum of rectangles to approximate the area under the curve over an interval[a, b].

-1 1 2 3 4

-15

-10

-5

5

Figure 15.1: A Riemann sum using 4 rectangles and right-hand endpoints.

A ≈ ∆x (f(x1) + f(x2) + · · ·+ f(xm))

=m∑i=1

f(xi)∆x, where ∆x =b− am

61

62 CHAPTER 15. MULTIPLE INTEGRALS

Now we have three dimensional surfaces, and instead of approximating area under a curve we’renow approximating under the surface over some rectangular area . We’llnow use three dimensional boxes to approximate the volume (think of using Legos R© to make the volume):

V ≈m∑i=1

n∑j=1

f(xij, yij)∆A

where (xij, yij) is some point in the rectangular base of the ijth box, f(xij, yij) is the height of the ijth

rectangular box, and ∆A represents the area of the base of the box (usually ∆A = ∆x∆y = b−am· d−c

n).

Figure 15.2: A double Riemann sum using 24 rectangles (m = 6 and n = 4) and lower left-hand corners.

In Calculus I

lim∆x→0

m∑i=1

f(x1)∆x =

b∫a

f(x) dx,

now we have

lim∆A→0

m∑i=1

n∑j=1

f(xij, yij)∆A =

∫∫R

f(x, y) dA.

For a function f(x, y) which is non-negative over the region R, thebut above the xy-plane is

V =

∫∫R

f(x, y) dA

≈m∑i=1

n∑j=1

f(xij, yij)∆x∆y.

15.1. VOLUMES AND DOUBLE INTEGRALS OVER RECTANGLES 63

Example 70. Estimate the volume of the solid that lies below the surface

z = 9− x2 − y2

and above the rectangle

R = [−1, 2]× [0, 1]

= {(x, y) | −1 ≤ x ≤ 2, 0 ≤ y ≤ 1}

using a Riemann sum with m = 3, n = 2, and

(a) the lower left-hand corners as the sample points.

(b) the midpoints as sample points


Double Iterated Integrals over Rectangular Regions

We’ve differentiated with respect to each variable, now we’ll look at integrating with respect to eachvariable which is called partial integration.

Example 71.

∫ 1

0

9− x2 − y2 dy =

In general: An iterated double integral is of the form∫ ba

[∫ dcf(x, y) dy

]dx =

∫ ba

∫ dcf(x, y) dy dx or∫ d

c

∫ baf(x, y) dx dy.

Note:

•

•

Theorem 15.1.1. (Fubini’s Theorem) If f is on the rectangle R = [a, b]×[c, d] then ∫∫

R

f(x, y) dA =

∫ b

a

∫ d

c

f(x, y) dy dx =

∫ d

c

∫ b

a

f(x, y) dx dy.

Example 72.

∫ 2

0

∫ 1

0

xe12x2−y dy dx

15.1. VOLUMES AND DOUBLE INTEGRALS OVER RECTANGLES 65

Example 73.

∫ 1

0

∫ 1

0

yexy dy dx

Theorem 15.1.2. If f(x, y) can be written as f(x, y) = g(x) · h(y) then∫∫R

f(x, y) dA =∫∫R

g(x) · h(y) dA =

Example 74. Find the volume of the solid that lies beneath the surface z = e−(x+y)/2 and above therectangle R = [0, 2]× [0, 4].


15.2 Double Integrals over General Regions

We’ve been looking at integrals with constant limits of integration, now what about something like thefollowing?

Example 75. (a)

∫ e

1

∫ x

lnx

x+ 1 dy dx

(b)

∫ 1

0

∫ y

0

1− xy dx dy

What do these represent? It gives us the of the solid under the surfacez = f(x, y) and above the (general) region D which lies between the graphs of the continuous limits.In the previous two problems we have the following regions D:

1.

1 2 3 4x

-1

1

2

3

4

y

2. 0.5 1.0 1.5 2.0x

0.5

1.0

1.5

2.0

y

15.2. DOUBLE INTEGRALS OVER GENERAL REGIONS 67

Note:

• Limits go from lower to upper and left to right.

• The order of integration is important!!!! Since the double integral geometrically represents avolume, we should always get a as the answer! The non-constant limits shouldalways go for these problems.

• It is possible to change the order of integration, but you cannot just switch the limits when theregion is not rectangular. Instead you must change how you describe the region.

Example 76. For the following questions, draw the region D over which we are integrating and thenset up the double integral to find the volume under f(x, y) and above D.

(a) D is the region bounded by y = −x, y = x3, x = 1 and x = 2.

(b) D is the region bounded by x = y, x = y3, and x ≥ 0.


(c) D is the circle y2 + x2 = 4

(d) D is the triangle with vertices (0, 0), (0, 1), and (2, 0).

(e) D is the region bounded by y = −x2 + 4 and y = x2 − 4.

15.2. DOUBLE INTEGRALS OVER GENERAL REGIONS 69

Properties of Double Integrals: (see the book page 1046-1047 for a full list)

9. If D = D1 ∪D2 and D1 and D2 don’t overlap (except maybe on the boundary) then∫∫D

f(x, y) dA =

∫∫D1

f(x, y) dA+

∫∫D2

f(x, y) dA

10.∫∫D

dA =

Example 77. Find the volume under the surface z = xy and above the triangle with vertices (1, 1), (4, 1)and (1, 2).

Example 78. Find the volume bounded by the cylinder x2 + y2 = 1 and the planes y = z, x = 0, andz = 0 in the first octant.


15.3 Double Integrals in Polar Coordinates

If f is continuous on a polar rectangle R given by 0 ≤ a ≤ r ≤ b, α ≤ θ ≤ β where 0 ≤ β−α ≤ 2π then

∫∫R

f(x, y) dA =

Example 79. Find the volume of the solid bounded above by the hemisphere z =√

16− x2 − y2 andbelow by the region R give by x2 + y2 = 4.

15.3. DOUBLE INTEGRALS IN POLAR COORDINATES 71

Example 80. Evaluate∫∫R

(x2 + y) dA where R is the region between the circles x2 + y2 = 1 and

x2 + y2 = 5.

Example 81. Find the volume of the solid bounded by the paraboloids z = 3x2+3y2 and z = 4−x2−y2.


15.6 Triple Integrals

With a triple integral we now integrate over a 3-dimensional solid E in R3:∫∫∫E

f(x, y, z) dV

Theorem 15.6.1. If f(x, y, z) is continuous on the rectangular box B = [a, b]× [c, d]× [r, s] then∫∫∫B

f(x, y, z) dV =

∫ s

r

∫ d

c

∫ b

a

f(x, y, z) dx dy dz

Note: With double integrals we only had two choices for the order of integration: dx dy or dy dx. Now

we have : . As

with double integrals, if all limits are constant then we can switch the order without any changes. If

all limits are not constant, then we cannot simple switch the limits without changing them!

Example 82. Evaluate the triple integral:∫ 3

0

∫ 1

−1

∫ ln(2)

0

ex(y + 2z) dx dy dz.

Example 83. Set up the integral to find the volume of the upper half of the sphere x2+y2+(z−2)2 = 4.

15.6. TRIPLE INTEGRALS 73

Setting up triple integrals:

Example 84. Set up all six orders of integration for∫∫∫E1

ex(y + 2z) dV,

where E1 is the region bounded by the planes

z = x+ y, z = 0, y = 0, y = x, and x = 2.

Then evaluate one of the six.

Figure 15.3: Two views of E1.

(a) xy-projection. (b) xz-projection. (c) yz-projection.

Figure 15.4: Projections of E1 into the three coordinate planes.


(a)

∫ ∫ ∫ex(y + 2z) dx dy dz

(b)

∫ ∫ ∫ex(y + 2z) dx dz dy

(c)

∫ ∫ ∫ex(y + 2z) dy dx dz


(d)

∫ ∫ ∫ex(y + 2z) dy dz dx

(e)

∫ ∫ ∫ex(y + 2z) dz dx dy

(f)

∫ ∫ ∫ex(y + 2z) dz dy dx


Example 85. Set up all six orders of integration for∫∫∫E2

f(x, y, z) dV,

where E2 is the region bounded by

z = 1− x2 , z = 0, y = 1− x , y = 0, and x = 0.

Figure 15.5: Two views of E2.

(a) xy-projection. (b) xz-projection. (c) yz-projection.

Figure 15.6: Projections of E2 into the three coordinate planes.


(a)

∫ ∫ ∫f(x, y, z) dx dy dz

(b)

∫ ∫ ∫f(x, y, z) dx dz dy

(c)

∫ ∫ ∫f(x, y, z) dy dx dz

(d)

∫ ∫ ∫f(x, y, z) dy dz dx

(e)

∫ ∫ ∫f(x, y, z) dz dx dy

(f)

∫ ∫ ∫f(x, y, z) dz dy dx


15.7 Triple Integrals in Cylindrical Coordinates

Cylindrical Coordinates: In R3 we have two alternative ways to get to the point (x, y, z) and the firstis using cylindrical coordinates. Here we will use polar coordinates r and θ to get to the (x, y) part ofthe coordinate and then move vertically the z amount. Thus the z coordinate is remaining unchanged.The notation is (r, θ, z) and the conversion equations are the same ones used with polar coordinates.

Example 86. 1. Plot (2, 3π/4, 3)

2. Convert (−1,−1,−3) to cylindrical coordinates.

Example 87. Describe the region bounded between the two cylinders x2 + y2 = 1, x2 + y2 = 9 andbetween the planes z = −1 and z = 4 in the first octant using cylindrical coordinates.

15.7. TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES 79

Triple Integrals in Cylindrical Coordinates:

∫∫∫E

f(x, y, z) dV =

∫ β

α

∫ h2(θ)

h1(θ)

∫ u2(r cos θ,r sin θ)

u1(r cos θ,r sin θ)

f(r cos θ, r sin θ, z) · r dz dr dθ

Example 88. Evaluate∫∫∫E

(x3 + xy2) dV where E is the solid under the paraboloid z = 1− x2− y2 in

the first octant.

Example 89. Find the volume of the solid that lies within both the cylinder x2 +y2 = 1 and the spherex2 + y2 + z2 = 4 using a triple integral.


15.8 Triple Integrals in Spherical Coordinates

Spherical coordinates are the second alternative way to represent a point (x, y, z) in R3. Here we willuse two angles and a radius/distance to describe the point. The angle from the positive z-axis is φ, theangle from the positive x-axis is θ, and the radius is ρ.

Standard to have:

0 ≤ θ ≤ 2π 0 ≤ φ ≤ π and ρ ≥ 0.

Spherical to Rectangular Conversion:

z =

x =

y =

Rectangular to Spherical Conversion:

ρ =

θ =

φ =

Example 90. (a) Convert the rectangular coordinates(−2, 2

√3, 4)

to spherical coordinates.

(b) Convert the spherical coordinates(3, 7π

4, 5π

6

)to rectangular coordinates.

15.8. TRIPLE INTEGRALS IN SPHERICAL COORDINATES 81

Example 91. Describe the following spherical regions where a is a constant:

(a) ρ = a

(b) θ = a

(c) φ = a

Triple Integrals in Spherical Coordinates:

∫∫∫E

f(x, y, z) dV =

∫ d

c

∫ β

α

∫ b

a

f(ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ) ρ2 sinφ dρ dθ dφ

Example 92. Evaluate∫∫∫H

(x2 + y2) dv where H is the solid above the xy-plane and below the sphere

x2 + y2 + z2 = 1.

Example 93. Find the volume of the solid that lies within the sphere x2 + y2 + z2 = 4, above thexy-plane, and below the cone z =

√x2 + y2 using a triple integral.


Chapter 16

Vector Calculus

16.1 Vector Fields

Definition 34. Let D be a subset of R2. A vector field on R2 is a function ~F that assigns a vector ~F (x, y)

to each point (x, y) in D. Written in terms of component functions we have ~F (x, y) = P (x, y)ı+Q(x, y).

(Similarly we can define a vector field in R3 with ~F (x, y, z) = P ı +Q +Rk.)

-2 -1 0 1 2

-2

-1

0

1

2

~Fx, y = 〈y, 1〉 ~F (x, y, z) = 〈x, y, z〉

Definition 35. If f is a scalar function of two (or three) variables then ∇f is a vector field on R2 (orR3) called a gradient vector field.

Example 94. Find the gradient vector field of f(x, y) = sin(x2 + y)

83

84 CHAPTER 16. VECTOR CALCULUS

-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

Definition 36. A vector field ~F is called a conservative vector field if it is the

of some scalar function f , i.e. if there exists a function f such that .

16.2 Line Integrals

A curve C has no sharp points or breaks. A curve is made

up of several smooth curves put together with no gaps/jumps, but it may have a sharp point where

they come together.

Recall from Section 13.3: dsdt

= |~r ′(t)| =√(

dxdt

)2+(dydt

)2

Definition 37. If f(x, y) is defined on a smooth curve C given by parametric equations x = x(t),y = y(t), a ≤ t ≤ b then the line integral of f along C is (line integral with respect to arc length):∫

C

f(x, y) ds =

Also if C = C1 ∪ C2 ∪ · · · ∪ Cn is piecewise smooth then∫C

f(x, y) ds =

(Similarly for a function f of three variables.)

Recall: The equation of the line between the tips of the vectors ~r0 and ~r1 is:

16.2. LINE INTEGRALS 85

~r(t) = (1− t)~r0 + t ~r1, 0 ≤ t ≤ 1.

Example 95. 1. Evaluate∫C

(x2 − y + 3z) ds where C is the line from (0, 0, 0) to (1, 2, 1).

2. Evaluate∫Cx ds where C consists of the line y = x and the curve y = x2 for 0 ≤ x ≤ 1.


Definition 38. The line integrals of f along C with respect to x and y are:∫C

f(x, y) dx =

∫ b

a

f(x(t), y(t)) x′(t) dt and

∫C

f(x, y) dy =

∫ b

a

f(x(t), y(t)) y′(t) dt

Note:

• Line integrals with respect to x and y often happen together so we’ll abbreviate it by writing∫CP (x, y) dx+Q(x, y) dy

• The orientation of the curve C is important when evaluating the line integrals with respect tox and y as it must be positive. Otherwise your answer will be the negative of the correct one:∫−C f(x, y) dx = −

∫Cf(x, y) dx and

∫−C f(x, y) dy = −

∫Cf(x, y) dy.

Example 96. Evaluate∫C

(x2y3 −√x) dy where C is the arc of the curve y2 = x from (4, 2) to (1, 1).

Defining the Line Integral of a Vector Field:

Definition 39. Let ~F be a continuous vector field defined on a smooth curve C given by a vectorfunction ~r(t), where a ≤ t ≤ b. The line integral of ~F along C is∫

C

~F q d~r =

16.2. LINE INTEGRALS 87

Example 97. 1. Evaluate∫C~F q d~r where ~F = 3x ı + 4y , ~r(t) = 2 cos t ı + 2 sin t , 0 ≤ t ≤ π/2.

2. Find the work done by the vector field ~F = −12x ı− 1

2y + 1

4k on a particle as it moves along the

path ~r(t) = cos t ı + sin t + t k as t moves from (1, 0, 0) to (−1, 0, 3π).


16.3 The Fundamental Theorem of Line Integrals

Theorem 16.3.1 (The Fundamental Theorem of Line Integrals). Let C be a smooth curve given by avector function ~r(t), where a ≤ t ≤ b, and let f be a differentiable function of two or three variableswhose gradient vector ∇f is continuous on C. Then∫

C

∇f · d~r =

Note that the answer doesn’t depend on the path, only the ! So, forany two paths C1 and C2 having the same initial point and same terminal point,∫

C1

∇f · d~r =

∫C2

∇f · d~r.

This leads to. . .

Definition 40. Let ~F be a continuous vector field with domain D. The line integral∫C~F · d~r is said

to be independent of path (or path independent if∫C1

~F · d~r =∫C2

~F · d~r for any two paths C1

and C2 in D that have the same initial point and same terminal point.

Definition 41. A curve C is said to be closed if its initial point and terminal point are the same.

Theorem 16.3.2.∫C~F · d~r is path independent in D if and only if

∫C~F · d~r = 0 for every closed path

C in D.

Recall that a vector field is conservative if there exists a scalar function f such that ∇f = ~F .

Theorem 16.3.3. Let ~F be a continuous vector field on an open connected region. Then the line integral∫C~F · d~r is path independent if and only if ~F is conservative.

Why do we care? If ~F is conservative, then that can make evaluating∫C~F · d~r much easier (if C

is closed we get 0 immediately, if not then we could still choose a ‘nicer’ path). But it would be nice to

have some way to check to see if ~F is conservative without going back to the definition:

Theorem 16.3.4. Let ~F = P ı +Q be a vector field on an open, simply-connected region D such thatP and Q have continuous first-order partial derivatives. Then ~F is conservative if and only if

∂P

∂y=∂Q

∂x

on D.

16.3. THE FUNDAMENTAL THEOREM OF LINE INTEGRALS 89

Curl and Divergence (Section 16.5)

For three-dimensional vector fields, we need another way to check for a conservative vector field. Recallthe “del” operator that we defined when we introduced the gradient:

∇ =∂

∂xı +

∂

∂y +

∂

∂zk

=

⟨∂

∂x,∂

∂y,∂

∂z

⟩We use this to define the curl and divergence of a three-dimensional vector field ~F = P ı +Q +Rk:

curl ~F = ∇× ~F

=

∣∣∣∣∣∣ı k∂∂x

∂∂y

∂∂z

P Q R

∣∣∣∣∣∣div ~F = ∇ · ~F

Theorem 16.3.5. Let ~F = P ı +Q +Rk be a vector field on an open, simply-connected region D suchthat P , Q, and R have continuous first-order partial derivatives. Then ~F is conservative if and only ifcurl ~F = ~0 on D.

Example 98. Determine whether or not the following vector fields are conservative. If so, find f(x, y)

such that ∇f = ~F .

(a) ~F (x, y) = 〈 y, 1 〉

(b) ~F (x, y) = 〈 y2, 2xy 〉

(c) ~F (x, y) = 〈 y3 + 1, 3xy2 + 1 〉


(d) ~F (x, y, z) = 〈 2xy, x2 + 2yz, y2 〉

(e) ~F (x, y, z) =⟨−1

2x, −1

2y, 1

4

⟩

We can combine these definitions and theorems as follows:

Theorem 16.3.6. Let ~F be a continuous vector field with continuous first partial derivatives in anopen connected region D, and let C be a [piecewise] smooth curve in D given by ~r(t). The following areequivalent:

1. ~F is conservative.

2.∂P

∂y=∂Q

∂x(for ~F = P ı +Q) or curl ~F = 0 (for ~F = P ı +Q +Rk) on D .

3. There exists f such that ∇f = ~F (which then allows us to use the Fundamental Theorem).

4.∫C~F · d~r is independent of path.

5.∫C~F · d~r = 0 for every closed curve C in D.

16.3. THE FUNDAMENTAL THEOREM OF LINE INTEGRALS 91

Example 99. Evaluate∫C~F · d~r, where ~F = 〈 y3 + 1, 3xy2 + 1 〉 and C is the semicircular path from

(0, 0) to (2, 0) given by ~r(t) = 〈 1− cos(t), sin(t) 〉 , 0 ≤ t ≤ π.

Example 100. Evaluate∫C~F · d~r, where ~F = 〈 y2, 2xy 〉 and C is the parabolic path from (4, 0) to

(1, 3) given by ~r(t) = 〈 4− t, 4t− t2 〉 , 0 ≤ t ≤ 3.

Example 101. Evaluate∫C~F · d~r, where ~F = 〈 2xy, x2 + 2yz, y2 〉 and C is the path given by

~r(t) =⟨t2, t cos(t), et sin(t)

⟩, 0 ≤ t ≤ 3π.


16.4 Green’s Theorem

Green’s Theorem relates a line integral around a simple closed curve C to a double integral over theregion in the plane D bounded by C:

Theorem 16.4.1 (Green’s Theorem). Let C be a positively oriented, piecewise-smooth, simple, closed

curve in R2 and let D be the region bounded by C, and let ~F = P ı + Q be a continuous vector fieldsuch that P and Q have continuous partial derivatives on an open region containing D. Then∫

C

~F · d~r(

=

∫C

P dx+Qdy

)=

∫∫D

(∂Q

∂x− ∂P

∂y

)dA

=

∫∫D

(curl ~F

)· k dA

Example 102. Evaluate∫Cx2y2 dx + xy dy where C is the positively oriented curve consisting of the

arc of the parabola y = x2 from (0, 0) to (1, 1) and the line segments from (1, 1) to (0, 1) and from (0, 1)to (0, 0).

Example 103. Use Green’s Theorem to evaluate∫C~F d~r where ~F = 〈e−x + y2, e−y + x2〉 where C

consists of the arc of the curve y = cosx from (−π/2, 0) to (π/2, 0) and the line segment from (π/2, 0)to (−π/2, 0).

16.6. PARAMETRIC SURFACES AND THEIR AREAS 93

16.6 Parametric Surfaces and Their Areas

Given a vector valued or parametric function in one variable, say t, we can trace out a curve in R2 orR3: ~r(t) = x(t)ı + y(t)[+z(t)k]. But with two parameters we can trace out a surface (in R3):

~r(u, v) = x(u, v)ı + y(u, v) + z(u, v)k.

Given a function z = f(x, y), one way to get a parameterization of the surface is to simply let u = x,v = y, and then let z = f(u, v) (similar construction for surfaces defined by y = f(x, z) or x = f(y, z)).

Tangent Planes to Parametric Surfaces

Recall that to find the equation of a plane, we need to know a normal vector to the plane ~n = 〈 a, b, c 〉and a point in the plane (x0, y0, z0). Then the equation of the plane is

a(x− x0) + b(y − y0) + c(z − z0) = 0.

Now, given a parameterization ~r(u, v) = x(u, v)ı + y(u, v) + z(u, v)k of a surface, to find anequation of a tangent plane we still need a normal vector and a point. The point will either be given as(x0, y0, z0), or we will be given (u0, v0) which we’ll then plug into ~r to find the coordinates of the point.Our normal vector is

~n = ~ru(u0, v0)× ~rv(u0, v0)

(or, for this type of problem, we could also use ~ru(u0, v0)× ~rv(u0, v0)) where

~ru =∂x

∂uı +

∂y

∂u +

∂z

∂uk and ~rv =

∂x

∂vı +

∂y

∂v +

∂z

∂vk

Example 104. Find the equation of the plane tangent to the surface

~r(u, v) = uı + v +√uvk

at the point (1, 1).


Example 105. Find the equation of the plane tangent to the surface

~r(u, v) = 3u cos(v)ı + 3u sin(v) + u2k

at the point (0, 6, 4).

Surface Area of Parametric Surfaces

Recall that the area of a parallelogram with sides given by ~a and ~b is |~a×~b|. This leads to the following:

Theorem 16.6.1. Let S is a smooth parametric surface given by ~r(u, v) = x(u, v)ı+y(u, v)+z(u, v)k,(u, v) ∈ D, such that S is covered just once as (u, v) ranges throughout D. Then the surface area of Sis

A(S) =

∫∫D

|~ru × ~rv| dA

If we have a surface given by z = f(x, y) and view x and y as the parameters, then we have~r(x, y) = 〈x, y, f(x, y) 〉 which means ~rx = 〈 1, 0, fx 〉 and ~ry = 〈 0, 1, fy 〉, so

~rx × ~ry =

∣∣∣∣∣∣ı k1 0 fx0 1 fy

∣∣∣∣∣∣= 〈−fx, fy, 1 〉

and

|~rx × ~ry| =√fx

2 + fy2 + 1

Thus the surface area is

A(S) =

∫∫D

√fx

2 + fy2 + 1 dA

(Similar formulae if the surface is given by y = f(x, z) or x = f(y, z).)

16.6. PARAMETRIC SURFACES AND THEIR AREAS 95

Example 106. Find the surface area of

~r(u, v) = 2u cos(v)ı + 2u sin(v) + u2k

for 0 ≤ u ≤ 1, 0 ≤ v ≤ 2π.

Example 107. Find the surface area of the portion of the hemisphere z =√

25− x2 − y2 that liesinside the cylinder x2 + y2 = 9.


16.7 Surface Integrals

Previously, we had that the line integral of f(x, y, z) along a curve C given by ~r(t) = 〈x(t), y(t), z(t) 〉,a ≤ t ≤ b, is

∫ b

a

f (x(t), y(t), z(t))

√(dx

dt

)2

+

(dy

dt

)2

+

(dz

dt

)2

dtor=

∫ b

a

f (~r(t)) |~r ′(t)| dt

Now we extend this definition to define the surface integral of f(x, y, z) over a surface S given by~r(u, v) = x(u, v)ı + y(u, v) + z(u, v)k where (u, v) ∈ D, as∫∫

S

f(x, y, z) dS =

∫∫D

f (~r(u, v)) |~ru × ~rv| dA

Example 108. Evaluate the surface integral∫∫S

xy dS where S is given by ~r(u, v) = 2 cos(u)ı+2 sin(u)+

vk with 0 ≤ u ≤ π2

and 0 ≤ v ≤ 2.

If S is a surface given by a function z = g(x, y) then we can let x and y be the parameters andfind |~rx × ~ry| as we did previously to get∫∫

S

f(x, y, z) dS =

∫∫D

f (x, y, g(x, y))√

(gx)2 + (gy)2 + 12 dA

(Similar formulae if the surface is given by y = g(x, z) or x = g(y, z).)

Example 109. Evaluate∫∫S

(x2 + y2 + z2) dS where S is the portion of the plane z = x + 2 that is

bounded by x = −1, x = 1, y = 0, and y = 2.

16.7. SURFACE INTEGRALS 97

Orientable Surfaces

Orientable surfaces have two sides, and we can distinguish between the two based on which way anormal vector is pointing. For a closed surface (such as a sphere, cube, or any region E over whichwe evaluated a triple integral) we can have a normal vector pointing outward from the surface (this iscalled positive orientation) or it can point inward (this is called negative orientation). Some surfacesare non-orientable, such as a Mobius strip.

Now that we have the notion of orientability, we can talk about. . .

Surface Integrals Over Oriented Surfaces

Let ~F = P ı +Q +Rk be a continuous vector field, defined on an oriented surface S with unit normalvector ~n. The surface integral of ~F over S is∫∫

S

~F · dS =

∫∫S

~F · ~n dS =

∫∫D

~F · (~ru × ~rv) dA

If S is given by the graph of a function z = g(x, y), then

∫∫S

~F · dS =

∫∫D

~F · 〈 −gx, −gy, 1 〉 dA =∫∫D

(−P gx −Qgy +R) dA (oriented upward)∫∫D

~F · 〈 gx, gy, −1 〉 dA =∫∫D

(P gx +Qgy −R) dA (oriented downward)

where D is the projection of z = g(x, y) onto the xy-plane.


~F · dS, where ~F = 〈x, y, z 〉 and S is the part of the

paraboloid z = 4− x2 − y2 that lies above the xy-plane, and has upward orientation.


~F · dS, where ~F = 〈 3z, −4, y 〉 and S is the part of

the plane x+ y + z = 1 in the first octant, with downward orientation.

Mathematics 261: Calculus III Guided Notes · 2018-10-29 · 4 CHAPTER 12. VECTORS AND THE GEOMETRY OF SPACE 12.2 Vectors De nition 1. A vector is A vector is represented by The book

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