Math 233 Calculus 3 - Fall 2016 - Linda Green · Math 233 Calculus 3 - Fall 2016. x12.1 - THREE-DIMENSIONAL COORDINATE SYSTEMS x12.1 - Three-Dimensional Coordinate Systems Deﬁnition.

Math 233 Calculus 3 - Fall 2016

§12.1 - THREE-DIMENSIONAL COORDINATE SYSTEMS

§12.1 - Three-Dimensional Coordinate Systems

Definition. R3 means

By convention, we graph points in R3 using a right-handed coordinate system. Right-handed means:

Question. For two points P1 = (x1, y1, z1) and P2 = (x2, y2, z2) inR3, what is the distancebetween P1 and P2?

Question. What is the equation of a sphere of radius r centered at the point (h, k, l)?

2


Circle the right-handed coordinate systems.

3


Example. (# 10) Find the distance from (4,−2, 6) to each of the following:

a. The point (9,−1,−4)

b. The xy-plane (where z = 0)

c. The xz-plane

d. The x-axis

4


Example. Describe the region (and draw a picture).

a. |z| ≤ 2

b. x2 + z2≤ 9

c. x2 + y2 + z2 > 2z

5

§12.6 QUADRIC SURFACES

§12.6 Quadric Surfaces

Match the equation to the graph using only your brain (no graphing software). Check your answer with graphing software.

A. B. C.

D. E. F.

G. H. I.

1. x2 + z2 = 4 2. 36x2 + 4y2 + 9z2 = 36 3. −x2− y2 + z = 0 4. −x2 + y2 + z = 0 5. x − y2

− z2 = 06. −x2

− 4y2 + z2 = 0 7. x2 + y2− z2 = 1 8. −x2

− y2 + z2 = 1 9. −x2 + y2 + z2 = 1 10. −x2 + z = 0

6

§12.2 VECTORS

§12.2 Vectors

Definition. A vector is a quantity with direction and magnitude (length).

Vectors are usually drawn with arrows.

Two vectors are considered to be the same if:

Example. Which pairs of vectors are equivalent?

Definition. A scalar is another word for .A scalar does not have a direction, in contrast to a vector.

7

§12.2 VECTORS

Vector addition: Draw ~a +~b

Multiplication of scalars and vectors: Draw 2~a and −3~a

8

§12.2 VECTORS

Note. In 2-dimensions, if we place the initial point of a vector ~a at the origin, and findits terminal point (a1, a2), then another way of representing the vector is using anglebracket notation:

The numbers a1, a2 are called the of the vector.

Example. The two equivalent vectors drawn above can both be written as < , > .

Note. In 3-d, vectors have 3 components < a1, a2, a3 >.

9

§12.2 VECTORS

Question. What are the components of the vector ~AB that starts at a point A = (3, 1)and ends at a point B = (−2, 5)?

Note. In general, the components of the vector that starts at a point A = (x1, y1) andends at a point B = (x2, y2) are:

10

§12.2 VECTORS

Note. Given two vectors ~a =< a1, a2 > and ~b =< b1, b2 >, and a scalar c

~a +~b =

~a −~b =

c~a =

Similarly, for three-dimensional vectors ~a =< a1, a2, a3 > and ~b =< b1, b2, b3 >,

~a +~b =

~a −~b =

c~a =

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§12.2 VECTORS

Properties of vectors:

1. Addition is commutative2. Addition is associative3. Additive identity (the zero vector)4. Additive inverses5. Distributive property6. Distributive property7. Associativity of scalar multiplication8. Multiplicative identity

Definition. The length of a vector ~v =< v1, v2 > is

|~v| =

Definition. The length of a vector ~w =< w1,w2,w3 > is

|~w| =

Note. The length of a vector is also called:

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§12.2 VECTORS

Definition. A unit vector is a vector

Question. How can we rescale any vector to make it a unit vector? (Rescale meansmultiply by a scalar.)

Note. Rescaling a vector to make it a unit vector is also called normalizing the vector.

Example. Find a unit vector that has the same direction as < 5, 1, 3 >.

Definition.~i =< 1, 0 >, ~j =< 0, 1 > are thein 2-dimensions.

Definition. The standard basis vectors in 3-dimensions are:

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§12.2 VECTORS

Review. Vectors can be represented with

•

•

•

Example. Which arrow(s) represent(s) the vector < 2, 1 >?

Example. Write < 2, 1 > as a sum of multiples of standard basis vectors.

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§12.2 VECTORS

Example. (#41 from book) Find the unit vectors that are parallel to the tangent line toy = x2 at (2, 4).

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§12.2 VECTORS

Example. The 2-d vector ~v has magnitude 9 and makes an angle of 5π6 with the positive

x-axis. Find the components of ~v and write ~v in terms of~i and ~j.

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§12.2 VECTORS

Example. Two forces act as shown. Find the magnitude of the resultant force and theangle it makes with the positive x-axis.

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§12.2 VECTORS

Example. Spiderman is suspended from two strands of spider silk as shown. Find thetension in each strand of spider silk. (You will need some additional information.)

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§12.2 VECTORS

Question. Is there a difference between 0 and ~0?

Question. There are two definitions of vector addition: the vector definition and thecomponent definition. Why are they the same?

19

§12.3 DOT PRODUCT

§12.3 Dot Product

Definition. If ~a =< a1, a2, a3 > and ~b =< b1, b2, b3 >, then the dot product of ~a and ~b isgiven by

~a ·~b =

Question. Is ~a ·~b a vector or a scalar?

Note. Dot product satisfies the following properties:

1. Commutative Property:

2. Distributive Property:

3. Associative Property:

4. Multiplication by ~0:

5. Relationship between dot product and norm:

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§12.3 DOT PRODUCT

Definition. Dot product can also be defined in terms of magnitudes and norms:

Question. If θ is the angle between ~a and~b, what is cos(θ) in terms of the dot product?

Question. Why are the two definitions of dot product the same?

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§12.3 DOT PRODUCT

Example. (# 11 and 12) If ~u is a unit vector, find ~u · ~v and ~u · ~w.

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§12.3 DOT PRODUCT

Example. (# 31) Find the acute angle between y = x2 and y = x3 at their intersectionpoint at (1, 1).

Example. (#13)

(a) Show that~i · ~j = ~j ·~k = ~k ·~i = 0

(b) Show that~i ·~i = ~j · ~j = ~k ·~k = 1

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§12.3 DOT PRODUCT

Example. (#27) Find a unit vector that is orthogonal to both~i + ~j and~i +~k.

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§12.3 DOT PRODUCT

Scalar and vector projections

Definition. The scalar projection of ~bonto ~a is given by

comp~a~b =

Definition. The vector projection of ~bonto ~a is given by

proj~a~b =

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§12.3 DOT PRODUCT

Example. (#42) Find the scalar and vector projections of ~b =< −2, 3,−6 > onto ~a =<5,−1, 4 >

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§12.3 DOT PRODUCT

Definition. The work done by a constant force ~F moving an object along a vector ~D isthe component of ~F in the direction of ~D, times the distance |~D|.Question. How can we write work in terms of the dot product?

Example. (# 51) A sled is pulled along a level path through snow by a rope. A 30-lbforce acting at an angle of 40◦ above the horizontal moves the sled 80 ft. Find the workdone by the force. (Round your answer to the nearest whole number.)

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§12.3 DOT PRODUCT

Definition. The direction angles are the angles that a vector makes with the positivex, y, and z-axes.

Example. Find the direction angles α, β, and γ of the vector ~v =< 3, 5, 4 >

Find cos2 α + cos2 β + cos2 γ. Why is this sum such a nice number?

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§12.3 DOT PRODUCT

Extra Example. (# 53) Use scalar projection to show that the distance from a pointP1(x1, y1) to the line ax + by + c = 0 is

|ax1 + by1 + c|√

a2 + b2

Use this formula to find the distance from the point (−2, 3) and the line 3x− 4y + 5 = 0.

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§12.3 DOT PRODUCT

Extra Example. (# 61) Prove the Cauchy-Schwarz Inequality:

|~a ·~b| ≤ ||~a||||~b||

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§12.3 DOT PRODUCT

Extra Example. (# 62) Prove the Triangle Inequality, which states that

||~a +~b|| ≤ ||~a|| + ||~b||

Hint: Use the fact that ||~a +~b||2 = (~a +~b) · (~a +~b)

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§12.4 CROSS PRODUCTS

§12.4 Cross Products

Definition. The cross-product of two vectors ~a =< a1, a2, a3 > and ~b =< b1, b2, b3 > isgiven by:

~a ×~b =

32


Example. Compute a cross product: for ~a =< 1, 2, 3 > and ~b =< 5,−1, 10 >, find ~a ×~b.

33


Theorem. The vector ~a ×~b is perpendicular to both ~a and ~b.

Note that the direction of ~a ×~b is given by the right hand rule:

34


Theorem. If θ is the angle between ~a and ~b, with 0 ≤ θ ≤ π, then

||~a ×~b|| = ||~a|| ||~b|| sinθ

Corollary. Two nonzero vectors ~a and ~b are parallel if and only if ~a ×~b = 0.

35


Example. For the two vectors shown, find ||~a × ~b|| and determine whether ~a × ~b isdirected into the page of out of the page.

36


Review. The cross-product of two vectors ~a and ~b is defined as:

or as

Question. How can you use cross product to:

• find a vector perpendicular to two vectors?

• determine if two vectors are parallel?

37


Example. Find a unit vector orthogonal to both~i + ~j and~i +~k.

38


Example. Use the cross product to write the area of the parallelogram in terms of ~aand ~b.

Example. Use the cross product to write the volume of the parallelpiped in terms of~a,~b, and ~c.

39


Example. (# 30) Find the area of the triangle with vertices P(0, 0,−3), Q(4, 2, 0), andR(3, 3, 1).

40


Example. Are these three vectors coplanar?

~u = 2~i + 3~j +~k

~v =~i − ~j

~w = 7~i + 3~j + 2~k

41


True or False

1. ~a ×~b is a scalar.

2. ~a × ~a = ~0.

3. For ~a,~b , ~0, if ~a ×~b = ~0 then ~a = ~b.

4. ~i × ~j = ~k

5. ~a ×~b = ~b × ~a

6. (~a ×~b) × ~c = ~a × (~b × ~c)

7. ~a × (~b + ~c) = ~a ×~b + ~a × ~c

42


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Extra Example. (#44)

1. Find all vectors ~v such that < 1, 2, 1 > ×~v =< 3, 1,−5 >.

2. Explain why there is no vector ~v such that < 1, 2, 1 > ×~v =< 3, 1, 5 >.

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§12.5 LINES AND PLANES

§12.5 Lines and Planes

Example. Is the line through (−4,−6, 1) and (−2, 0,−3) parallel to the line through(10, 18, 4) and (5, 3, 14)?

What is the equation of the line through the origin, that is parallel to the line through(−4,−6, 1) and (−2, 0,−3)?

What is the equation of the line through (−4,−6, 1) and (−2, 0,−3)?

45


How else could you write an equation of the line through (−4,−6, 1) and (−2, 0,−3)?

46


Note. The equation of a line through the point (x0, y0, z0) in the direction of the vector< a, b, c > can be described:

with the parametric equations:

or, with ”symmetric equations”:

or, with the vector equation:

47


Example. Find the equation of the plane through the point (−3, 2, 0) and perpendicularto the vector < 1,−2, 5 >

Note. The plane through the point (x0, y0, z0) and perpendicular to the vector < a, b, c >is given by the equation

48


Review. Which of these equations represents a line in 3-dimensional space?

1. 3x + 5y = 2

2. 3x + 5y − 8z = 2

3. 3(x − 1) + 5(y − 3) − 8(z − 2) = 0

4. x = 7 + 4t, y = 5 − 3t, z = 7t

5. x−74 =

−y+53 = z

7

6. 5(x − 1) = 7(y + 4) = −6(z + 3)

49


Example. (# 19) Determine whether the lines L1 and L2 are parallel, skew, or intersect-ing.

L1 : x = 3 + 2t, y = 4 − t, z = 1 + 3t

L2 : x = 1 + 4s, y = 3 − 2s, z = 4 + 5s

50


Example. Find the line of intersection of the planes 2x − y + z = 5 and x + y − z = 1.

51


Example. (# 33) Find an equation for the plane through the points (3,−1, 2), (8, 2, 4),and (−1,−2,−3).

52


Example. (# 56) Determine whether the planes are parallel, perpendicular or neither.If neither, find the angle between them.

x + 2y + 2z = 1, 2x − y + 2z = 1

53


Example. (# 73) Show that the distance between the parallel planes ax + by + cz + d1 = 0and ax + by + cz + d2 = 0 is

D =|d1 − d2|√

a2 + b2 + c2

54


Example. Show that the distance between the point (x1, y1, z1) and the plane ax + by +

cz + d = 0 is given by

D =|ax1 + by1 + cz1 + d|√

a2 + b2 + c2

55


Example. Find the distance between the skew lines

L1 : x = 3 + 2t, y = 4 − t, z = 1 + 3t

L2 : x = 1 + 4s, y = 3 − 2s, z = 4 + 5s

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§13.1 VECTOR FUNCTIONS

§13.1 Vector Functions

Definition. A vector function or vector-valued function is:

If we think of the vectors as position vectors with their initial points at the endpoints,then the endpoints of ~v(t) trace out a in R3 (or in R2).

57


Example. Sketch the curve defined by the vector function ~r(t) =< t, sin(5t), cos(5t) >.

58


Example. Consider the vector function ~r(t) =t2− t

t − 1~i +√

t + 8~j +sin(πt)

ln t~k

1. What is the domain of ~r(t)?

2. Find limt→1

~r(t)

3. Is ~r(t) continuous on (0,∞)? Why or why not?

59


Review. Match the vector equations with the curves.

(a) ~r(t) =< t2, t4, t6 >

(b) ~r(t) =< t + 2, 3 − t, 2t − 1 >

(c) ~s(t) =< cos(t),− cos(t), sin(t) >

60


Example. (# 30) At what points does the helix ~r(t) =< sin(t), cos(t), t > intersect thesphere x2 + y2 + z2 = 5?

61


Example. (#46) Find parametric equations for the curve of intersection of theparabolic cylinder y = x2 and the top half of the ellipsoid x2 + 4y2 + 4z2 = 16.

62


Example. (# 43) Find a vector function that represents the curve of intersection ofthe hyperboloid z = x2

− y2 and the cylinder x2 + y2 = 1.

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§13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS

§13.2 Derivatives and Integrals of Vector Functions

Suppose a particle is moving according to the vector equation ~r(t). How can we find atangent vector that gives the direction and speed that the particle is traveling?

64


Definition. The derivative of the vector function ~r(t) is the same thing as the tangentvector, defined as

d~rdt

= ~r ′(t) =

If ~r(t) =< f (t), g(t), h(t) >, then

~r ′(t) =

The derivative of a vector function is a (circle one) vector / scalar.

The unit tangent vector is:~T(t) =

The tangent line at t = a is:

65


Example. For the vector function ~r(t) =< t2, t3 >

1. Find ~r ′(1).

2. Sketch ~r(t) and ~r ′(1).

3. Find ~T(1).

4. Find the equation for the tangent line at t = 1.

66


Definition. If ~r(t) =< f (t), g(t), h(t) >, then∫~r(t) dt =

and

∫ b

a~r(t) dt =

Example. Compute∫ 2

1

1t~i + et~j + tet~k.

67


Review. If ~r(t) represents the position of a particle at time t,

(a) what does the direction of the tangent vector signify?

(b) what does the magnitude of the tangent vector signify?

68


Example. Find the tangent vector, the unit tangent vector, and the tangent line for thefollowing curves at the point given

1. ~r(t) =< t, t2, t3 > at t = 1

2. ~r(t) =< t2, t4, t6 > at t = 1

3. ~p(t) =< t + 2, 3 − t, 2t − 1 > at t = 0

69


Example. (# 34) At what point do the curves ~r1(t) =< t, 1 − t, 3 + t2 > and ~r2(t) =<3 − t, t − 2, t2 > intersect? Find their angle of intersection correct to the nearest degree.

70


Derivative rules - see textbook

• Is there a product rule for derivatives of vector functions?

• Is there a quotient rule for derivatives of vector functions?

• Is there a chain rule for derivatives of vector functions?

71


Example. Show that if ||~r(t)|| = c (a constant), then ~r ′(t) is orthogonal to ~r(t) for all t.

72


Review. If ~r(t) =< f (t), g(t), h(t) >, then∫~r(t) dt =

and

∫ b

a~r(t) dt =

Example. Find ~p(t) if ~p ′(t) = cos(πt)~i + sin(πt)~j + t~k and ~p(1) = 6~i + 6~j + 6~k.

73


Extra Example. (# 51) Show that if ~r is a vector function such that ~r ′′ exists, then

ddt

[~r(t) × ~r ′(t)] = ~r(t) × ~r ′′(t)

Extra Example. (#55 in book) If ~u(t) = ~r(t) ◦ [~r ′(t) × ~r ′′(t)], show that

~u ′(t) = ~r(t) · [~r ′(t) × ~r ′′′(t)]

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§13.3 ARCLENGTH AND CURVATURE

§13.3 Arclength and Curvature

To find the arc length of a curve ~r(t) =< f (t), g(t), h(t) >, we can approximate it withstraight line segments.

75


Note. The arc length of a curve ~r(t) =< x(t), y(t), z(t) > between t = a and t = b is givenby

Definition. The arc length function (starting at t = a) is

s(t) =

Note. If s(t) is the arc length function, then s′(t) =

76


Example. (#2) Consider the two curves:

1. ~r(u) =< 2u,u2, 13u3 > for 0 ≤ u ≤ 1

2. ~q(t) =< 2 ln(t), (ln(t))2, 13(ln(t))3 > for 1 ≤ t ≤ e

How are the curves related?

We say that ~q(t) is a reparametrization of ~r(u) because:

Also ~r(u) is a reparametrization of ~q(t) because:

You can think of a reparametrization of a curve as the same curve, traveled at a differentspeed. In our case, ~q moves along the curve (circle one) slower / faster than ~r.

In mathematical notation, ~q(t) is a reparametrization of ~r(u) if ~q(t) = ~r(φ(t)) for somestrictly increasing (and therefore invertible) function u = φ(t).

77


Find the arc length of each curve.

~r(u) =< 2u,u2,13

u3 >

for 0 ≤ u ≤ 1~q(t) =< 2 ln(t), (ln(t))2,

13

(ln(t))3 >

for 1 ≤ t ≤ e

78


Fact. Arc length does not depend on parametrization.

Proof:

Is there a natural, best way to parametrize a curve?

79


Example. Reparametrize by arc length:

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

Original function:

t 0 1 2 3 4 5~v(t) < −3, 1 > < −2, 1 > < 0, 0 > < 1, 1 > < 2, 0 > < 3, 2 >

arclength

Reparametrized by arclength:

s~v(s) < −3, 1 > < −2, 1 > < 0, 0 > < 1, 1 > < 2, 0 > < 3, 2 >

arclength

80


Example. Reparametrize by arc length:

~p(t) = 3 sin(t)~i + 4t~j + 3 cos(t)~kfor t ≥ 0

~r(t) = e3t~i + e3t~j + 3~kfor t ≥ 0

81


Fact. For any curve ~q parametrized by arc length, ||q′(s)|| = 1 for all s.

Proof:

Note. For any curve ~q parametrized by arc length, ~T(s) = ~q ′(s)

82


CurvatureNote. In the rest of this section only, s denotes arc length and writing a curve withparameter s like ~r(s) implies ~r is parametrized by arc length.

For the rest of this section, we will only work with smooth parametrizations of curves;i.e. we will assume ~r ′ exists, is continuous, and is never zero. Smooth curves have nosharp corners or cusps.

83


Intuition about curvature:

1. Graph the parabola y = x2. What part of the parabola has the biggest curvature?The smallest curvature?

2. Graph the function y = sin(x) and describe its curvature at various points on thecurve.

3. Draw a curve that has the same curvature at all points. How could you quantifyits curvature as a number?

4. Extend this idea to define the curvature at points of other curves.

84


Principles:

• To define curvature, we want to quantify how fast ~r ′(t) changes direction.

• We use a unit tangent vector ~T ′(t) =~r ′(t)||~r ′(t)|| because we are only interested in changes

in direction, not length.

• We start with a curve parametrized by arc length so that the answer will not dependon parametrization.

Definition. The curvature of a curve ~q(s) which is parametrized by arc length, at thepoint ~q(s0) is defined as

κ(s0) = ||~q ′′(s0)||

Equivalently,

κ(s0) =

∣∣∣∣∣∣∣∣∣∣∣∣d~Tds

∣∣∣∣∣∣∣∣∣∣∣∣ at s = s0

where ~T is the unit tangent vector, since ~T = ~q ′ when q is parametrized by arc length.

Equivalently, if the curve ~r(t) is not parametrized by arc length,d~Tdt

=d~Tds

dsdt

so

85


Example. Find the curvature of a circle of radius a.

Example. Find the curvature of the helix ~r(t) = cos(t)~i + sin(t)~j + t~k

86


Theorem. The curvature of the curve given by the vector function r (not nec. parametrizedby arclength) is

κ(t) =||~r ′(t) × ~r ′′(t)||||~r ′(t)||3

87


Proof. Corollary: if y = f (x) is a plane curve, then

κ)x) =

Example. Find the curvature of the parabola y = x2.

88


Note. A helix has the same curvature as a circle. So κ(t) = ||~T ′(s)|| =||~T ′(t)||||~r′(t)|| doesn’t

distinguish a helix from a circle.

What expression can we use to measure the difference between how a helix and a circlebend through space?

89


Definition. The unit normal to a curve ~q(s) parametrized by arclength is

~N(t) =~q ′′(s)||~q ′′(s)||

=~T ′(s)

||~T ′(s)||

In other words, curvature measures the length and the unit normal measures thedirection of ~q ′′(s) = ~T ′(s)

Note. If the curve is not parametrized by arc length,

d~Tdt

=d~Tds

dsdt

so

~N(t) =~T′(s)

||~T ′(s)||=

~T ′(t)dsdt

||~T ′(t)

dsdt||

=~T ′(t)

||~T ′(t)||

so unlike curvature, the definition of the unit vector is the same whether or not thecurve is parametrized by arc length.

Note. ~N(t) is perpendicular to ~T(t):

90


Definition. The binormal vector ~B(t) is defined as ~T(t) × ~N(t)

Note. ||~B(t)|| = ||~N(t)|| = ||~T(t)|| = 1 and ~B, ~N, and ~T are all perpendicular to each other.

Definition. The plane defined by ~N(t) and ~B(t) at a point P = ~r(t) is called the n¯ormal

plane of the curve ~r at P.

The plane defined by ~T(t) and ~N(t) at a point P = ~r(t) is called the o¯scullating plane of

the curve ~r at P.

Example. Find the normal plane and the osculating plane for the helix ~r(t) = cos(t)~i +

sin(t)~j + t~k at P(0, 1, π2 ).

91

§13.4 VELOCITY AND ACCELERATION

§13.4 Velocity and Acceleration

The language of physics and the language of curvature:

PHYSICS:

position vector:

velocity vector:

acceleration vector:

speed ν:

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PHYSICS, CURVATURE, and the TNB frame:

1) relate ~T(t) to velocity and speed

2) relate ~a to speed, curvature, and ~T and ~N.

3) if a particle’s path is parametrized by arc length (i.e. it is traveling at a constantspeed), then ~a(t) =

93


Example. A projectile is fired from a height of y0 meters at an angle of θ to thehorizontal, with initial speed of v0 m/s. Find a vector equation for its position.

94


Example. A projectile is fired from a height of 10 meters at an angle of 30◦ to thehorizontal, with initial speed of 500 m/s. Find a vector equation for its position. Findits range, maximum height, and speed at impact.

95

§14.1 FUNCTIONS OF TWO OR MORE VARIABLES

§14.1 Functions of Two or More Variables

Example. Consider the function of two variables f (x, y) =√

xy

1. What is its domain?

2. What are its level curves?

3. What does its graph look like?

96


1. z = sin√

x2 + y2 2. z = 1x2+4y2 3. z = sin(x) sin(y)

4. z = x2y2e−(x2+y2) 5. z = x3− 3xy2 6. z = sin2(x) +

y2

4

I. II. III.

IV. V. VI.

A. B. C.

D. E. F.

97


Functions of 3 or more variables

To visualize functions f (x, y, z) of three variables, it is handy to look at level surfaces.

Example. f (x, y, z) = x2 + y2 + z2

(a) Guess what the level surfaces should look like.

(b) Graph a few level surfaces (e.g. x2 + y2 + z2 = 10, x2− y2 + z2 = 20, x2

− y2 + z2 = 30)on a 3-d plot.

Example. f (x, y, z) = x2− y2 + z2

1. Guess what the level surfaces should look like.

2. Graph a few level surfaces (e.g. x2− y2 + z2 = 0, x2

− y2 + z2 = 10, x2− y2 + z2 = 20)

on a 3-d plot.

98

§14.2 LIMITS AND CONTINUITY

§14.2 Limits and Continuity

Recall LIMITS from Calculus 1:

Informally, limx→a

f (x) = L if the y-values f (x) get closer and closer to the same number Lwhen x approaches a from either the left of the right.

Does limx→0

f (x) exist for these functions?

99


LIMITS for functions of two variables:

Informally, lim(x,y)→(a,b) f (x, y) = L if the z-values f (x, y) ...

100


For each function, decide if lim(x,y)→(0,0)

f (x, y) exists. The color is based on height: low is

blue and high is red.

A) f (x, y) =x2− y2

x2 + y2 B) f (x, y) =1 − x2

x2 + y2 + 1

C) f (x, y) =x

x2 + y2 D) f (x, y) =x3

x2 + y2

101


ε − δ definition of limit from Calculus 1:

limx→a

f (x) = L if for every small number ε > 0 there is a number δ > 0 so that wheneverx is within δ of a, f (x) is within ε of L.

ε − δ definition of limit for functions of two variables:

Definition. For a function f of two variables and a point (a, b)

lim(x,y)→(a,b)

f (x, y) = L

means for every number there is a corresponding number such that

102


Does lim(x,y)→(0,0)

f (x, y) exist? The color is based on height: low is blue and high is red.

A) lim(x,y)→(0,0)

2xyx2 + y2 B) lim

(x,y)→(0,0)

xy2

x2 + y4

C) f (x, y) =10 sin(xy)x + y + 20

D) f (x, y) =3x2y − 3x2

− 3y2

x2 + y2

103


Example. Show that lim(x,y)→(0,0)

2xyx2 + y2 does not exist.

104



xy2

x2 + y4 does not exist.

105



f (x, y) =10 sin(xy)x + y + 20

does exist.

106



3x2y − 3x2− 3y2

x2 + y2 does exist.

107


Summary: for practical purposes, the best way to show that a limit does not exist is to:

For practical purposes, the best ways to show that a limit exists is to:

•

•

•

•

108


Recall CONTINUITY from Calculus 1: A function f is continuous at the point x = a if:

1.

2.

3.

Recall from Calculus 1: common functions that are continuous include:

109


CONTINUITY for functions of of two (or more) variables:

A function f (x, y) is continuous at the point (x, y) = (a, b) if:

1.

2.

3.

Common functions that are continuous include:

•

•

•

•

110


Example. Where is f (x, y) =

√

4 − x2 +√

4 − y2

1 − x2 − y2 continuous?

111

§14.3 PARTIAL DERIVATIVES

§14.3 Partial Derivatives

Example. (problem # 4 in book) The wave heights h in the open sea depend on thespeed ν of the wind and the length of time t that the wind has been blowing at thatspeed. So we write h = f (ν, t).

1. What is f (40, 20)?

2. If we fix duration at t = 20 hours andthink of g(ν) = f (ν, 20) as a functionof ν, what is the approximate valueof the derivative dg

dν

∣∣∣∣ν=40

?

3. If we fix wind speed at 40 knots, andthink of k(t) = f (40, t) as a function ofduration t, what is the approximatevalue of the derivative dk

dt

∣∣∣t=20

?

112


Definition. For a function f (x, y) defined near (a, b), the partial derivatives of f at (a, b)are:

fx(a, b) = the derivative of f (x, b) with respect to x when x = a, and

fy(a, b) = the derivative of f (a, y) with respect to y when y = b.

In terms of the limit definition of derivatives, we have:

fx(a, b) = limh→0

fy(a, b) = limh→0

Geometrically, f (x, b) can be thought of as

So fx(a, b) = ddx f (x, b)|x=a can be thought of as

113


Note. To compute fx, we just take the derivative with x as our variable, holding allother variables constant. Similarly for the partial derivative with respect to any othervariable.

Example. f (x, y) = xy. Find fx(1, 2) and fy(1, 2).

Notation. There are many notations for partial derivatives, including the following:

fx∂ f∂x

∂z∂x f1 D1 f Dx f

Note. Partial derivatives can also be taken for functions of three or more variables. Forexample, if f (x, y, z,w) is a function of 4 variables, then D3 f (x, y, z,w) means:

114


Review. For f (x, y, z) = 3x2yz + 5y2z3 + cos(xyz), find ∂ f∂z .

115


Example. Find the specified partial derivatives

1. f (x, y) =ax+bycx+dy, find ∂ f

∂x and ∂ f∂y

2. f (x, y, z) = ex sin(y−z), find Dx f , Dy f and Dz f

3. z = f (x)g(y), find ∂z∂x and ∂z

∂y

4. x2− y2 + z2

− 2z = 4, find ∂z∂x and ∂z

∂y

116


Notation. Second derivatives are written using any of the following notations:

Example. For f (x, y) = x2 + x2y2− 2y2, calculate fxx, fxy, fyx, and fyy.

117


Theorem. (Clairout’s Thm) Let f (x, y) be defined on a disk D containing (a, b). If fxy

and fyx are both defined and on D, then fxy(a, b) = fyx(a, b).

Example. The function

f (x, y) =

xyx2−y2

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

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

has D2 f (x, 0) = x for all x and D1 f (0, y) = −y for all y, and therefore, D2,1 f (0, 0) = 1while D1,2 f (0, 0) = −1.

118


Extra Example. The wave equation is given by

∂2u∂t2 = a2∂

2u∂x2

where u(x, t) represents displacement, t represents time, x represents the distance fromone end of the wave, and a is a constant.

Verify that the function u(x, t) = sin(x − at) satisfies the wave equation.

119

§14.4 TANGENT PLANES AND LINEAR APPROXIMATIONS

§14.4 Tangent Planes and Linear Approximations

Example. Given a surface z = f (x, y), what is the intuitive idea of a tangent plane atthe point (x0, y0, z0)?

Example. Find the tangent plane to the surface z = y2− x2 at the point (1, 2, 3). Hint:

look at the curves of intersection of the surface with two vertical planes through thepoint. Find the tangent vectors of these curves. These tangent vectors should lie in thetangent plane for the surface.

120


Find a general formula for the tangent plane of z = f (x, y) at the point (x0, y0, z0).

Notice that this formula is analogous to the formula for a tangent line for a function ofone variable.

121


Usually, the tangent plane at a point (x0, y0, z0) is a good approximation of the surfacenear that point. (We will see later some situations when it is not.)

Example. Find the tangent plane for f (x, y) = 1 − xy cos(πy) at (x, y) = (1, 1) and use itto approximate f (1.02, 0.97).

122


This method of approximating a function’s value with the height of the tangent planecan be written in the language of differentials.

Recall: In Calc 1, for a function y = f (x), the differential was defined as d f = f ′(x)dx.

Definition. The differential for the function y = f (x, y) is:

123


∆ f represents the actual change in a function f (x, y) − f (a, b). d f represents the corre-sponding change in the tangent plane between (a, b) and (x, y).

Differentials are useful for estimating errors and making approximations, because thedifferential is linear in all its variables, and linear functions are easier to calculate.

124


Example. (# 34) Use differentials to estimate the amount of metal in a closed cylindricalcan that is 10 cm high and 4 cm in diameter if the metal in the top and bottom is 0.1cm thick and the metal in the sides is 0.05 cm thick.

125


Example. (# 40) Four positive numbers, each less than 50, are rounded to the firstdecimal place and multiplied together. Use differentials to estimate the maximumpossible error in the computed product that might result from the rounding.

126


When the tangent plane approximation breaks down.

Note. Anytime fx and fy exist at a point (x0, y0), it is possible to write down an equationfor a tangent plane at the point (x0, y0, z0).

Usually, this plane is a good approximation to the function.

Occasionally it is not.

Example. f (x, y) =

xy

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

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

Compute fx and fy at (0, 0) to find the tangent plane. Why is it a poor approximationof the surface near (0, 0, 0)?

127


Extra Example. f (x, y) = (xy)1/3

Compute fx and fy at (0, 0) to find the tangent plane. Why is it a poor approximationof the surface near (0, 0, 0)?

128


Definition. (Informal definition) If the tangent plane is a “good approximation” to thesurface near a point (a, b), then the function f is called differentiable at (a, b).

More formally, f (x, y) is differentiable at (a, b) if

lim(∆x,∆y)→(0,0)

∆ f − d f√(∆x)2 + ∆y)2

= 0

Equivalently,∆ f = fx(a, b)∆x + fy(a, b)∆y + ε(∆x,∆y)

where ε√(∆x)2+(∆y)2

→ 0 as (∆x,∆y)→ (0, 0).

Theorem. If fx and fy exist in near (a, b) and are continuous at (a, b), then f is differen-tiable at (a, b).

129


Theorem. If f is differentiable at (a, b), then f is continuous at (a, b).

130

§14.5 THE CHAIN RULE

§14.5 The Chain Rule

Remember the Chain Rule for functions of 1 variable:

dydt

=dydx·

dxdt

Theorem. (Chain Rule, Case 1) Suppose z = f (x, y) is a differentiable function of x andy and x = g(t) and y = h(t) are both differentiable functions of t. Then

dzdt

=

131


Informal proof of the Chain Rule:dzdt

=∂z∂x

dxdt

+∂z∂y

dydt

132


Example. w = x cos(4y2z), x = t2, y = 1 − t, z = 1 + 2t. Find dzdt .

133


Theorem. (Chain Rule, Case 2) Suppose z = f (x, y) is a differentiable function of x andy and x = g(s, t) and y = h(s, t) are both differentiable functions of t. Then

∂z∂s

=

∂z∂t

=

134


Example. z = er cos(θ), r = st, θ =√

s2 + t2. Find ∂z∂s and ∂z

∂t .

135


Review. The Chain Rule If u is a differentiable function of n variables x1, x2, x3, . . . xn,and each xi is a differentable function of m variables t1, t2, . . . tm, then

∂udti

=

136


Example. (#6) w = ln√

x2 + y2 + z2, x = sin t, y = cos t, z = tan t. Finddwdt

.

137


Example. (# 24) u = xety, x = α2β, y = β2γ, t = γ2α. Find∂u∂α

,∂u∂β

,∂u∂γ

when α = −1,

β = 2, γ = 1.

138


Example. (# 42) A manufacturer modeled its yearly production P in millions of dollarsas a Cobb-Douglas function

P(L,K) − 1.47L0.65K0.35

where L is the number of labor hours (in thousands) and K is the invested capital (inmillions of dollars). Suppose that when L is 30 and K is 8, the labor force is decreasingat a rate of 2000 labor hours per year and capital is increasing at a rate of $500,000 peryear. Find the rate of change of production.

139


Extra Example. Suppose f (x, y) = h(x2− y2, 3xy2

− 4), where h(u, v) is a differentiablefunction.

1. Find equations for fx and fy in terms of hu and hv.

2. Use the contour graph for h to estimate fx and fy at the point (x, y) = (2, 1).

This is a contour graph for h not f .

140


The Implicit Function Theorem

Previously, we calculated ∂z∂x, where z is defined implicitly as a function of x and y by

x2− y2 + z2

− 2z = 4 .

The Implicit Function Theorem formalizes this process.

If z is given implicitly as a function z = f (x, y) by an equation for the form F(x, y, f (x, y)) =

0, then we can use the Chain Rule to take the derivative of F and solve for ∂z∂x.

141


Theorem. If F is defined on an open sphere containing (a, b, c), where F(a, b, c) = 0,Fz(a, b, c) , 0, and Fx, Fy, and Fz are continuous inside the sphere, then the equationF(x, y, z) = 0 defines z as a function of x and y near the point (a, b, c) and the function isdifferentiable with derivatives given by :

Theorem. If F is defined on an open disk containing (a, b), where F(a, b) = 0, Fy(a, b) , 0,and Fx, Fy are continuous inside the disk, then the equation F(x, y) = 0 defines y asa function of x near the point (a, b) and the function is differentiable with derivativegiven by :

142

§14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT

§14.6 Directional Derivatives and the Gradient

Example. What is the (approximate) directional derivative of the temperature functionf (x, y) at Dubbo, New South Wales, in the direction of Sydney? What are the units?

143


Definition. The directional derivative of f at (x0, y0) in the direction of the unit vector~u =< a, b > is given by:

D~u f (x0, y0) =

if this limit exists.

Equivalently,

What does the Chain Rule tell us about this derivative?

Theorem. For the unit vector ~u =< a, b >,

D~u f (x0, y0) = fx(x0, y0) + fy(x0, y0)

144


Question. What would happen if we did the same computation of D~v f where ~v wasnot a unit vector but instead some multiple of a unit vector c · ~u?

By convention, the directional derivative in the direction of ~v means:

Question. What are D~i f and D~j f ?

145


Definition. The gradient of the function f (x, y) at (x0, y0) is defined as:

∇ f (x0, y0) =

Note. For a unit vector ~u =< a, b >, D~u f (x0, y0) can be written in terms of the gradientas:

D~u f (x0, y0) =

146


Example. For f (x, y) = xe−y− y,

1. Find ∇ f at the point (2, 0).

2. Find the directional derivative at (2, 0), in the direction of ~v =< 3,−4 >.

147


Question. For a general function f , at a point (x0, y0), in what direction is f increasingmost steeply? (i.e. for what unit vector ~u is D~u f maximal?) Answer in terms of thegradient.

Theorem. The maximum value of D~u is and this maximum occurs in thedirection of .

The minimum value of D~u is and this minimum occurs in the direction of.

148


Example. For f (x, y) = xe−y− y at (2, 0), what is the maximum directional derivative

and what direction does it occur in?

Note. Consider the graph of ∇ f on the x-y plane drawn below.

Since ∇ f points in the direction of greatest increase, where do you expect the levelcurves to be?

f (x, y) = xe−y− y

-4.8 -4 -3.2 -2.4 -1.6 -0.8 0 0.8 1.6 2.4 3.2 4 4.8

149


Theorem. For a function f of two variables, the level curves of f are to ∇ f .

Proof:

150


Example. Use the gradient to find the equation of the tangent line to the level curvefor f (x, y) = x2

− y2 at (2, 3).

Note. The tangent line for a level curve f (x, y) = k at the point (a, b) is given by theequation:

151


Everything we have done so far can also be done for functions of three or morevariables! For f (x, y, z), and unit vector ~u =< a, b, c >

• ∇ f (x0, y0, z0) =

• D~u f (x0, y0, z0) =

• The direction of greatest increase at the point (x0, y0, z0) is , and thelargest directional derivative has magnitude .

• ∇ f (x, y, z) is perpendicular to .

• The equation for a tangent plane to a level surface f (x, y, z) = k at the point (x0, y0, z0)is given by the equation:

152


Example. (# 44) Find the equation of (a) the tangent plane, and (b) the normal line tothe surface xy+ yz+zx = 5 at the point (1, 2, 1). (The normal line is a line perpendicularto the surface at the given point.)

153


Note. The tangent plane for the surface z = f (x, y) is the same as the tangent plane forthe level surface g(x, y, z) = 0, where g(x, y, z) = .

Use the equation for a tangent plane to a level surface for g(x, y, z) at (x0, y0, z0) to findan equation for the tangent plane of z = f (x, y). Compare the equation to the equationwe found in Section 14.4.

154


Extra Example. Here is an example of a function that is not continuous at (0, 0) eventhough the directional derivatives exist in every direction!

f (x, y) =xy2

x2 + y4

155

§14.7 MAXIMUMS AND MINIMUMS

§14.7 Maximums and Minimums

Note. Recall: For a function f (x) of one variable, a critical point is:

Note. Recall: For a function f (x) of one variable, max and min points are related tocritical points.

• If f (x) has a local max or min at x = x0, then x0 (circle one) is / is not / may or maynot be a critical point.

• If x0 is a critical point, then f (x) (circle one) does / does not / may or may nothave a maximum or minimum at x0.

156


Definition. For a function f (x, y) of two variables, (x0, y0) is a critical point if

Theorem. If f (x, y) has a local max or local min at (x0, y0) and fx and fy exist at (x0, y0),then:

1. fx(x0, y0) = and fy(x0, y0) = .

2. D~u f (x0, y0) = for all unit vectors ~u.

3. The tangent plane to f at (x0, y0) is

Proof:

Note. If (x0, y0) is a critical point, then f (circle one) does / does not / may or may nothave a maximum or minimum at (x0, y0).

157


Note. In Calc 1, we had the 2nd derivative test:

If f ′(x0) = 0 and f ′′(x0) > 0, then f has a local at x0.

If f ′(x0) = 0 and f ′′(x0) < 0, then f has a local at x0.

Definition. The discriminant of f at (x0, y0) is given by

D =

Theorem. (Second Derivatives Test) Suppose that the second partial derivatives off exist and are continuous on a disk around (x0, y0), and suppose that fx(x0, y0) =

fy(x0, y0) = 0. Then

(a) If D > 0 and fxx(x0, y0) > 0, then f has a at (x0, y0).

(b) If D > 0 and fxx(x0, y0) < 0, then f has a at (x0, y0).

(c) If D < 0 then f has a at (x0, y0).

(d) If D = 0, then .

(e) If D > 0 and fxx(x0, y0) = 0, then .

158


Example. Find the local maxes and mins for

f (x, y) = x2 + y2 + x2y + 4

.

159


Example. (# 40) Find the point on the plane x − 2y + 3z = 6 that is closest to the point(0, 1, 1).

160


Note. Recall, in Calc 1:

• If f is continuous and [a, b] is a closed interval, then f (x) (circle one) must / may ormay not achieve a maximum and minimum value on [a, b].

• The maximum value of f on [a, b] occurs (circle one) at one of the end points a or b/ at a critical point / either at a critical point or at one of the end points a or b.

Question. How do we generalize the idea of a closed interval to R2?

On which of these would you expect a continuous function to always have a max anda min?

Definition. (Informal definition) A closed set is one that contains all of its boundarypoints.

Definition. A bounded set is one that is contained inside some large disk.

161


Theorem. (Extreme Value Theorem) If f (x, y) is continuous on a closed bounded setD, then f attains (at least one) absolute max and abs min.

The absolute max and min occur either at critical points or on the boundary of D.

Notation. ∂D means the boundary of a region D.

Note. To find the abs max and min of f on D

1.

2.

3.

162


Example. Find the abs max and min of f (x, y) = x2− 2xy + 2y on the triangular region

with vertices (0, 2), (3, 0), (3, 2).

163


Theorem. (Second Derivatives Test) Suppose that the second partial derivatives off exist and are continuous on a disk around (x0, y0), and suppose that fx(x0, y0) =

fy(x0, y0) = 0. Then

(a) If D > 0 and fxx(x0, y0) > 0, then f has a at (x0, y0).

(b) If D > 0 and fxx(x0, y0) < 0, then f has a at (x0, y0).

(c) If D < 0 then f has a at (x0, y0).

Proof of the Second Derivatives Test (if time)

164

§14.8 LAGRANGE MULTIPLIERS

§14.8 Lagrange Multipliers

Example. Find the rectangle of largest area that can be inscribed in the ellipsex2

4+

y2

9= 1

(with sides parallel to the x and y axes)

165


Consider the graph of f (x, y) = 4xy and the graph ofx2

4+

y2

9= 1 in 3-dimensions (on

the left).

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

Consider the graph of the level curves of f (x, y) = 4xy and the graph ofx2

4+

y2

9= 1 in

2-dimensions (on the right).

166


Theorem. (Lagrange multipliers) To find the max and min values of f (x, y) subjectto the constraint g(x, y) = k (assuming these extreme values exist and ∇g , 0 on theconstraint curve g(x, y) = k), we need to:

1. find all values (x0, y0) such that ∇ f is parallel to ∇g, i.e. where ∇ f = λ∇g for someλ ∈ R.

2. compare the size of f (x0, y0) on all these candidate points.

Note. In practice, we setfx(x, y) = λgx(x, y)

fy(x, y) = λgy(x, y)

g(x, y) = k

and solve this system of equations for x, y, and λ.

Note. For this to work, g must be ”smooth” (have continuous gx and gy) Otherwise wemust also check ”corners” and ”cusps” where gx or gy don’t exist or have discontinu-ities.

167


Note. Method of Lagrange multipliers also works to maximize or minimize functionsf (x, y, z) of 3 variables, subject to a constraint surface g(x, y, z) = k.

We setfx(x, y, z) = λgx(x, y, z)

fy(x, y, z) = λgy(x, y, z)

fz(x, y, z) = λgz(x, y, z)

g(x, y, z) = k

and solve this system of equations for x, y, z, and λ.

168


Example. Find the rectangle of largest area that can be inscribed in the ellipsex2

4+

y2

9= 1,

using Lagrange multipliers.

169


Example. Find the point on the plane x− 2y + 3z = 6 that is closest to the point (0, 1, 1).

170


Example. Maximize f (x, y) = 2x3 + y4 on the disk D = {(x, y)|x2 + y2≤ 1}.

171

§15.1 DOUBLE INTEGRALS

§15.1 Double Integrals

In Calculus 1, we defined the integral of f (x) over an interval [a, b] as the limit of aRiemann sum:

172


For a function f (x, y) over a rectangular region R = [a, b] × [c, d] we can define anintegral similarly:

∫∫R

f (x, y)dA =

173


Example. For each problem, estimate the integral of f (x, y) over the rectangle specified:

1. (# 8) The figure shows the level curves of a function f . Estimate∫∫

R f (x, y) dA whereR = [0, 2] × [0, 2].

2. (# 14) Find∫∫

R

√9 − y2 dA, where R = [0, 8] × [0, 3]. Hint: draw a sketch.

3. (# 5) Find∫∫

R f (x, y) dA, where R = [0, 4] × [2, 4], using the midpoint rule withn = m = 2.

174

§15.2 ITERATED INTEGRALS OVER RECTANGULAR REGIONS

§15.2 Iterated Integrals over Rectangular Regions

To compute the double integrals of a function f (x, y) over a rectangle, it is easiest tothink in terms of volume.

If R = [a, b] × [c, d], then

Volume =∫ b

a A(x) dx , where A(x) =

So∫∫

R f (x, y) dA = .

175


Similarly, if we look at cross-sectional area in the other direction A(y), we can computethe volume over the rectangle R = [a, b] × [c, d] as:

∫∫R f (x, y) dA = .

This is known as Fubini’s Theorem.

Theorem. Fubini’s Theorem If f is continuous on the rectangle R = [a, b] × [c, d], then thedouble integral equals the iterated integrals, i.e.∫∫

R f (x, y) dA =

This theorem still holds true if we just assume that f is bounded on R, discontinuous only ona finite number of smooth curves, and the iterated integrals exist.

176


Example. (# 3) Calculate∫∫

R(6x2y − 2x) dA, where

R = (x, y)|1 ≤ x ≤ 4, 0 ≤ y ≤ 2

177


Example. (# 28) Find the volume enclosed by the surfacez = 1 + ex sin y and the planes x = ±1, y = 0, y = π, and z = 0.

178

§15.3 DOUBLE INTEGRALS OVER GENERAL REGIONS

§15.3 Double Integrals over General Regions

Example. Calculate∫∫

D(x2 + 2y) dA for the region D bounded by the parabolas y = 2x2

and y = 1 + x2.

179


Example. Calculate∫∫

Dy dA for the region D bounded by the parabolas y = x−1 and

y2 = 2x + 6.

180


These two regions are examples of Type I and Type II regions.

Type I Region Type II Region

For a Type I region,∫∫D

f (x, y) dA =

For a Type II region,∫∫D

f (x, y) dA =

Note. If a region is both a Type I region and a Type II region, sometimes, it is easier toevaluate the integral in one way instead of the other.

181


Example. (# 54) Evaluate

∫ 8

0

∫ 2

3√yex4

dx dy

182


Example. Find an upper and lower bound for∫∫D

e−(x2+y2) dx dy

where D is the disk {(x, y)|x2 + y2≤

14}

183


Left-over from Section 15.2:

Question. True or False:∫ 6

1f (x)g(x) dx =

∫ 6

1f (x) dx ·

∫ 6

1g(x) dx


1

∫ 9

5f (x)g(y) dx dy =

∫ 9

5f (x) dx ·

∫ 6

1g(y) dy


1

∫ y2

5f (x)g(y) dx dy =

∫ y2

5f (x) dx ·

∫ 6

1g(y) dy

184

§15.4 INTEGRATION USING POLAR COORDINATES

§15.4 Integration using Polar Coordinates

Note. Recall: The polar coordinates of a point in the plane are given by (r, θ), where r isthe distance from the origin (radius) and θ is the angle made with the positive x-axis.

Note. A negative angle means to go clockwise from the positive x-axis. A negativeradius means jump to the other side of the origin, that is, (−r, θ) means the same pointas (r, θ + π)

185


Note. Recall: To convert between polar and Cartesian coordinates, note that:

• x =

• y =

• r2 =

• tan(θ) =

186


Example. (# 5 and # 6) Sketch the region described.

1. {(r, θ)|π4 ≤ θ ≤3π4 , 1 ≤ r ≤ 2} This is an example of a polar rectangle.

2. {(r, θ)|π2 ≤ θ ≤ π, 0 ≤ r ≤ 2 sinθ}

187


Example. (# 7) Evaluate∫∫

Dx2y dA, where D is the top half of the disk with center the

origin and radius 5.

188


Theorem. If f (x, y) is continuous on a polar rectangle R = {(r, θ)|α ≤ θ ≤ β, a ≤ r ≤ b},then ∫∫

Rf (x, y) dA =

Proof. (Where does the extra r come from?)

189


Example. (# 21) Find the volume enclosed by the hyperboloid −x2− y2 + z2 = 1 and

the plane z = 2.

190


Integration over more general regions:

Theorem. If f (x, y) is continuous on the region

R = {(r, θ)|α ≤ θ ≤ β, h1(θ) ≤ r ≤ h2(θ)},

then ∫∫R

f (x, y) dA =

191


Example. (# 16) Find the area of the region enclosed by both of the cardioids r = 1+cosθand r = 1 − cosθ.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

192


Example. The equation of the standard normal curve (with mean 0 and standarddeviation 1) is

f (x) =1√

2πe−

x22

Prove that the area under the standard normal curve is 1.

193

§15.5 CENTER OF MASS

§15.5 Center of Mass

Example. Suppose we have 4 children Abe, Beatrice, Catalonia, and Drew sitting ona see-saw. The children have masses 25 kg, 30 kg, 10 kg, and 35 kg, respectively, andare at distances 1 ft, 3 ft, 5 ft, and 7 ft from the left end of the see-saw. To balance theseesaw perfectly, where should the fulcrum be? (Ignore the mass of the seesaw itself.)

194


Now suppose we have a thin rod lying along the x-axis between x = a and x = b.The rod is made of a non-uniform material whose density at position x is given by afunction ρ(x). If we want to balance the rod, where should we place the fulcrum?

195


Now suppose we have a lamina (thin sheet) in the shape of a region D. The densityof the lamina at point (x, y) is given by the function ρ(x, y). If we want to balance thelamina, where should we place the fulcrum?

Definition. The balance point (x, y) is called the .The numerator of the expression for x is called the moment about theand is written as .The numerator of the expression for y is called the moment about theand is written as .The denominator of the expressions for x and y is the .

196


Example. SET UP the equations to find the mass and the center of mass of a regionenclosed by the curves x = y2 and x = 4, if ρ(x, y) = x + y2.

197


Example. A lamina occupies the part of the first quadrant that is inside a disk of radius5 and outside a disk of radius 1. Find its center of mass if the density at any point isinversely proportional to its distance from the origin.

198

§15.6 SURFACE AREA

§15.6 Surface Area

To find the surface afea of a surface z = f (x, y), we approximate the surface with littlepieces of planes that lie above little rectangles on the x-y plane.

What is the area of a parallelogram formed by two vectors?

What are the vectors that run along the sides of the little pieces of planes?

199

§15.6 SURFACE AREA

What is the area of a little piece of plane in the figure?

What is the area of the surface z = f (x, y), for (x, y) in some region D?

200

§15.6 SURFACE AREA

Example. (#12) Find the surface area of the part of the sphere x2 + y2 + z2 = 4z that liesinside the paraboloid z = x2 + y2.

201

§15.7 TRIPLE INTEGRALS

§15.7 Triple Integrals

The integral of f (x, y, z) over a rectangular box B = [a, b] × [c, d] × [r, s] can be definedas a limit of a Riemann sum:∫∫∫

Bf (x, y, z) dV =

Theorem. (Fubini’s Theorem for Triple Integrals) If f (x, y, z) is continuous over the boxB, then ∫∫∫

Bf (x, y, z) dV =

202


Integrals over general regions.

Example. (# 10) Evaluate the triple integral:∫∫∫E

ez/y dV

where E = {(x, , y, z)|0 ≤ y ≤ 1, y ≤ x ≤ 1, 0 ≤ z ≤ xy}

203


Example. (# 18) Set up the bounds of integration for∫∫∫

Ef (x, y, z)dV, where E is

bounded by the cylinder y2 + z2 = 9 and the planes x = 0, y = 3x, and z = 0 in the firstoctant.

204


Extra Example. (# 34) Rewrite the integral by changing the order of integration in asmany ways as possible.

∫ 1

0

∫ 1−x2

0

∫ 1−x

0f (x, y, z) dy dz dx

205


Example. Suppose I wanted to find the volume of this solid.

Could I compute it using a double integral?

Could I compute it using a triple integral?

206

§15.8 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES

§15.8 Triple Integrals in Cylindrical Coordinates

Note. The cylindrical coordinates of a point in the plane are given by (r, θ, z), where zis the height and r and θ are the polar coordinates of the projection of the point ontothe x-y plane.

As with polar coordinates, r can be positive or negative.

Note. Cartesian coordinates and cylindrical coordinates are related by:

• x =

• y =

• z =

• r2 =

• tanθ =

207


Example. What surfaces are described by these equations?

1. r = 5

2. θ = π3

3. z = r

4. z2 = 4 − r2

208


Note. Using cylindrical coordinates, if E is a region of space that can be described by:α ≤ θ ≤ β, h1(θ) ≤ r ≤ h2(θ),u1(r, θ) ≤ z ≤ u2(r, θ), then∫∫∫

Ef (x, y, z) dV =

209


Example. Find the mass of the solid cone bounded by z = 2r and z = 6, if the densityat any point is proportional to its distance from the z-axis.

210


Example. (#30) Rewrite the integral in cylindrical coordinates.

∫ 3

−3

∫ √

9−x2

0

∫ 9−x2−y2

0

√x2 + y2 dz dy dx

Hint: project the region onto the x-y plane and write this in polar coordinates.

211


Extra Example. (#31) Consider a mountain that is in the shape of a right circular cone.Suppose that the weight density of the material in the vicinity of a point P is g(P) andthe height is h(P) .

1. Find a definite integral that represents the total work done in forming the mountain.

2. Assume that Mount Fiji is in the shape of a right circular cone with radius 19,000meters, height 3,800 meters, and density a constant 3,200 kg/m3. How much workwas done in forming Mount Fiji if the land was initially at sea level?

212

§15.9 TRIPLE INTEGRALS IN SPHERICAL COORDINATES

§15.9 Triple Integrals in Spherical Coordinates

Note. The spherical coordinates of a point in the plane are given by (ρ, θ, φ), whereρ is the distance from the origin, θ is the angle with the positive x-axis, and φ is theangle with the positive z-axis.

Note. For spherical coordinates, ρ ≥ 0 always. Also, 0 ≤ φ ≤ π always.

Note. Cartesian coordinates and spherical coordinates are related by:

• x =

• y =

• z =

• ρ =

• tanθ =

• tanφ =

213


Example. What surfaces are described by these equations?

1. ρ = 5

2. θ = π4

3. φ = π6 .

214


Note. Using spherical coordinates, for a ≤ ρ ≤ b, α ≤ θ ≤ β, γ ≤ φ ≤ δ,∫∫∫E

f (x, y, z) dV =

Proof. (Informal Justification)

215


Example. (# 30) Find the volume of the solid that lies within the sphere x2 + y2 + z2 = 4,above the x-y plane, and below the cone z =

√x2 + y2.

216


Example. (# 41) Change the integral to spherical coordinates.∫ 2

−2

∫ √

4−x2

−

√

4−x2

∫ 2+√

4−x2−y2

2−√

4−x2−y2(x2 + y2 + z2)3/2 dz dy dx

217


Extra Example. (#28) Find the average distance from a point in a ball of radius a to itscenter.

218

§15.10 CHANGE OF VARIABLES

§15.10 Change of Variables

In Calc 1, we use u-substitution to do a change of variables.

Example.∫ π

0sin(x2)2x dx =

General formula for change of variables in 1-dimension:

∫ b

af (g(x))g′(x) dx =

To agree with the notation in the book, reverse the roles of x and u to get:

∫ b

af (g(u))g′(u) du =

Goal: generalize this change of variables formula to 2 and 3 dimensions. First, needto define what we mean by a change of variables.

219


Definition. A C1 transformation is a function (”mapping”)

T : R ×R −→ R ×R

by

T(u, v) = (g(u, v), h(u, v))

for functions g and h with continuous first derivatives.

Definition. A transformation T is one − to − one if ...

Example.T(r, θ) = (r cosθ, r sinθ)

g(r, θ) = h(r, θ) =

T : [0, 1] × [0, π] −→

220


Example.

T(u, v) =

(vu,u2

v

)x(u, v) = , y(u, v) =

Where does T map the unit square S = {(u, v)|1 ≤ u ≤ 2, 1 ≤ v ≤ 2}?

221


Definition. The Jacobian J of a transformation T is given by T(u, v) = (x(u, v), y(u, v)) is

∂(x, y)∂(u, v)

=

∣∣∣∣∣∣∣∣∣∂x∂u

∂x∂v

∂y∂u

∂y∂v

∣∣∣∣∣∣∣∣∣ =∂x∂u∂y∂v−∂y∂u∂x∂v

Example. Compute the Jacobian for the transformation x =vu, y =

u2

v.

222


Theorem. Suppose T is a C1 transformation whose Jacobian is non-zero and maps aregion S in the u-v plane to a region R in the x-y plane. Suppose that T is one-to-one,except possibly on the boundary. Then for a continuous function f ,

∫∫S

f (x(u, v), y(u, v)) abs(∣∣∣∣∣∣ ∂x

∂u∂x∂v

∂y∂u

∂y∂v

∣∣∣∣∣∣)

du dv =

∫∫R

f (x, y) dx dy

223


For transformations in 3-dimensions:

Definition. ForT : R ×R ×R −→ R ×R ×R

T(u, v,w) = (x(u, v,w), y(u, v,w), z(u, v,w))

the Jacobian J is given by

∂(x, y, z)∂(u, v,w)

=

∣∣∣∣∣∣∣∣xu xv xw

yu yv yw

zu zv zw

∣∣∣∣∣∣∣∣and the conclusion of the theorem is that:

∫∫S

f (x(u, v,w), y(u, v,w), z(u, v,w)) abs

∣∣∣∣∣∣∣∣∣∂x∂u

∂x∂v

∂x∂w

∂y∂u

∂y∂v

∂y∂w

∂z∂u

∂z∂v

∂z∂w

∣∣∣∣∣∣∣∣∣ du dv dw =

∫∫R

f (x, y, z) dx dy dz

224


Example. (Like #20) Use a change of variables to evaluate∫ ∫R

x2 dA

where R is the region bounded by the curves xy = 1, xy = 2, x2y = 1, and x2y = 2.

225


Example. (# 25) Make a change of variables to evaluate the integral∫∫R

cos(

y − xy + x

)dA

where R is the trapezoidal region with vertices (1, 0), (2, 0), (0, 2), and (0, 1).

226


Theorem. (Change to polar coordinates) If f (x, y) is continuous on some region R givenby R = {(r, θ)|a ≤ r ≤ b, α ≤ θ ≤ β}, where a, b > 0 and 0 ≤ β − α ≤ 2π, then∫ ∫

Rf (x, y) dA =

∫ β

α

∫ b

af (r cosθ, r sinθ) r dr dθ

227


Theorem. (Change to spherical coordinates) If f (x, y, z) is continuous on some regionR given by R = {(ρ, θ, φ)|a ≤ ρ ≤ b, α ≤ θ ≤ β, γ ≤ φ ≤ δ}, then∫ ∫ ∫

Rf (x, y, z) dA =

∫ δ

γ

∫ β

α

∫ b

af (ρ sinφ cosθ, ρ sinφ sinθ, ρ cosφ) ρ2 sinφ dρ dθdφ

228


Note. Explanation for why the Jacobian is the right thing to multiply by:

229

§16.1 VECTOR FIELDS

§16.1 Vector Fields

Definition. A vector field on R2 is a function~F(x, y) : D ⊂ R2

−→ R2

where D is a subset ofR2. We usually think of the output as a two-dimensional vector.

230


Definition. A vector field on R3 is a function

~F(x, y, z) : E ⊂ R3−→ R3 .

where E is a subset ofR3 We usually think of the output as a three-dimensional vector.

Example. The gradient ∇ f of a function or two or three variables is a vector field.

231


Example. Match the vector field with the plot.

A) B) C)

D)1) ~F(x, y) = cos(x + y)~i + x~j

2) ~G(x, y) =~i + x~j

3) ~H(x, y) = y~i + (x − y)~j

4) ~J(x, y) =y√

x2 + y2~i +

x√x2 + y2

~j

232


Example. The force of gravity can be written as a vector field. Let

M = mass of earthm = mass of objectG = gravitational constantF = force of gravity

Newton’s law says ||~F|| =mMG

r2 , where r is the diestance between the object and thecenter of the earth.

Write ~F as a vector field. Hint: put the origin at the center of the earth.

233


Definition. A vector field ~F is conservative if it is the gradient of some function. Thatis, there exists a function f such that ~F = ∇ f .

Definition. If ~F = ∇ f , then f is called the for ~F.

Question. Is the force of gravity a conservative vector field?

Question. Are their vector fields that aren’t conservative? Give an example.

234

§16.2 LINE INTEGRALS

§16.2 Line Integrals

Final Goal:

• define an integral of a vector field over a curve

• important applications to fluid flows, electricity and magnetism, etc.

First step:

• define the integral of a function over a curve, for a function with input in R2 or R3

and output in R.

• note: old definition of∫ ∫

R f dA cannot be used for an integral over a curve, why?

• start with functions of 2-dimensions and then extend the ideas and formulas to3-dimensions.

We start with a curve C parametrized by ~r(t) =< x(t), y(t) > for a ≤ t ≤ b

In this section, we will work only with curves that are piecewise smooth, whichmeans that the curve can be divided into finitely many pieces, and each for each piece:

235


We want to define∫

C f (x, y) ds, the integral of a function of 2 variables over a curve“with respect to arclength”. Think of the function as the height of a vertical ”curtain”over the curve, and think of the integral as the area of this curtain.

Special case: If f (x, y) = 1, then∫

C f (x, y) ds should equal ...

Definition. The line integral of f over C with respect to arc length is given by:∫C

f (x, y) ds =

236


Note. It is also possible to define∫

C f (x, y) ds as the limit of a Riemann sum, which isequivalent.

Note. The line integral of f over C with respect to arc length does not depend on theparametrization of C as long as the parametrization goes from t = a to t = b with a < b.

237


Example. (# 2) Calculate∫

Cxy ds, for C : x = t2, y = 2t, 0 ≤ t ≤ 1.

238


Some other kinds of line integrals:

Definition. The line integral of f with respect to x is∫C

f (x, y) dx =

The line integral of f with respect to y is∫C

f (x, y) dy =

The line integral∫C

f (x, y) dx + g(x, y) dy =

Note. The line integral with respect to x and the line integral with respect to y areindependent of parametrization provided that the parametrizations traverse C in thesame direction.

239


Example. (#8) Find∫

Cx2 dx + y2 dy where C consists of the arc of the circle x2 + y2 = 4

from (2, 0) to (0, 2) followed by the line segment from (0, 2) to (4, 3).

240


Finally, the last kind of line integrals: the line integral of a vector field over a curve.

We want to define∫

C

~F dsomething , and get a scalar answer.

Definition. For a vector field ~F(x, y) over a curve C parametrized by~r(t) =< x(t), y(t) >,define the line integral of ~F over C by:∫

C

~F ◦ d~r =

Equivalently, this can be written in terms of the components of ~r(t).

Equivalently, this can be written in terms of a line integral of a function with respectto arc length (hint: first write it in terms of the unit tangent vector):

241


Equivalently, this can be written in terms of a line integrals of a function with respectto x and with respect to y.

First, write ~F in components as ~F(x, y) =< P(x, y),Q(x, y) >. Then:

242


Motivation for this definition comes from the physics concept of work.

Work = force · distance

Find an expression for the work done by a vector field ~F(x, y) as it pushes a particlealong a curve C.

243


Example. (# 39) Find the work done by the force field

~F(x, y) = x~i + (y + 2)~j

in moving an object along an arch of the cycloid

~r(t) = (t − sin t)~i + (1 − cos t)~j

for 0 ≤ t ≤ 2π.

244


There are analogous definitions for functions and vector fields of 3-variables:

For a function f (x, y, z), and a curve C in R3 parametrized by ~r(t) =< x(t), y(t), z(t) >,

Definition. ∫C

f (x, y, z) ds =

The line integral of f (x, y, z) with respect to x is∫C

f (x, y, z) dx =

The line integral of f (x, y, z) with respect to y is∫C

f (x, y, z) dy =

The line integral of f (x, y, z) with respect to z is∫C

f (x, y, z) dz =

The line integral∫C

f (x, y, z) dx + g(x, y, z) dy + h(x, y, z) dz =

245


For a vector function ~F(x, y, z) in R3∫C

~F(x, y, z) ◦ d~r =

246


Note. All of these line integrals are independent of the parametrization of C up to apositive or negative sign. There are some subtleties as far as which parametrizationsgive the exact same value and which parametrizations give opposite values.

Example. Let f (x, y) = 3. For the following five curves, sketch the curve and calculate∫C

f (x, y) ds

1. C1 : ~r1(t) =< cos t, sin t >, 0 ≤ t ≤ π2

2. C2 : ~r2(t) =< cos t, sin t >, t runs from π2 to 0.

3. C3 : ~r3(t) =< sin t, cos t >, 0 ≤ t ≤ π2

4. C4 : ~r4(t) =< sin t, cos t >, t runs from π2 to 0.

5. C5 : ~r5(t) =< cos 2t, sin 2t >, 0 ≤ t ≤ π4

Conclusion:∫

Cf (x, y) ds does not depend on parametrization, as long as ...

247


Example. Let f (x, y) = 3. For the following five curves, sketch the curve and calculate∫C

f (x, y) dx

1. C1 : ~r1(t) =< cos t, sin t >, 0 ≤ t ≤ π2

2. C2 : ~r2(t) =< cos t, sin t >, t runs from π2 to 0.

3. C3 : ~r3(t) =< sin t, cos t >, 0 ≤ t ≤ π2

4. C4 : ~r4(t) =< sin t, cos t >, t runs from π2 to 0.

5. C5 : ~r5(t) =< cos 2t, sin 2t >, 0 ≤ t ≤ π4

Conclusion:∫

Cf (x, y) dx does not depend on parametrization, as long as ...

248


Question. What can we say about∫

C

~F(x, y) ◦ d~r and the effect of parametrization?

249

§16.3 FUNDAMENTAL THEOREM FOR LINE INTEGRALS

§16.3 Fundamental Theorem for Line Integrals

We will state and prove results for vector fields in R2, but they also hold for vectorfields in R3.

We will only work with piecewise smooth curves.

Recall: in Calc 1, the Fundamental Theorem of Calculus (FTC) says:

∫ b

aF′(x) dx =

For line integrals, we have a similar theorem where plays the role ofF′.

250


Theorem. (Fundamental Theorem for Line Integrals - FTLI) Let f (x, y) be a differen-tiable function whose gradient is continuous. Let C be a smooth curve parametrizedby ~r(t) for a ≤ t ≤ b. Then∫

C∇ f ◦ d~r =

Proof. Use the chain rule in reverse:

251


Example. (#1) The figure shows a curve C and a contour map of a function

f (x, y) = y sin(x + y) + x

whose gradient ∇ f is continuous. Find∫

C∇ f ◦ d~r

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

252


Example. Find the work done by gravity to move an object of mass 100 kg from thepoint (3.1, 3, 5) (in millions of meters) to the point (3, 3, 5).

Note: the force of gravity is given by

~F(x, y, z) = −mMGx

(x2 + y2 + z2)3/2~i −

mMGy(x2 + y2 + z2)3/2

~j −mMGz

(x2 + y2 + z2)3/2~k

Recall: ~F = ∇ f , where f = M = 5.97219 × 1024kg, G = 6.67384 × 10−11 m3

kg s2

253


Note. 1. The Fundamental Theorem of Line Integrals (FTLI) only applies to vectorfields that are gradients, i.e. to conservative vector fields.

2. The FTLI says that we can evaluate the line integral of a conservative vector field∇ f if we only know the value of f at .

3. In the language of physics, the FTLI says that the work done in moving an objectfrom one point to another by a conservative force (like gravity) does not dependon , only on .

4. If C1 and C2 are two paths connecting points A and B, then∫C1

∇ f ◦ d~r −∫

C2

∇ f ◦ d~r =

254


Definition. For continuous vector field ~F(x, y), we say that∫

C

~F(x, y) ◦ d~r is indepen-

dent of path if

for any two curves C1 and C2 with the same initial and terminal end points.

Question. Are there any vector fields that have line integrals that are NOT independentof path?

True or False: If a vector field is conservative, then it has line integrals that areindependent of path.

True or False: If a vector field has line integrals that are independent of path, then it isconservative.

255


Need a few definitions to make this and subsequent facts precise.

Definition. A curve C is closed if the initial point and terminal point of C coincide.

Definition. A curve C is simple if C does not intersect itself, except possibly at itsendpoints.

256


Definition. (Informal definition) A region D is open if it doesn’t contain any of itsboundary points.

257


Definition. A region D is connected if any two points in D can be joined by a path inD.

Definition. A region D is simply connected if it is connected and for every simpleclosed curve C in D, C encloses only points of D. Infomally, simply connected meansconnected with no holes.

258


Classify each curve: is it simple? closed?

Classify each region: is it open? connected? simply connected?

259


Theorem. Suppose ~F is a continuous vector field on an open, connected region D. Then∫C

~F ◦ d~r is independent of path ⇐⇒

260


Another characterization of independence of path:

Theorem. Suppose ~F is a continuous vector field on a region D, then∫

C

~F ◦ d~r is

independent of path ⇐⇒

for every closed path C in D.

Proof:

261


Theorem. For a continuous vector field ~F on an open, connected region D, the followingare equivalent:

1.

2.

3.

4.

Note. So far all of this is true for vector fields in R3 as well.

262


Goal: find a more practical condition to decide when a vector field is the gradient of afunction and when it could not be the gradient of a function.

Theorem. If ~F(x, y) = P(x, y)~i + Q(x, y)~j is ∇ f for some function f , where P(x, y) andQ(x, y) have continuous first partial derivatives, then

263


Theorem. If ∂P∂y ,

∂Q∂x on a region D, then

Question. Does the converse hold? Is it true that if ∂P∂y = ∂Q

∂x on a region D, then ~F = ∇ ffor some function f ?

Theorem. If ~F(x, y) = P(x, y)~i + Q(x, y)~j is a vector field on an open, simply connectedregion D, and P and Q have continuous first partial derivatives, then

264


Example. Determine if ~F = (3 + 2xy)~i + (x2− 3y2)~j is conservative. If it is conservative,

find f such that ∇ f = ~F.

265


We don’t yet have an analogous, easy method to determine whether a vector field inR3 is conservative. (Wait till section 16.5). But if we know that ~F is conservative, thenthe method for finding f with ∇ f = ~F is the same.

Example. (#18) ~F(x, y, z) = sin y~i + (x cos y + cos z)~j − y sin z ~k is conservative. Find fsuch that ∇ f = ~F.

266


Note. If the region D is not simply connected, then Py = Qx on D does not guaranteethat ~F is conservative.

Example. Is this vector field conservative? Why or why not?

~F(x, y) =−y

x2 + y2~i +

xx2 + y2

~j

267

§16.4 GREEN’S THEOREM

§16.4 Green’s Theorem

Green’s Theorem relates a line integral over simple closed curve C to a double integralover the region that C encloses.

Definition. A simple closed curve C that forms the boundary of a region D is calledpositively oriented if, as you traverse C, the region D is always to the left.

A collection of curves C1 ∪ C2 ∪ · · ·Cn that together form the boundary of a region Dare positively oriented if each curve is positively oriented.

268


Theorem. Green’s Theorem Let ~F(x, y) = P(x, y)~i + Q(x, y)~j be a vector field and suppose thatP and Q have continuous first partial derivatives. Let ∂D be a positively oriented, piecewisesmooth curve or collection of curves that bound a region D. Then∫

∂D=

∫ ∫D

Note. Alternative notations for∫∂D P dx + Q dy are:

∫C P dx + Q dy OR

∮C P dx + Q dy ,

where C = ∂D.

269


Note. .

1. Green’s theorem works for any vector field ~F (unlike the Fundamental Theorem ofLine Integrals which only applies to conservative vector fields).

2. Green’s theorem is analogous to the fundamental theorem of calculus: both relatethe integral of a derivative of a function to a function on the boundary of the region.

3. Green’s theorem applies only to 2-dimensional vector fields (but there is a moregeneral theory that applies in n dimensions).

4. Green’s theorem can be used to prove that when Py = Qx on an open, simplyconnected region, then ~F = P~i + Q~j is conservative.

270


Example. (# 9) Use Green’s Theorem to evaluate the line integral∫

Cy3 dx − x3 dy

along the positively oriented circle x2 + y2 = 4.

271


Green’s theorem can be used to to convert an integral over a region to a lineintegral. This is useful when computing area:

Since∫ ∫

D1 dA = area(D), to use Green’s Theorem, we need to find P and Q such

that

∂Q∂x−∂P∂y

= 1

What can we use for P and Q?

Note. area(D) =∫∂D

272


Extra Example. (# 21)

(a) If C is the line segment connecting the point (x1, y1) to (x2, y2), show that∫

C x dy−y dx = x1y2 − x2y1.

(b) Find the area of a pentagon with vertices (0, 0), (2, 1), (1, 3). (0, 2), (−1, 1).

273


Extra Example. (Example 5 from book) If ~F(x, y) =<−y

x2 + y2 ,x

x2 + y2 >, show that∫C~F◦d~r = 2π for every positively oriented simple closed curve that encloses the origin.

274


PROOFS

Theorem. Green’s Theorem Let ~F(x, y) = P(x, y)~i + Q(x, y)~j be a vector field and suppose thatP and Q have continuous first partial derivatives. Let ∂D be a positively oriented, piecewisesmooth curve or collection of curves that bound a region D. Then∫

∂DP dx + Q dy =

∫ ∫D

(∂Q∂x−∂P∂y

)dA

Proof. .

Step 1: Prove that if D is a Type I region, then∫∂D

P dx =

∫ ∫D

(−∂P∂y

)dA.

Step 2: Prove that if D is a Type II region, then∫∂D

Q dy =

∫ ∫D

(∂Q∂y

)dA.

Step 3: Prove Green’s Theorem for regions that are both Type I and Type II .

Step 4: Prove Green’s Theorem in general.

275

§16.5 DIVERGENCE AND CURL

§16.5 Divergence and Curl

Note. Divergence (div) and curl apply to vectors in 3-dimensions.

Definition. (Informal definition) If ~F represents the velocity of a fluid, then div ~F(x, y, z)represents the net flow from the point (x, y, z).

~F =< −x,−y, 0 > ~F =< −2y, 2x, 0 > ~F =<y3 + 1, 0, 0 > ~F =< x + 2, y + 2, 0 >

Example. (#9 - 11) The vector field ~F = P~i + Q~j + R~k is shown in the xy plane andlooks the same in all other horizontal planes. Is div ~F positive, negative, or zero at theorigin?

276


Definition. For ~F = P~i + Q~j + R~k, the divergence div ~F is defined as:

Note that div ~F is a (circle one) scalar / vector.

277


Definition. (Informal definition) If ~F represents the velocity of a fluid, then curl ~F(x, y, z)represents the net rotation of the fluid around the point (x, y, z). The direction of thecurl is the axis of rotation, as determined by the right-hand rule, and the magnitudeof the curl is the magnitude of rotation.

~F(x, y, z) =< −x,−y, 0 > ~F(x, y, z) =< −2y, 2x, 0 > ~F(x, y, z) =<y3 + 1, 0, 0 > ~F(x, y, z) =< x + 2, y + 2, 0 >

Example. (#9 - 11) The vector field ~F = P~i + Q~j + R~k is shown in the xy plane and looksthe same in all other horizontal planes. Determine whether curl ~F = ~0 at the origin. Ifnot, in which direction does curl ~F point?

278


Definition. For ~F = P~i + Q~j + R~k, the curl ~F is defined as:

Note that curl ~F is a (circle one) scalar / vector.

279


Example. Compute the divergence and curl of the previous 4 examples:

1. ~F(x, y, z) =< −x,−y, 0 >

2. ~F(x, y, z) =< −2y, 2x, 0 >

3. ~F(x, y, z) =<y3 + 1, 0, 0 >

4. ~F(x, y, z) =< x + 2, y + 2, 0 >

280


Example. (# 4) For the vector field ~F(x, y, z) = sin yz~i + sin zx ~j + sin xy ~k.

1. Find div(~F)

2. Find curl(~F)

3. Find div(curl(~F))

281


Theorem. If ~F = P~i + Q~j + R~k is a vector field on R3 and P, Q, and R have continuoussecond-order partial derivatives, then

div curl ~F =

Proof:

282


Theorem. For any function f onR3 with continuous second-order derivatives, curl(∇ f ) =

.

Proof:

Question. Is ~F(x, y, z) = sin yz~i + sin zx ~j + sin xy ~k conservative?

283


We have that if ~F is conservative, then curl ~F = ~0. What about the converse?

Theorem. If ~F is a vector field defined on a simply connected open region of R3.Assume all the component functions of ~F have continuous partial derivatives andsuppose that curl ~F = ~0. Then ...

Proof is a consequence of Stoke’s Thm in Section 16.8.

284


Review. In 3-dimensions, we have

~F = ∇ f ⇐⇒ curl ~F = ~0

In 2-dimensions, we saw~F = ∇ f ⇐⇒

∂Q∂y−∂P∂x

= 0

The 2-d fact is a special case of the 3-d fact:

285


Vector forms of Green’s Theorem

Green’s Theorem (which relates to 2-dim vector fields) can also be stated in terms ofcurl and divergence for associated 3-dim vector fields.

286

§16.6 PARAMETRIC SURFACES

§16.6 Parametric Surfaces

Review. A parametric curve ~r(t) =< x(t), y(t), z(t) > gives the x, y, and z coordinates ofpoints on the curve in terms of a third variable t.

Definition. A parametric surface ~r(u, v) =< x(u, v), y(u, v), z(u, v) > gives the x, y, and zcoordinates of points on the surface in terms of two parameters u and v.

Example. Graph the parametric surface

~r(u, v) =< cos(u), sin(u) + cos(v), sin(v) >

for 0 ≤ u ≤ 2π, 0 ≤ v ≤ 2π.

287


Example. (# 19) Parametrize the plane through the point (0,−1, 5) that contains thevectors < 2, 1, 4 > and < −3, 2, 5 >.

288


Example. (#22) Parametrize the part of the ellipsoid x2 + 2y2 + 3z2 = 1 that lies in thehalf-space where y ≤ 0.

289


Tangent Planes:

Find the tangent plane to the surface ~r(u, v) =< cos(u), sin(u) + cos(v), sin(v) > at thepoint (1, 1, 0)

290


Goal: Find the surface area of a parametric surface ~r(u, v) =< x(u, v), y(u, v), z(u, v) >

291


Example. (# 48) Find the surface area of the helicoid (spiral ramp) x = u cos(v), y =

u sin(v), z = v. for 0 < u < 5, 0 < v < 4π.

292


Note. There are alternative formulas for surface area in special cases.

1. If the surface is of the form z = f (x, y), then surface area is given by:

2. If the surface is a surface of revolution, formed by rotating the curve y = g(x)around the x-axis for a ≤ x ≤ b, then surface area is given by:

Both of these formulas can be derived from the general formula for the area of aparametric surface.

293

§16.7 SURFACE INTEGRALS

§16.7 Surface Integrals

Preliminaries:

Definition. An orientation of a surface S inR3 is a choice of unit normal vector ~n(x, y, z)at every point (x, y, z) ∈ S, in such a way that ~n varies continuously.

Example. A hemisphere has two orientations:

Definition. A surface is orientable if it is possible to find an orientation for it.

Question. Are all surfaces orientable?

294


Note. A surface in R3 is orientable if and only if it has “two sides”. Any orientablesurface has two orientations.

Definition. For a closed surface in R3 (finite with no boundary), the positive orientationis the orientation for which the normal vectors point . The negative orientationis the orientation for which the normal vectors point .

295


A surface integral is analogous to a line integral.

Review. The line integral of f (x, y) with respect to arclength over the curve C is definedas: ∫

Cf (x, y, z)ds =

and represents:

296


Definition. The surface integral of f (x, y, z) over the parametrized surface S is definedas: ∫ ∫

Sf (x, y, z)dS =

and represents:

If the surface is described as z = g(x, y) for some function g, then∫ ∫S

f (x, y, z)dS =

297


Review. The line integral of a vector field ~F(x, y, z) is defined as∫C

~F ◦ d~r =

and represents:

298


Definition. The surface integral of a function ~F(x, y, z) over a parametrized surface Sis: ∫ ∫

S

~F(x, y, z) ◦ d~S =

This represents:

Note. The surface integral of ~F(x, y, z) over the surface S can be calculated as:∫ ∫S

~F(x, y, z) ◦ d~S =

299


Why does∫ ∫

S

~F ◦ ~n dS represent flux (rate of fluid flow)?

300


If the surface is described as z = g(x, y) and ~F = P~i + Q~j + R~k for (x, y) ∈ D, then∫ ∫S

~F(x, y, z) ◦ d~S =

∫ ∫D−P∂g∂x−Q

∂g∂y

+ R dA

Proof:

301


Question. Are surface integrals well-defined?

If we use two different parametrizations of a surface, will we still get the same answerfor the surface integral?

302


Example. (# 46) Gauss’s Law says that the net charge enclosed by a closed surface S is

q = ε0

∫ ∫S

~E ◦ d~S

where ε0 is a constant.

Use Gauss’s Law to find the charge enclosed by the cube with vertices (±1,±1,±1) ifthe electric field is ~E(x, y, z) = x~i + y~j + z~k.

303


Extra Example. (# 47) The temperature at the point (x, y, z) in a substance with conduc-tivity K = 6.5 is u(x, y, z) = 2y2 + 2z2. The heat flow is defined as

~F = −K∇u

Find the rate of heat flow inward across the cylindrical surface y2 + z2 = 6, 0 ≤ x ≤ 4.

304


Extra Example. (#44) Seawater has density 1025 kg/m3 and flows in a velocity field~v = y~i + x~j, where x, y, and z are in meters and the components of ~v are in m/sec. Findthe rate of flow outward through the hemisphere x2 + y2 + z2 = 9, z ≥ 0.

305


Extra Example. (# 7) Set up the equation to find∫ ∫

S y dS, where S is the helicoid withvector equation ~r(u, v) =< u cos(v),u sin(v), v > with 0 ≤ u ≤ 1, 0 ≤ v ≤ π.

306


Extra Example. (#22) Set up the equation to find∫ ∫

S~F◦ ~dS, where ~F(x, y, z) = z~i+ y~j+x~k

and S is the helicoid again: ~r(u, v) =< u cos(v),u sin(v), v > with 0 ≤ u ≤ 1, 0 ≤ v ≤ π.

307

§16.8 STOKES THEOREM

§16.8 Stokes Theorem

Preliminaries: If S is an oriented surface with boundary ∂S, a positive orientation on∂S means that if you walk around ∂S with your head in the direction of the normalvector for S, the surface S lies on your left.

Review. Green’s Theorem says that for ~F =< P,Q > (assuming P,Q have continuouspartials and ∂D is piecewise smooth)∫

∂DP dx + Q dy =

∫ ∫D

∂Q∂x−∂P∂y

dA

.

This can also be written as ∫∂D

~F ◦ d~r =

∫ ∫D

curl(~F) ◦~k dA

.

308


Theorem. (Stoke’s Theorem) Let S be an oriented piecewise smooth surface boundedby a simple, closed, piecewise smooth boundary curve ∂S with positive orientation.Let~(F) be a vector field whose components have continuous partial derivatives in anopen region of R3 around S. Then∫

∂S

~F ◦ d~r =

∫ ∫S

curl(~F) ◦ ~n dS

This can also be written as ∫∂S

~F ◦ d~r =

∫ ∫S

curl(~F) ◦ d~S

In words this means:

309


Intuition:

310


Example. Find∫

C~F ◦ d~r where ~F(x, y, z) = −y2~i + x~j + z2~k where C is the intersection

curve of the plane y + z = 2 and x2 + y2 = 1.

311


Consequences of Stoke’s Theorem:

1. Stokes Theorem implies: If curl(~F) = ~0 on a simply connected regoin, then ~F isconservative.

2. Stokes Theorem implies:∫ ∫

S1curl(~F) ◦ d~S =

∫ ∫S2

curl(~F) d~S if S1 and S2 are twosurfaces with the same boundary.

3. Stokes Theorem implies: Green’s Theorem.

312


Extra Example. (# 6) Use Stokes’ Theorem to evaluate∫ ∫

S curl(~F)◦d~S, where ~F(x, y, z) =

exy~i + exz~j + x2z~k and S is the half of the ellipsoid 4x2 + y2 + 4z2 = 4 that lies to the rightof the xz-plane, oriented in the direction of the positive y axis.

313

§16.9 DIVERGENCE THEOREM

§16.9 Divergence Theorem

Review. Green’s Theorem says that for ~F =< P,Q > (assuming P,Q have continuouspartials and ∂D is piecewise smooth)∫

∂DP dx + Q dy =

∫ ∫D

∂Q∂x−∂P∂y

dA

.

This can also be written as∫∂D

~F ◦ ~n ds =

∫ ∫D

div ~F(x, y) dA

.

314


Theorem. (Divergence Theorem) Let E be a solid region and let S be the boundarysurface of D, given positive (outward) orientation. Let ~F be a vector field whose com-ponent functions have continuous partial derivatives on an open region that containsE. Then

In words this means:

315


Intuition

316


Example. Find the flux of the vector field ~F(x, y, z) = x2yz~i + xy2z~j + xyz2~k over thesurface that is the box bounded by the coordinate planes and the planes x = a, y = b,z = c.

317


Extra Example. Find the flux of the vector field ~F(x, y, z) = z~i + y~j + x~k over the unitsphere x2 + y2 + z2 = 1.

318


Extra Example. Consider the electric field ~E(x, y, z) =εQ||~x||

~x, where the electric charge

Q is located at the origin and ~x =< x, y, z > is a position vector. Use the DivergenceTheorem to show that the electric fluz of ~E through any closed surface S that enclosesthe origin is

∫ ∫S~E ◦ d~S = 4πεQ.

319

Math 233 Calculus 3 - Fall 2016 - Linda Green · Math 233 Calculus 3 - Fall 2016. x12.1 - THREE-DIMENSIONAL COORDINATE SYSTEMS x12.1 - Three-Dimensional Coordinate Systems Deﬁnition.

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