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3: Double Integrals over General Regions

3: Double Integrals over General Regions

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1142

CHAPTER 16

VECTOR CALCULUS

and, by Formula 9, the flux is

yy F ؒ dS ෇ yy F ؒ ͑r␾ ϫ r␪ ͒ dA

S

D

෇y

2␲

y

0

͑2 sin 2␾ cos ␾ cos ␪ ϩ sin 3␾ sin 2␪ ͒ d␾ d␪

0

෇ 2 y sin2␾ cos ␾ d␾

y

0

෇ 0 ϩ y sin 3␾ d␾

0

y

2␲

0

2␲

0

cos ␪ d␪ ϩ y sin3␾ d␾

0

ͩ

sin 2␪ d␪

y

2␲

sin2␪ d␪

0

ͪ

2␲

since y cos ␪ d␪ ෇ 0

0

4␲

3

by the same calculation as in Example 1.

If, for instance, the vector field in Example 4 is a velocity field describing the flow of a

fluid with density 1, then the answer, 4␲͞3, represents the rate of flow through the unit

sphere in units of mass per unit time.

In the case of a surface S given by a graph z ෇ t͑x, y͒, we can think of x and y as parameters and use Equation 3 to write

ͩ

F ؒ ͑rx ϫ ry͒ ෇ ͑P i ϩ Q j ϩ R k͒ ؒ Ϫ

ͪ

Ѩt

Ѩt

jϩk

Ѩx

Ѩy

Thus Formula 9 becomes

yy F ؒ dS ෇ yy

10

S

D

ͩ

ϪP

ͪ

Ѩt

Ѩt

ϪQ

ϩ R dA

Ѩx

Ѩy

This formula assumes the upward orientation of S; for a downward orientation we multiply by Ϫ1. Similar formulas can be worked out if S is given by y ෇ h͑x, z͒ or x ෇ k͑y, z͒.

(See Exercises 37 and 38.)

v EXAMPLE 5 Evaluate xxS F ؒ dS, where F͑x, y, z͒ ෇ y i ϩ x j ϩ z k and S is the

boundary of the solid region E enclosed by the paraboloid z ෇ 1 Ϫ x 2 Ϫ y 2 and the

plane z ෇ 0.

z

SOLUTION S consists of a parabolic top surface S1 and a circular bottom surface S2. (See

Figure 12.) Since S is a closed surface, we use the convention of positive (outward)

orientation. This means that S1 is oriented upward and we can use Equation 10 with

D being the projection of S1 onto the xy-plane, namely, the disk x 2 ϩ y 2 ഛ 1. Since

S™

y

P͑x, y, z͒ ෇ y

x

FIGURE 12

on S1 and

Q͑x, y, z͒ ෇ x

Ѩt

෇ Ϫ2x

Ѩx

R͑x, y, z͒ ෇ z ෇ 1 Ϫ x 2 Ϫ y 2

Ѩt

෇ Ϫ2y

Ѩy

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SECTION 16.7

we have

ͩ

yy F ؒ dS ෇ yy

S1

D

ϪP

SURFACE INTEGRALS

1143

ͪ

Ѩt

Ѩt

ϪQ

ϩ R dA

Ѩx

Ѩy

෇ yy ͓Ϫy͑Ϫ2x͒ Ϫ x͑Ϫ2y͒ ϩ 1 Ϫ x 2 Ϫ y 2 ͔ dA

D

෇ yy ͑1 ϩ 4xy Ϫ x 2 Ϫ y 2 ͒ dA

D

෇y

2␲

y

0

෇y

2␲

y

0

෇y

1

0

2␲

0

1

0

͑1 ϩ 4r 2 cos ␪ sin ␪ Ϫ r 2 ͒ r dr d␪

͑r Ϫ r 3 ϩ 4r 3 cos ␪ sin ␪ ͒ dr d␪

( 14 ϩ cos ␪ sin ␪) d␪ ෇ 14 ͑2␲͒ ϩ 0 ෇

2

The disk S2 is oriented downward, so its unit normal vector is n ෇ Ϫk and we have

yy F ؒ dS ෇ yy F ؒ ͑Ϫk͒ dS ෇ yy ͑Ϫz͒ dA ෇ yy 0 dA ෇ 0

S2

S2

D

D

since z ෇ 0 on S2 . Finally, we compute, by definition, xxS F ؒ dS as the sum of the surface integrals of F over the pieces S1 and S2 :

yy F ؒ dS ෇ yy F ؒ dS ϩ yy F ؒ dS ෇

S

S1

S2

ϩ0෇

2

2

Although we motivated the surface integral of a vector field using the example of fluid

flow, this concept also arises in other physical situations. For instance, if E is an electric

field (see Example 5 in Section 16.1), then the surface integral

yy E ؒ dS

S

is called the electric flux of E through the surface S. One of the important laws of electrostatics is Gauss’s Law, which says that the net charge enclosed by a closed surface S is

11

Q ෇ ␧0 yy E ؒ dS

S

where ␧0 is a constant (called the permittivity of free space) that depends on the units used.

(In the SI system, ␧0 Ϸ 8.8542 ϫ 10 Ϫ12 C 2͞Nؒm2.) Therefore, if the vector field F in

Example 4 represents an electric field, we can conclude that the charge enclosed by S is

Q ෇ 43 ␲ ␧0.

Another application of surface integrals occurs in the study of heat flow. Suppose the

temperature at a point ͑x, y, z͒ in a body is u͑x, y, z͒. Then the heat flow is defined as the

vector field

F ෇ ϪK ∇u

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1144

CHAPTER 16

VECTOR CALCULUS

where K is an experimentally determined constant called the conductivity of the substance. The rate of heat flow across the surface S in the body is then given by the surface

integral

yy F ؒ dS ෇ ϪK yy ∇u ؒ dS

S

S

v EXAMPLE 6 The temperature u in a metal ball is proportional to the square of the

distance from the center of the ball. Find the rate of heat flow across a sphere S of

radius a with center at the center of the ball.

SOLUTION Taking the center of the ball to be at the origin, we have

u͑x, y, z͒ ෇ C͑x 2 ϩ y 2 ϩ z 2 ͒

where C is the proportionality constant. Then the heat flow is

F͑x, y, z͒ ෇ ϪK ٌu ෇ ϪKC͑2x i ϩ 2y j ϩ 2z k͒

where K is the conductivity of the metal. Instead of using the usual parametrization of

the sphere as in Example 4, we observe that the outward unit normal to the sphere

x 2 ϩ y 2 ϩ z 2 ෇ a 2 at the point ͑x, y, z͒ is

n෇

1

͑x i ϩ y j ϩ z k͒

a

Fؒn෇Ϫ

and so

2KC 2

͑x ϩ y 2 ϩ z 2 ͒

a

But on S we have x 2 ϩ y 2 ϩ z 2 ෇ a 2, so F ؒ n ෇ Ϫ2aKC. Therefore the rate of heat

flow across S is

yy F ؒ dS ෇ yy F ؒ n dS ෇ Ϫ2aKC yy dS

S

S

S

෇ Ϫ2aKCA͑S͒ ෇ Ϫ2aKC͑4␲ a 2 ͒ ෇ Ϫ8KC␲ a 3

16.7

Exercises

1. Let S be the boundary surface of the box enclosed by the

planes x ෇ 0, x ෇ 2, y ෇ 0, y ෇ 4, z ෇ 0, and z ෇ 6. Approximate xxS eϪ0.1͑xϩyϩz͒ dS by using a Riemann sum as in Definition 1, taking the patches Sij to be the rectangles that are the

faces of the box S and the points Pij* to be the centers of the

rectangles.

2. A surface S consists of the cylinder x 2 ϩ y 2 ෇ 1, Ϫ1 ഛ z ഛ 1,

3. Let H be the hemisphere x 2 ϩ y 2 ϩ z 2 ෇ 50, z ജ 0, and

suppose f is a continuous function with f ͑3, 4, 5͒ ෇ 7,

f ͑3, Ϫ4, 5͒ ෇ 8, f ͑Ϫ3, 4, 5͒ ෇ 9, and f ͑Ϫ3, Ϫ4, 5͒ ෇ 12.

By dividing H into four patches, estimate the value of

xxH f ͑x, y, z͒ dS.

4. Suppose that f ͑x, y, z͒ ෇ t (sx 2 ϩ y 2 ϩ z 2 ), where t is a

function of one variable such that t͑2͒ ෇ Ϫ5. Evaluate

xxS f ͑x, y, z͒ dS, where S is the sphere x 2 ϩ y 2 ϩ z 2 ෇ 4.

together with its top and bottom disks. Suppose you know that

f is a continuous function with

f ͑Ϯ1, 0, 0͒ ෇ 2

f ͑0, Ϯ1, 0͒ ෇ 3

f ͑0, 0, Ϯ1͒ ෇ 4

Estimate the value of xxS f ͑x, y, z͒ dS by using a Riemann sum,

taking the patches Sij to be four quarter-cylinders and the top

and bottom disks.

CAS Computer algebra system required

5–20 Evaluate the surface integral.

5.

xxS ͑x ϩ y ϩ z͒ dS,

S is the parallelogram with parametric equations x ෇ u ϩ v,

y ෇ u Ϫ v, z ෇ 1 ϩ 2u ϩ v, 0 ഛ u ഛ 2, 0 ഛ v ഛ 1

1. Homework Hints available at stewartcalculus.com

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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SECTION 16.7

6.

S is the helicoid of Exercise 7 with upward orientation

23. F͑x, y, z͒ ෇ x y i ϩ yz j ϩ zx k,

S is the part of the

paraboloid z ෇ 4 Ϫ x 2 Ϫ y 2 that lies above the square

0 ഛ x ഛ 1, 0 ഛ y ഛ 1, and has upward orientation

xxS y dS,

S is the helicoid with vector equation

r͑u, v͒ ෇ ͗u cos v, u sin v, v ͘ , 0 ഛ u ഛ 1, 0 ഛ v ഛ ␲

8.

xxS ͑x 2 ϩ y 2 ͒ dS,

S is the surface with vector equation

r͑u, v͒ ෇ ͗2uv, u 2 Ϫ v 2, u 2 ϩ v 2 ͘ , u 2 ϩ v 2 ഛ 1

24. F͑x, y, z͒ ෇ Ϫx i Ϫ y j ϩ z 3 k,

xxS x 2 yz dS,

25. F͑x, y, z͒ ෇ x i Ϫ z j ϩ y k,

xxS xz dS,

26. F͑x, y, z͒ ෇ xz i ϩ x j ϩ y k,

xxS x dS,

27. F͑x, y, z͒ ෇ y j Ϫ z k,

S is the part of the cone z ෇ sx 2 ϩ y 2 between the planes

z ෇ 1 and z ෇ 3 with downward orientation

S is the part of the plane z ෇ 1 ϩ 2x ϩ 3y that lies above the

rectangle ͓0, 3͔ ϫ ͓0, 2͔

10.

S is the part of the sphere x 2 ϩ y 2 ϩ z 2 ෇ 4 in the first octant,

with orientation toward the origin

S is the part of the plane 2x ϩ 2y ϩ z ෇ 4 that lies in the first

octant

11.

S is the hemisphere x 2 ϩ y 2 ϩ z 2 ෇ 25, y ജ 0, oriented in the

direction of the positive y-axis

S is the triangular region with vertices ͑1, 0, 0͒, ͑0, Ϫ2, 0͒,

and ͑0, 0, 4͒

12.

28. F͑x, y, z͒ ෇ xy i ϩ 4x 2 j ϩ yz k,

S is the surface z ෇ ͑x

14.

S consists of the paraboloid y ෇ x 2 ϩ z 2, 0 ഛ y ഛ 1,

and the disk x 2 ϩ z 2 ഛ 1, y ෇ 1

xxS y dS,

2

3

13.

xxS x

3͞2

ϩy

3͞2

S is the surface z ෇ xe y,

0 ഛ x ഛ 1, 0 ഛ y ഛ 1, with upward orientation

͒ , 0 ഛ x ഛ 1, 0 ഛ y ഛ 1

29. F͑x, y, z͒ ෇ x i ϩ 2y j ϩ 3z k,

2 2

z dS,

S is the part of the cone z 2 ෇ x 2 ϩ y 2 that lies between the

planes z ෇ 1 and z ෇ 3

S is the cube with vertices ͑Ϯ1, Ϯ1, Ϯ1͒

30. F͑x, y, z͒ ෇ x i ϩ y j ϩ 5 k,

S is the boundary of the region

enclosed by the cylinder x 2 ϩ z 2 ෇ 1 and the planes y ෇ 0

and x ϩ y ෇ 2

xxS z dS,

S is the surface x ෇ y ϩ 2z 2, 0 ഛ y ഛ 1, 0 ഛ z ഛ 1

15.

31. F͑x, y, z͒ ෇ x 2 i ϩ y 2 j ϩ z 2 k,

xxS y dS,

S is the boundary of the solid

half-cylinder 0 ഛ z ഛ s1 Ϫ y 2 , 0 ഛ x ഛ 2

S is the part of the paraboloid y ෇ x ϩ z that lies inside the

cylinder x 2 ϩ z 2 ෇ 4

xxS y

2

2

32. F͑x, y, z͒ ෇ y i ϩ ͑z Ϫ y͒ j ϩ x k,

S is the surface of the tetrahedron with vertices ͑0, 0, 0͒,

͑1, 0, 0͒, ͑0, 1, 0͒, and ͑0, 0, 1͒

2

16.

dS,

S is the part of the sphere x 2 ϩ y 2 ϩ z 2 ෇ 4 that lies

inside the cylinder x 2 ϩ y 2 ෇ 1 and above the xy-plane

17.

xxS ͑x 2 z ϩ y 2 z͒ dS,

CAS

CAS

xxS xz dS,

S is the boundary of the region enclosed by the cylinder

y 2 ϩ z 2 ෇ 9 and the planes x ෇ 0 and x ϩ y ෇ 5

19.

20.

xxS ͑x

ϩ y ϩ z ͒ dS,

S is the part of the cylinder x 2 ϩ y 2 ෇ 9 between the planes

z ෇ 0 and z ෇ 2, together with its top and bottom disks

2

2

2

21–32 Evaluate the surface integral xxS F ؒ dS for the given vector

field F and the oriented surface S. In other words, find the flux of F

across S. For closed surfaces, use the positive (outward) orientation.

21. F͑x, y, z͒ ෇ ze xy i Ϫ 3ze xy j ϩ xy k,

S is the parallelogram of Exercise 5 with upward orientation

34. Find the exact value of xxS x 2 yz dS, where S is the surface

z ෇ xy, 0 ഛ x ഛ 1, 0 ഛ y ഛ 1.

CAS

35. Find the value of xxS x 2 y 2z 2 dS correct to four decimal places,

where S is the part of the paraboloid z ෇ 3 Ϫ 2x 2 Ϫ y 2 that

lies above the x y-plane.

xxS ͑z ϩ x 2 y͒ dS,

S is the part of the cylinder y 2 ϩ z 2 ෇ 1 that lies between the

planes x ෇ 0 and x ෇ 3 in the first octant

33. Evaluate xxS ͑x 2 ϩ y 2 ϩ z 2 ͒ dS correct to four decimal places,

where S is the surface z ෇ xe y, 0 ഛ x ഛ 1, 0 ഛ y ഛ 1.

S is the hemisphere x 2 ϩ y 2 ϩ z 2 ෇ 4, z ജ 0

18.

1145

22. F͑x, y, z͒ ෇ z i ϩ y j ϩ x k,

xxS x yz dS,

S is the cone with parametric equations x ෇ u cos v,

y ෇ u sin v, z ෇ u, 0 ഛ u ഛ 1, 0 ഛ v ഛ ␲͞2

7.

9.

SURFACE INTEGRALS

CAS

36. Find the flux of

F͑x, y, z͒ ෇ sin͑x yz͒ i ϩ x 2 y j ϩ z 2e x͞5 k

across the part of the cylinder 4y 2 ϩ z 2 ෇ 4 that lies above

the xy-plane and between the planes x ෇ Ϫ2 and x ෇ 2 with

upward orientation. Illustrate by using a computer algebra system to draw the cylinder and the vector field on the same

screen.

37. Find a formula for xxS F ؒ dS similar to Formula 10 for the case

where S is given by y ෇ h͑x, z͒ and n is the unit normal that

points toward the left.

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1146

VECTOR CALCULUS

CHAPTER 16

38. Find a formula for xxS F ؒ dS similar to Formula 10 for the case

44. Seawater has density 1025 kg͞m3 and flows in a velocity field

39. Find the center of mass of the hemisphere x 2 ϩ y 2 ϩ z 2 ෇ a 2,

45. Use Gauss’s Law to find the charge contained in the solid

where S is given by x ෇ k͑ y, z͒ and n is the unit normal that

points forward (that is, toward the viewer when the axes are

drawn in the usual way).

v ෇ y i ϩ x j, where x, y, and z are measured in meters and the

components of v in meters per second. Find the rate of flow

outward through the hemisphere x 2 ϩ y 2 ϩ z 2 ෇ 9 , z ജ 0 .

z ജ 0, if it has constant density.

hemisphere x 2 ϩ y 2 ϩ z 2 ഛ a 2, z ജ 0, if the electric field is

E͑x, y, z͒ ෇ x i ϩ y j ϩ 2z k

40. Find the mass of a thin funnel in the shape of a cone

z ෇ sx 2 ϩ y 2 , 1 ഛ z ഛ 4, if its density function is

␳ ͑x, y, z͒ ෇ 10 Ϫ z.

46. Use Gauss’s Law to find the charge enclosed by the cube

with vertices ͑Ϯ1, Ϯ1, Ϯ1͒ if the electric field is

41. (a) Give an integral expression for the moment of inertia I z

E͑x, y, z͒ ෇ x i ϩ y j ϩ z k

about the z-axis of a thin sheet in the shape of a surface S if

the density function is ␳.

(b) Find the moment of inertia about the z-axis of the funnel in

Exercise 40.

47. The temperature at the point ͑x, y, z͒ in a substance with con-

ductivity K ෇ 6.5 is u͑x, y, z͒ ෇ 2y 2 ϩ 2z 2. Find the rate of

heat flow inward across the cylindrical surface y 2 ϩ z 2 ෇ 6,

0 ഛ x ഛ 4.

42. Let S be the part of the sphere x 2 ϩ y 2 ϩ z 2 ෇ 25 that lies

above the plane z ෇ 4. If S has constant density k, find

(a) the center of mass and (b) the moment of inertia about

the z-axis.

48. The temperature at a point in a ball with conductivity K is

inversely proportional to the distance from the center of the

ball. Find the rate of heat flow across a sphere S of radius a

with center at the center of the ball.

43. A fluid has density 870 kg͞m3 and flows with velocity

Խ Խ

49. Let F be an inverse square field, that is, F͑r͒ ෇ cr͞ r

v ෇ z i ϩ y 2 j ϩ x 2 k , where x, y, and z are measured in

meters and the components of v in meters per second. Find the

rate of flow outward through the cylinder x 2 ϩ y 2 ෇ 4 ,

0 ഛ z ഛ 1.

16.8

Stokes’ Theorem

z

n

n

S

C

0

x

FIGURE 1

3

for

some constant c, where r ෇ x i ϩ y j ϩ z k. Show that the flux

of F across a sphere S with center the origin is independent of

the radius of S.

y

Stokes’ Theorem can be regarded as a higher-dimensional version of Green’s Theorem.

Whereas Green’s Theorem relates a double integral over a plane region D to a line integral

around its plane boundary curve, Stokes’ Theorem relates a surface integral over a surface

S to a line integral around the boundary curve of S (which is a space curve). Figure 1 shows

an oriented surface with unit normal vector n. The orientation of S induces the positive

orientation of the boundary curve C shown in the figure. This means that if you walk in

the positive direction around C with your head pointing in the direction of n, then the surface will always be on your left.

Stokes’ Theorem Let S be an oriented piecewise-smooth surface that is bounded

by a simple, closed, piecewise-smooth boundary curve C with positive orientation.

Let F be a vector field whose components have continuous partial derivatives on

an open region in ‫ ޒ‬3 that contains S. Then

y

C

F ؒ dr ෇ yy curl F ؒ dS

S

Since

y

C

F ؒ dr ෇ y F ؒ T ds

C

and

yy curl F ؒ dS ෇ yy curl F ؒ n dS

S

S

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_16_ch16_p1138-1147.qk_97817_16_ch16_p1138-1147 11/9/10 9:25 AM Page 1147

STOKES’ THEOREM

SECTION 16.8

George Stokes

Stokes’ Theorem is named after the Irish mathematical physicist Sir George Stokes (1819–1903).

Stokes was a professor at Cambridge University

(in fact he held the same position as Newton,

Lucasian Professor of Mathematics) and was

especially noted for his studies of fluid flow

and light. What we call Stokes’ Theorem was

actually discovered by the Scottish physicist

Sir William Thomson (1824–1907, known as

Lord Kelvin). Stokes learned of this theorem

in a letter from Thomson in 1850 and asked

students to prove it on an examination at

Cambridge University in 1854. We don’t know

if any of those students was able to do so.

1147

Stokes’ Theorem says that the line integral around the boundary curve of S of the tangential component of F is equal to the surface integral over S of the normal component of the

curl of F.

The positively oriented boundary curve of the oriented surface S is often written as

ѨS, so Stokes’ Theorem can be expressed as

yy curl F ؒ dS ෇ y

1

ѨS

S

F ؒ dr

There is an analogy among Stokes’ Theorem, Green’s Theorem, and the Fundamental

Theorem of Calculus. As before, there is an integral involving derivatives on the left side

of Equation 1 (recall that curl F is a sort of derivative of F ) and the right side involves the

values of F only on the boundary of S.

In fact, in the special case where the surface S is flat and lies in the xy-plane with

upward orientation, the unit normal is k, the surface integral becomes a double integral,

and Stokes’ Theorem becomes

y

C

F ؒ dr ෇ yy curl F ؒ dS ෇ yy ͑curl F͒ ؒ k dA

S

S

This is precisely the vector form of Green’s Theorem given in Equation 16.5.12. Thus we

see that Green’s Theorem is really a special case of Stokes’ Theorem.

Although Stokes’ Theorem is too difficult for us to prove in its full generality, we can

give a proof when S is a graph and F, S, and C are well behaved.

z

PROOF OF A SPECIAL CASE OF STOKES’ THEOREM We assume that the equation of S is

z ෇ t͑x, y͒, ͑x, y͒ ʦ D, where t has continuous second-order partial derivatives and D

is a simple plane region whose boundary curve C1 corresponds to C. If the orientation of

S is upward, then the positive orientation of C corresponds to the positive orientation of

C1. (See Figure 2.) We are also given that F ෇ P i ϩ Q j ϩ R k, where the partial derivatives of P, Q, and R are continuous.

Since S is a graph of a function, we can apply Formula 16.7.10 with F replaced by

curl F. The result is

n

z=g(x, y)

S

0

x

C

D

FIGURE 2

y

2

yy curl F ؒ dS

S

yy

D

ͫͩ

Ϫ

ѨR

ѨQ

Ϫ

Ѩy

Ѩz

ͪ ͩ

Ѩz

Ϫ

Ѩx

ѨP

ѨR

Ϫ

Ѩz

Ѩx

ͪ ͩ

Ѩz

ϩ

Ѩy

ѨQ

ѨP

Ϫ

Ѩx

Ѩy

ͪͬ

dA

where the partial derivatives of P, Q, and R are evaluated at ͑x, y, t͑x, y͒͒. If

x ෇ x͑t͒

y ෇ y͑t͒

aഛtഛb

is a parametric representation of C1, then a parametric representation of C is

x ෇ x͑t͒

y ෇ y͑t͒

z ෇ t ( x͑t͒, y͑t͒)

aഛtഛb

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97817_16_ch16_p1148-1157.qk_97817_16_ch16_p1148-1157 11/9/10 9:25 AM Page 1148

1148

CHAPTER 16

VECTOR CALCULUS

This allows us, with the aid of the Chain Rule, to evaluate the line integral as follows:

y

C

F ؒ dr ෇

ͩ

ͪ

ͩ

ͪͬ

ͪ ͩ ͪ ͬ

y ͫͩ

ͪ ͩ ͪ

ͪ ͩ ͪͬ

yy ͫ ͩ

y

P

dx

dy

dz

ϩQ

ϩR

dt

dt

dt

P

dx

dy

Ѩz dx

Ѩz dy

ϩQ

ϩR

ϩ

dt

dt

Ѩx dt

Ѩy dt

b

a

b

a

b

PϩR

a

PϩR

C1

D

Ѩ

Ѩx

Ѩz

Ѩx

Ѩz

Ѩx

dt

dx

Ѩz

ϩ QϩR

dt

Ѩy

dx ϩ Q ϩ R

QϩR

Ѩz

Ѩy

Ϫ

Ѩ

Ѩy

Ѩz

Ѩy

dy

dt

dt

dt

dy

PϩR

Ѩz

Ѩx

dA

where we have used Green’s Theorem in the last step. Then, using the Chain Rule again

and remembering that P, Q, and R are functions of x, y, and z and that z is itself a function

of x and y, we get

y

C

F ؒ dr ෇

yy

D

ͫͩ

ѨQ

ѨQ Ѩz

ѨR Ѩz

ѨR Ѩz Ѩz

Ѩ2z

ϩ

ϩ

ϩ

ϩR

Ѩx

Ѩz Ѩx

Ѩx Ѩy

Ѩz Ѩx Ѩy

Ѩx Ѩy

Ϫ

ͩ

ͪ

ѨP

ѨP Ѩz

ѨR Ѩz

ѨR Ѩz Ѩz

Ѩ2z

ϩ

ϩ

ϩ

ϩR

Ѩy

Ѩz Ѩy

Ѩy Ѩx

Ѩz Ѩy Ѩx

Ѩy Ѩx

ͪͬ

dA

Four of the terms in this double integral cancel and the remaining six terms can be

arranged to coincide with the right side of Equation 2. Therefore

y

C

F ؒ dr ෇ yy curl F ؒ dS

S

v

curve of intersection of the plane y ϩ z ෇ 2 and the cylinder x 2 ϩ y 2 ෇ 1. (Orient C to

be counterclockwise when viewed from above.)

z

S

Խ

i

Ѩ

curl F ෇

Ѩx

Ϫy 2

D 0

y

FIGURE 3

xC F ؒ dr could be

evaluated directly, it’s easier to use Stokes’ Theorem. We first compute

SOLUTION The curve C (an ellipse) is shown in Figure 3. Although

C

y+z=2

x

EXAMPLE 1 Evaluate xC F ؒ dr, where F͑x, y, z͒ ෇ Ϫy 2 i ϩ x j ϩ z 2 k and C is the

j

Ѩ

Ѩy

x

Խ

k

Ѩ

෇ ͑1 ϩ 2y͒ k

Ѩz

z2

Although there are many surfaces with boundary C, the most convenient choice is the

elliptical region S in the plane y ϩ z ෇ 2 that is bounded by C. If we orient S upward,

then C has the induced positive orientation. The projection D of S onto the xy-plane is

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_16_ch16_p1148-1157.qk_97817_16_ch16_p1148-1157 11/9/10 9:26 AM Page 1149

SECTION 16.8

STOKES’ THEOREM

1149

the disk x 2 ϩ y 2 ഛ 1 and so using Equation 16.7.10 with z ෇ t͑x, y͒ ෇ 2 Ϫ y, we have

y

C

F ؒ dr ෇ yy curl F ؒ dS ෇ yy ͑1 ϩ 2y͒ dA

S

෇y

D

2␲

0

y

2␲

0

y

1

0

ͫ

͑1 ϩ 2r sin ␪ ͒ r dr d␪

ͬ

r2

r3

ϩ2

sin ␪

2

3

1

d␪ ෇ y

2␲

0

0

( 12 ϩ 23 sin ␪) d␪

෇ 12 ͑2␲͒ ϩ 0 ෇ ␲

z

≈+¥+z@=4

S

C

v EXAMPLE 2 Use Stokes’ Theorem to compute the integral xxS curl F ؒ dS, where

F͑x, y, z͒ ෇ xz i ϩ yz j ϩ xy k and S is the part of the sphere x 2 ϩ y 2 ϩ z 2 ෇ 4 that

lies inside the cylinder x 2 ϩ y 2 ෇ 1 and above the xy-plane. (See Figure 4.)

SOLUTION To find the boundary curve C we solve the equations x 2 ϩ y 2 ϩ z 2 ෇ 4 and

x 2 ϩ y 2 ෇ 1. Subtracting, we get z 2 ෇ 3 and so z ෇ s3 (since z Ͼ 0). Thus C is the

circle given by the equations x 2 ϩ y 2 ෇ 1, z ෇ s3 . A vector equation of C is

0

y

x

FIGURE 4

0 ഛ t ഛ 2␲

r͑t͒ ෇ cos t i ϩ sin t j ϩ s3 k

≈+¥=1

rЈ͑t͒ ෇ Ϫsin t i ϩ cos t j

so

Also, we have

F͑r͑t͒͒ ෇ s3 cos t i ϩ s3 sin t j ϩ cos t sin t k

Therefore, by Stokes’ Theorem,

yy curl F ؒ dS ෇ y

C

F ؒ dr ෇ y

2␲

0

F͑r͑t͒͒ ؒ rЈ͑t͒ dt

S

෇y

2␲

0

෇ s3

(Ϫs3 cos t sin t ϩ s3 sin t cos t) dt

y

2␲

0

0 dt ෇ 0

Note that in Example 2 we computed a surface integral simply by knowing the values

of F on the boundary curve C. This means that if we have another oriented surface with

the same boundary curve C, then we get exactly the same value for the surface integral!

In general, if S1 and S2 are oriented surfaces with the same oriented boundary curve C

and both satisfy the hypotheses of Stokes’ Theorem, then

3

yy curl F ؒ dS ෇ y

C

F ؒ dr ෇ yy curl F ؒ dS

S1

S2

This fact is useful when it is difficult to integrate over one surface but easy to integrate

over the other.

We now use Stokes’ Theorem to throw some light on the meaning of the curl vector.

Suppose that C is an oriented closed curve and v represents the velocity field in fluid flow.

Consider the line integral

y

C

v ؒ dr ෇ y v ؒ T ds

C

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_16_ch16_p1148-1157.qk_97817_16_ch16_p1148-1157 11/9/10 9:26 AM Page 1150

1150

CHAPTER 16

VECTOR CALCULUS

and recall that v ؒ T is the component of v in the direction of the unit tangent vector T.

This means that the closer the direction of v is to the direction of T, the larger the value of

v ؒ T. Thus xC v ؒ dr is a measure of the tendency of the fluid to move around C and is

called the circulation of v around C. (See Figure 5.)

T

C

T

v

C

FIGURE 5

v

(a) jC v и dr>0, positive circulation

(b) jC v и dr<0, negative circulation

Now let P0͑x 0 , y0 , z0 ͒ be a point in the fluid and let Sa be a small disk with radius a and

center P0. Then (curl F͒͑P͒ Ϸ ͑curl F͒͑P0͒ for all points P on Sa because curl F is continuous. Thus, by Stokes’ Theorem, we get the following approximation to the circulation

around the boundary circle Ca :

y

Ca

v ؒ dr ෇ yy curl v ؒ dS ෇ yy curl v ؒ n dS

Sa

Sa

Ϸ yy curl v͑P0 ͒ ؒ n͑P0 ͒ dS ෇ curl v͑P0 ͒ ؒ n͑P0 ͒␲ a 2

Sa

Imagine a tiny paddle wheel placed in the

fluid at a point P, as in Figure 6; the paddle

wheel rotates fastest when its axis is parallel

to curl v.

curl v

FIGURE 6

This approximation becomes better as a l 0 and we have

curl v͑P0 ͒ ؒ n͑P0 ͒ ෇ lim

4

al0

1

␲a 2

y

Ca

v ؒ dr

Equation 4 gives the relationship between the curl and the circulation. It shows that

curl v и n is a measure of the rotating effect of the fluid about the axis n. The curling effect

is greatest about the axis parallel to curl v.

Finally, we mention that Stokes’ Theorem can be used to prove Theorem 16.5.4 (which

states that if curl F ෇ 0 on all of ‫ ޒ‬3, then F is conservative). From our previous work

(Theorems 16.3.3 and 16.3.4), we know that F is conservative if xC F ؒ dr ෇ 0 for every

closed path C. Given C, suppose we can find an orientable surface S whose boundary is

C. (This can be done, but the proof requires advanced techniques.) Then Stokes’ Theorem

gives

y

C

F ؒ dr ෇ yy curl F ؒ dS ෇ yy 0 ؒ dS ෇ 0

S

S

A curve that is not simple can be broken into a number of simple curves, and the integrals

around these simple curves are all 0. Adding these integrals, we obtain xC F ؒ dr ෇ 0 for

any closed curve C.

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_16_ch16_p1148-1157.qk_97817_16_ch16_p1148-1157 11/9/10 9:26 AM Page 1151

SECTION 16.8

16.8

STOKES’ THEOREM

1151

Exercises

10. F͑x, y, z͒ ෇ xy i ϩ 2z j ϩ 3y k,

1. A hemisphere H and a portion P of a paraboloid are shown.

C is the curve of intersection of the plane x ϩ z ෇ 5 and the cylinder x 2 ϩ y 2 ෇ 9

Suppose F is a vector field on ‫ޒ‬3 whose components have continuous partial derivatives. Explain why

11. (a) Use Stokes’ Theorem to evaluate xC F ؒ dr, where

yy curl F ؒ dS ෇ yy curl F ؒ dS

H

P

F͑x, y, z͒ ෇ x 2 z i ϩ x y 2 j ϩ z 2 k

z

z

4

4

;

P

H

;

and C is the curve of intersection of the plane

x ϩ y ϩ z ෇ 1 and the cylinder x 2 ϩ y 2 ෇ 9 oriented

counterclockwise as viewed from above.

(b) Graph both the plane and the cylinder with domains

chosen so that you can see the curve C and the surface

that you used in part (a).

(c) Find parametric equations for C and use them to graph C.

12. (a) Use Stokes’ Theorem to evaluate xC F ؒ dr, where

x

2

2

x

y

2

2

y

2–6 Use Stokes’ Theorem to evaluate xxS curl F ؒ dS.

2. F͑x, y, z͒ ෇ 2y cos z i ϩ e x sin z j ϩ xe y k,

S is the hemisphere x ϩ y ϩ z ෇ 9, z ജ 0, oriented

upward

2

2

3. F͑x, y, z͒ ෇ x z i ϩ y z j ϩ xyz k,

2 2

4. F͑x, y, z͒ ෇ tanϪ1͑x 2 yz 2 ͒ i ϩ x 2 y j ϩ x 2 z 2 k,

S is the cone x ෇ sy 2 ϩ z 2 , 0 ഛ x ഛ 2, oriented in the direction of the positive x-axis

5. F͑x, y, z͒ ෇ x yz i ϩ x y j ϩ x yz k,

2

S consists of the top and the four sides (but not the bottom)

of the cube with vertices ͑Ϯ1, Ϯ1, Ϯ1͒, oriented outward

6. F͑x, y, z͒ ෇ e

;

2 2

S is the part of the paraboloid z ෇ x 2 ϩ y 2 that lies inside the

cylinder x 2 ϩ y 2 ෇ 4, oriented upward

i ϩ e j ϩ x z k,

S is the half of the ellipsoid 4x 2 ϩ y 2 ϩ 4z 2 ෇ 4 that lies to

the right of the xz-plane, oriented in the direction of the

positive y-axis

xy

;

2

F͑x, y, z͒ ෇ x 2 y i ϩ 13 x 3 j ϩ x y k and C is the curve of

intersection of the hyperbolic paraboloid z ෇ y 2 Ϫ x 2 and

the cylinder x 2 ϩ y 2 ෇ 1 oriented counterclockwise as

viewed from above.

(b) Graph both the hyperbolic paraboloid and the cylinder

with domains chosen so that you can see the curve C and

the surface that you used in part (a).

(c) Find parametric equations for C and use them to graph C.

xz

2

7–10 Use Stokes’ Theorem to evaluate xC F ؒ dr. In each case C is

oriented counterclockwise as viewed from above.

7. F͑x, y, z͒ ෇ ͑x ϩ y 2 ͒ i ϩ ͑ y ϩ z 2 ͒ j ϩ ͑z ϩ x 2 ͒ k,

13–15 Verify that Stokes’ Theorem is true for the given vector

field F and surface S.

13. F͑x, y, z͒ ෇ Ϫy i ϩ x j Ϫ 2 k,

S is the cone z 2 ෇ x 2 ϩ y 2, 0 ഛ z ഛ 4, oriented downward

14. F͑x, y, z͒ ෇ Ϫ2yz i ϩ y j ϩ 3x k,

S is the part of the paraboloid z ෇ 5 Ϫ x 2 Ϫ y 2 that lies

above the plane z ෇ 1, oriented upward

15. F͑x, y, z͒ ෇ y i ϩ z j ϩ x k,

S is the hemisphere x 2 ϩ y 2 ϩ z 2 ෇ 1, y ജ 0, oriented in the

direction of the positive y-axis

16. Let C be a simple closed smooth curve that lies in the plane

x ϩ y ϩ z ෇ 1. Show that the line integral

xC z dx Ϫ 2x dy ϩ 3y dz

depends only on the area of the region enclosed by C and not

on the shape of C or its location in the plane.

C is the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1)

8. F͑x, y, z͒ ෇ i ϩ ͑x ϩ yz͒ j ϩ ( xy Ϫ sz ) k,

17. A particle moves along line segments from the origin to the

C is the boundary of the part of the plane 3x ϩ 2y ϩ z ෇ 1

in the first octant

9. F͑x, y, z͒ ෇ yz i ϩ 2 xz j ϩ e xy k,

C is the circle x 2 ϩ y 2 ෇ 16, z ෇ 5

;

Graphing calculator or computer required

points ͑1, 0, 0͒, ͑1, 2, 1͒, ͑0, 2, 1͒, and back to the origin

under the influence of the force field

F͑x, y, z͒ ෇ z 2 i ϩ 2xy j ϩ 4y 2 k

Find the work done.

1. Homework Hints available at stewartcalculus.com

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