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, 4͞3, 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
iϪ
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¡
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 ͑2͒ 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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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
FؒnϪ
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.
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.
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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.
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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
C¡
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
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.
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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
ͩ
ͪ
ͩ
ͪͬ
yͫ
ͪ ͩ ͪ ͬ
y ͫͩ
ͪ ͩ ͪ
yͩ
ͪ ͩ ͪͬ
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
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_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 ͑2͒ 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.
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_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
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.