James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

section 16.8 Stokes’ Theorem 1137
the disk x 2 1 y 2 < 1 and so using Equation 16.7.10 with z − tsx, yd − 2 2 y, we have
y F dr −
C
yy curl F dS − y
S
D
y s1 1 2yd dA
− y 2
0
y 1
0
s1 1 2r sin d r dr d
2
− y F r 2
0 2 1 2 r 3 1
3 sin d − y 2
( 1 2
0
G0
1 2 3 sin ) d
− 1 2 s2d 1 0 − ■
z
0
S
C
≈+¥+z@=4
ExamplE 2 Use Stokes’ Theorem to compute the integral yy S
curl F dS, where
Fsx, y, zd − xz i 1 yz j 1 xy k and S is the part of the sphere x 2 1 y 2 1 z 2 − 4 that
lies inside the cylinder x 2 1 y 2 − 1 and above the xy-plane. (See Figure 4.)
SOLUTIon To find the boundary curve C we solve the equations x 2 1 y 2 1 z 2 − 4 and
x 2 1 y 2 − 1. Subtracting, we get z 2 − 3 and so z − s3 (since z . 0). Thus C is the
circle given by the equations x 2 1 y 2 − 1, z − s3 . A vector equation of C is
x
y
≈+¥=1
rstd − cos t i 1 sin t j 1 s3 k
0 < t < 2
FIGURE 4
so
r9std − 2sin t i 1 cos t j
Also, we have
Fsrstdd − s3 cos t i 1 s3 sin t j 1 cos t sin t k
Therefore, by Stokes’ Theorem,
yy curl F dS − y F dr −
C
y 2
Fsrstdd r9std dt
0
S
− y 2
(2s3 cos t sin t 1 s3 sin t cos t) dt
0
− s3 y 2
0 dt − 0 ■
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 S 1 and S 2 are oriented surfaces with the same oriented boundary curve C
and both satisfy the hypotheses of Stokes’ Theorem, then
3 y
S 1
y curl F dS − y C
F dr − y
S 2
y curl F dS
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 C
v T ds
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