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

section 16.7 Surface Integrals 1129
If we write F − v, then F is also a vector field on R 3 and the integral in Equation 7
becomes
yy F n dS
S
A surface integral of this form occurs frequently in physics, even when F is not v, and
is called the surface integral (or flux integral) of F over S.
8 Definition If F is a continuous vector field defined on an oriented surface S
with unit normal vector n, then the surface integral of F over S is
y
S
y F dS − y
S
y F n dS
This integral is also called the flux of F across S.
In words, Definition 8 says that the surface integral of a vector field over S is equal to
the surface integral of its normal component over S (as previously defined).
If S is given by a vector function rsu, vd, then n is given by Equation 6, and from Definition
8 and Equation 2 we have
y
S
y F dS − y
S
y F
r u 3 r v
| r u 3 r v | dS
− y
Dy FFsrsu, vdd
where D is the parameter domain. Thus we have
r u 3 r v
| r u 3 r v |G| r u 3 r v | dA
Compare Equation 9 to the similar
expression for evaluating line integrals
of vector fields in Definition 16.2.13:
y C
F dr − y b
a
Fsrstdd r9std dt
9
y
S
y F dS − y
D
y F sr u 3 r v d dA
Figure 11 shows the vector field F in
Example 4 at points on the unit sphere.
z
ExamplE 4 Find the flux of the vector field Fsx, y, zd − z i 1 y j 1 x k across the unit
sphere x 2 1 y 2 1 z 2 − 1.
SOLUTIon As in Example 1, we use the parametric representation
rs, d − sin cos i 1 sin sin j 1 cos k 0 < < 0 < < 2
Then
Fsrs, dd − cos i 1 sin sin j 1 sin cos k
and, from Example 16.6.10,
x
y
Therefore
r 3 r − sin 2 cos i 1 sin 2 sin j 1 sin cos k
FIGURE 11
Fsrs, dd sr 3 r d − cos sin 2 cos 1 sin 3 sin 2 1 sin 2 cos cos
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