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

1128 Chapter 16 Vector Calculus
the parametric representation
rs, d − a sin cos i 1 a sin sin j 1 a cos k
for the sphere x 2 1 y 2 1 z 2 − a 2 . Then in Example 16.6.10 we found that
and
r 3 r − a 2 sin 2 cos i 1 a 2 sin 2 sin j 1 a 2 sin cos k
| r 3 r | − a 2 sin
So the orientation induced by rs, d is defined by the unit normal vector
n −
r 3 r
| r 3 r | − sin cos i 1 sin sin j 1 cos k − 1 rs, d
a
Observe that n points in the same direction as the position vector, that is, outward from the
sphere (see Figure 8). The opposite (inward) orientation would have been obtained (see
Figure 9) if we had reversed the order of the parameters because r 3 r − 2r 3 r .
z
z
0
y
y
x
x
FIGURE 8
Positive orientation
FIGURE 9
Negative orientation
For a closed surface, that is, a surface that is the boundary of a solid region E, the
convention is that the positive orientation is the one for which the normal vectors point
outward from E, and inward-pointing normals give the negative orientation (see Figures
8 and 9).
z
0
x
FIGURE 10
S ij
n
S
F=∏v
y
Surface Integrals of Vector Fields
Suppose that S is an oriented surface with unit normal vector n, and imagine a fluid with
density sx, y, zd and velocity field vsx, y, zd flowing through S. (Think of S as an imaginary
surface that doesn’t impede the fluid flow, like a fishing net across a stream.) Then the
rate of flow (mass per unit time) per unit area is v. If we divide S into small patches S ij ,
as in Figure 10 (compare with Figure 1), then S ij is nearly planar and so we can approximate
the mass of fluid per unit time crossing S ij in the direction of the normal n by the
quantity
sv ndAsS ij d
where , v, and n are evaluated at some point on S ij . (Recall that the component of
the vector v in the direction of the unit vector n is v n.) By summing these quantities
and taking the limit we get, according to Definition 1, the surface integral of the function
v n over S:
7 y
S
y v n dS − y
S
y sx, y, zdvsx, y, zd nsx, y, zd dS
and this is interpreted physically as the rate of flow through S.
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