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

1144 Chapter 16 Vector Calculus
n
n¡
S¡
S _n¡
FIGURE 3
Although we have proved the Divergence Theorem only for simple solid regions, it
can be proved for regions that are finite unions of simple solid regions. (The procedure is
sim ilar to the one we used in Section 16.4 to extend Green’s Theorem.)
For example, let’s consider the region E that lies between the closed surfaces S 1 and
S 2 , where S 1 lies inside S 2 . Let n 1 and n 2 be outward normals of S 1 and S 2 . Then the
boundary surface of E is S − S 1 ø S 2 and its normal n is given by n − 2n 1 on S 1 and
n − n 2 on S 2 . (See Figure 3.) Applying the Divergence Theorem to S, we get
7 y yy div F dV − yy F dS − y
E
S
S
y F n dS
− y
S 1
y F s2n 1 d dS 1 y
S 2
y F n 2 dS
− 2y
S 1
y F dS 1 y
S 2
y F dS
ExamplE 3 In Example 16.1.5 we considered the electric field
Esxd − «Q
| x | 3 x
where the electric charge Q is located at the origin and x − kx, y, zl is a position vector.
Use the Divergence Theorem to show that the electric flux of E through any closed
surface S 2 that encloses the origin is
y
S 2
y E dS − 4«Q
SOLUTION The difficulty is that we don’t have an explicit equation for S 2 because
it is any closed surface enclosing the origin. The simplest such surface would be a
sphere, so we let S 1 be a small sphere with radius a and center the origin. You can
verify that div E − 0. (See Exercise 23.) Therefore Equation 7 gives
yy E dS − yy E dS 1 y yy div E dV − yy E dS − y
S 2 S 1 E
S 1
S 1
y E n dS
The point of this calculation is that we can compute the surface integral over S 1 because
S 1 is a sphere. The normal vector at x is xy| x | . Therefore
since the equation of S 1 is | x |
E n − «Q
| x |
S x x
3 | |D x − «Q
| x | x x − «Q
4 | x | − «Q
2 a 2
− a. Thus we have
yy E dS − yy E n dS − «Q
yy dS − «Q
a 2 a AsS 1d − «Q
2 a 4a 2 − 4«Q
2
S 2 S 1 S 1
This shows that the electric flux of E is 4«Q through any closed surface S 2 that contains
the origin. [This is a special case of Gauss’s Law (Equation 16.7.11) for a single charge.
The relationship between « and « 0 is « − 1ys4« 0 d.]
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