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

A five star textbook for college calculus

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section 16.9 The Divergence Theorem 1143

Notice that the method of proof of the

Divergence Theorem is very similar to

that of Green’s Theorem.

Comparison with Equation 5 shows that

y

S

y R k n dS − y y

E

y −R

−z dV

Equations 2 and 3 are proved in a similar manner using the expressions for E as a

type 2 or type 3 region, respectively.

ExamplE 1 Find the flux of the vector field Fsx, y, zd − z i 1 y j 1 x k over the unit

sphere x 2 1 y 2 1 z 2 − 1.

SOLUtion First we compute the divergence of F:

div F − − −x szd 1 − −y syd 1 − −z sxd − 1

The solution in Example 1 should

be compared with the solution in

Example 16.7.4.

z

The unit sphere S is the boundary of the unit ball B given by x 2 1 y 2 1 z 2 < 1. Thus

the Divergence Theorem gives the flux as

y

S

y F dS − y yy div F dV − y y 1 dV − VsBd −

B

By 4 3 s1d3 − 4 3

ExamplE 2 Evaluate yy S

F dS, where

(0, 0, 1)

0

(1, 0, 0)

x

FIGURE 2

y=2-z

(0, 2, 0) y

z=1-≈

Fsx, y, zd − xy i 1 (y 2 1 e xz2 ) j 1 sinsxyd k

and S is the surface of the region E bounded by the parabolic cylinder z − 1 2 x 2 and

the planes z − 0, y − 0, and y 1 z − 2. (See Figure 2.)

SOLUtion It would be extremely difficult to evaluate the given surface integral

directly. (We would have to evaluate four surface integrals corresponding to the four

pieces of S.) Furthermore, the divergence of F is much less complicated than F itself:

div F − − −x sxyd 1 − −y (y2 1 e ) xz2 1 − ssin xyd − y 1 2y − 3y

−z

Therefore we use the Divergence Theorem to transform the given surface integral into a

triple integral. The easiest way to evaluate the triple integral is to express E as a type 3

region:

Then we have

y

S

E − hsx, y, zd | 21 < x < 1, 0 < z < 1 2 x 2 , 0 < y < 2 2 zj

y F dS − y yy div F dV − y y

E

E

y 3y dV

− 3 y 1 y 22z

y dy dz dx − 3 y 1 s2 2 zd 2

dz dx

21 y12x2 0 0 21 y12x2 0 2

− 3 2 y21F2 1 12x

s2 2 2 zd3

dx − 2 1 2 y 1 fsx

3 21

G0

2 1 1d 3 2 8g dx

− 2y 1

sx 6 1 3x 4 1 3x 2 2 7d dx − 184

0 35

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