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Thomas Calculus 13th [Solutions]

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1234 Chapter 16 Integrals and Vector Fields<br />

15. Assume that S is a surface to which Stokes Theorem applies. Then E dr ( E)<br />

n d<br />

C<br />

S<br />

B n d B n d . Thus the voltage around a loop equals the negative of the rate of change of<br />

t<br />

t<br />

S<br />

S<br />

magnetic flux through the loop.<br />

16. According to Gausss Law, F n d 4 GmM for any surface enclosing the origin. But if F H then<br />

S<br />

the integral over such a closed surface would have to be 0 by the Divergence Theorem since div F 0.<br />

17.<br />

C<br />

S<br />

S<br />

S<br />

f g dr f g n d<br />

(Stokes Theorem)<br />

S<br />

f g f g n d<br />

(Section 16.8,Exercise 19b)<br />

( f )( 0)<br />

f g n d<br />

(Section 16.7, Equation 8)<br />

f g n d<br />

18. F1 F2 F2 F1 0 F2 F 1 is conservative F2 F1 f ; also, F1 F2<br />

2<br />

F2 F 1 0 f 0 (so f is harmonic). Finally, on the surface S,<br />

f n F2 F1<br />

n<br />

2<br />

F2 n F1 n 0. Now, f f f f f f so the Divergence Theorem gives<br />

2 2<br />

f dV f f dV f f dV f f n d 0, and since<br />

D D D S<br />

2 2<br />

f dV 0 0 F2 F1<br />

dV 0 f f n d 0, as claimed.<br />

D D S<br />

2<br />

f 0 we have<br />

i j k<br />

19. False; let F yi xj 0 F ( y) x<br />

( x) y<br />

0 and F<br />

x y z<br />

0i 0j 0k 0<br />

x y 0<br />

20.<br />

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2<br />

ru rv ru rv sin ru rv 1 cos ru rv ru rv cos ru rv ru rv<br />

2 2 2<br />

ru rv EG F d ru rv<br />

du dv EG F du dv<br />

21. r xi yj zk r 1 1 1 3 r dV 3 dV 3 V V 1 rdV 1 r n d , by<br />

3 3<br />

D D D S<br />

the Divergence Theorem<br />

Copyright<br />

2014 Pearson Education, Inc.

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