Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

dS
SECTION 16.9 THE DIVERGENCE THEOREM 0 335
16.9 The Divergence Theorem
1. div F = 3 + x + 2x = 3 + 3x, so
J I IE div F dV = J: J: f 0
1
(3x + 3) dx dy dz = ~ (notice the triple integral is
three times the volume of the cube plus three times X).
To compute Jfs F · dS , on
S 1 : n = i, F = 3 i +yj + 2z k, and Jf 51 F · dS = ffsi 3 dS = 3;
X
s2: F = 3x i + X j + 2xz k , n = j and I fs2 F
0
= I f s2 X dS = ~;
S3: F = 3xi + xyj + 2x k, n = k and f f 53
F · dS = fj 53
2xdS = 1;
S4 : F = 0 , ffs,, F · dS = 0; S:;: F = 3xi +2x k , n = - j and ff 56
F · dS = ffs
5
OdS = 0;
S 6 : F = 3xi + xyj , n = - kand JJ 56
F · dS = JJ 56
OdS = 0. Thus ff 5
F · dS = ~ -
3. div F = 0 + 1 + 0 = 1, so JJJ E div F dV = JJJ E l dV = V(E) = ~ 7T · 4 3 = 2 ~ 6 7T. S is a sphere of radius 4 centered at
the origin which can be parametrized by r (¢>, 0) = (4sin ¢>cos 0, 4 sin sin 0, 4 cos¢>), 0 $ ¢ $ 1r, 0 $ 0 $ 27T (similar to
Example 16.6.10). Then
r .p x r 9 = (4 cos ¢>cos 0, 4 cos sine, - 4sin ) x (- 4sin sin8,4 sin cos 0, 0)
= (16sin 2 ¢>cos 0, 16 sin 2 ¢sin 0, 16 cos ¢>sin¢)
and F (r (¢,0)) = (4cos¢,4sin ¢sin 0,4sincos0). Thus
F · (r .p x r o) = 64 cos ¢sin 2 cos 0 + 64sin 3 ¢sin 2 B + 64cos ¢sin 2 ¢>cos() = 128 cos ¢sin 2 q> cos () + 64 sin 3 ¢> sin 2 ()
and
'
f fs F · dS = JJ.; F · {r ,p x r o) dA = J~" J; (128cos ¢sin 2 ¢>cosO + 64sin 3 sin 2 8) ddB
= J~" [ 1 ~ 8 sin 3 ¢cos0+64(- ~{2+sin 2 )cos¢)sin 2 8J !:~ dO
0
f27r 256 ° 2 0 d() = ill (l.() - l 28] 2 " - Jo
ill
3 Sill 3 2 4 Sill 0 - 3 7T
5. div F = /}., (xye") + !Ju (xy 2 z 3 ) + //:: ( - ye" ) = ye" + 2xyz 3 - ye" = 2xyz 3 , so by the Divergence Theorem,
Jfs F · dS = JJJ E div F dV = f 0
3 j0
2 f0
1 2xyz 3 dzdydx = 2 J~ xdx J 0
2
ydy J 0
1
z 3 dz
= 2 (~ x 2 ]~ [h 2 ]~ (tz"]~ = 2 (~) (2) {i) = ¥
7. div F = 3y 2 + 0 + 3z 2 , so using cylindrical coordinates with y = r cos B; z = r sin B, x = x we have
ffs F · dS = J/J E(3y 2 + 3z 2 ) dV = J~" J: f~ 1 {3r 2 cas 2 B + 3r 2 sin 2 B) r dx dr dB
~ 3 f 0
2
" dB ] 0
1 r
3
dr f~ 1 dx = 3{27r) (i) (3) = 9 2 "
9. div F = 2x sin y - x sin y - x sin y = 0, so by the Divergence Th~orem , J f s F · dS = J J EO dV = 0.
11. div F = y 2 + 0 + x 2 = x 2 + y 2 so
Jj~ F · dS = JJJ E(x 2 + y 2 ) dV = J;" f~ .J;~ r 2 • r dz dr d(J ~ J~" .[ 0
2 r 3 (4- r 2 ) dr dB
= J~" dB J~ (4r 3 - r 5 ) dr = 27r [r 4 - 'kr 6 ) ~ = ~7T
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