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

334 D CHAPTER 16 VECTOR CALCULUS
(b) (c) One possible parametrization is x = 3 cost, y = 3 sin t,
z = 1 - 3 cos t - 3 sin t, 0 ~ t ~ 271'.
2
-2
13. The boundary curve Cis the circle x 2 + y 2 = 16, z = 4 oriented in the clockwise direction as viewed from above (since Sis
oriented downward). We can parametrize C by r(t) = 4costi - 4sintj + 4k, 0 ~ t ~ 21r, and then
r'(t) = - 4sint i-4costj. Thus F (r(t)) = 4sint i + 4cos t j - 2 k , F (r (t)) · r ' (t) = -16sin 2 t- 16cos 2 t = -16, and
fcF · dr = J; .,. F(r(t)) · r'(t) dt = J;"(-16) dt = -16 (27r) = -327!'
Now curl F = 2 k, and the projection D of Son the xy-plane is the disk x 2 + y 2 ~ 16, so by Equation 16.7.10 with
z = g(x, y) = .Jx 2 + y 2 [and multiplying by - 1 for the downward orientation] we have
ffs curl F · dS = - ffv( -0 - 0 + 2) dA = -2 · A(D) = -2 · 7r(4 2 ) = - 327!'
15. The boundary curve C is the circle x 2 + z 2 = 1, y = 0 oriented in the counterclockwise direction as viewed from the positive
y-axis. Then C can be described by r(t) = cost i - sin t k, 0 ~ t ~ 21r, and r' (t) = -sin t i- cost k. Thus
F (r (t)) = -sint j +costk, F (r (t)) · r ' (t) = - cos 2 t, and .fc F · dr = J;"(-cos 2 t).dt = -~t- ~ sin2t]~.,. = -1r.
Now curl F = -i- j - k , and S can be parametrized (see Example 16.6.1 0) by
r (rfJ, 9) = sin rfJ cos9i+ sin rfJ sinBj + cosq'> k,O ~ B ~ 7!',0 ~ rp ~ 7!'. Then
r .p x ro = sin 2 4> cos B i + sin 2 4> sin B j +sin 1p cos rp k and
JJ 5
curlF·dS= JJ cur!F·(r,.;x ro)dA=f 0
" J 0
" (-sin 2 ¢cosB - sin 2 ¢sinB - sin¢cos¢)dBd¢
~ +~ S l .
17. lt is 'easier to use Stokes' Theorem than to compute the work directly. LetS be the planar region enclosed by the path of the
particle, so Sis the portion of the plane z = h for 0 ~ x ~ 1, 0 ~ y ~ 2, with upward orientation.
curl F =By i + 2z j + 2y k and
fcF · dr = ff 5
curlF · dS = ff 0
[-By (0) - 2z (~ )
= f 0
1
J; ~Y dy dx = J; [1Y 2 ]~=~ dx = f 0
1
3 dx = 3
+ 2y] dA = J 0
1
J 0
2
(2y- h) dydx
19. Assume S is centered at the origin with radius a and let Ht and H 2 be the upper and lower hemispheres, respectively, of S.
I
F · dr by Stokes' Theorem. But C1 is the
Then .ff 5
curlF · dS = JJH 1
curlF · dS + JJH 2
curlF · dS = J~ 1
F · dr + J~ 2
circle x 2 + y 2 = a 2 oriented in the counterclockwise direction while C 2 is the same circle oriented in the clockwise direction.
Hence fc 2
F · dr = - fc 1
F · dr so f fs curl F · dS = 0 as desired.
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