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

332 0 CHAPTER 16 VECTOR CALCULUS
39. m =· ffs }( dS = }( · 47r(~a 2 ) = 21ra 2 K; by symmetry M xz = lvf 11 z = 0, and
M:cv = Jj~ z }( dS = }( f~" f 0 1f12 (a cos ¢)(a 2 sin¢) d¢ d() = 2711< a 3 [- ~cos 2¢] ~ 12 = 1r }( a 3 •
Hence (x,y,z) = (0,0, ~ a).
(b) 1:: = ffs(x 2 + y 2 )(10- Jx 2 +y 2 ) dS = ff (x 2 + y 2 )(10- jx 2 +y 2 ) ../2dA
. 1 s :r 2 + v 2 s 16 •
I
43. The rate of flow through the cylinder is the flux JJ~ pv · n dS = ffs pv · dS . We use the parametric representation
r (u,v) = 2cosui + 2 sinuj + v k for S, where 0 $ u $ 27r, 0 $ v $ 1, so r ,. = - 2sinu i + 2cosuj, r v = k , and the .
outward orientation is given by ru x r v = 2 cos u i + 2 sin u j . Then
1
ffs pv · dS = p ]~" f 0
(vi+ 4sin 2 u j + 4 cos 2 u k) · (2cosui+ 2sinuj ) dvdu
2
= p f 0 " . J: (2v casu+ 8sin 3 u) dv du = p J~" (casu+ 8 sin 3 u) du
= p[sin u + 8(-i)(2 + sin 2 u) casu] ~" = 0 kg/s
45. S consists of the hemisphere 8 1 given by z = ja 2 - x 2 - y 2 and the disk 8 2 given by 0 $ x 2 + y 2 $ a 2 , z = 0.
On 81: E = asin¢ cosB i +a sin ¢ sinBj + 2acos¢k ,
T q, x To = a 2 sin 2 ¢ cosB i +a 2 sin 2 sinl'.lj +a 2 sin ¢ costpk. Thus
ffs 1 E · dS = .r;" f 0 " 1 \a 3 sin 3 q) + 2a 3 sin