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

SECTION 16.7 SURFACE INTEGRALS 0 329
13. S is the portion of the cone z 2 = x 2 + y 2 for 1 $ z $ 3, or equivalently, Sis the part of the surface z = ...jx 2 + y2 over the
region D = { (x, y) Jl $ x 2 + y 2 $ 9 }. Thus
r---------~----------~--
·Jfs x2z2dS= Jfv x 2 (x 2 +y 2 )
(~Y +(~Y + l dA
15. Using x and z as parameters, we have r (x, z) = x i+ (x 2 + z 2 )j + z k 1 x 2 + z 2 :::; 4. Then
r x x r : = (i + 2xj ) X (2zj + k) = 2x i - j +2z kand Jr, X r, J = J4x 2 + 1 + 4z 2 = ) 1 +4(x 2 +z 2 ). Thus
JJ (x 2 + z 2 ))1 + 4(x 2 + z 2 ) dA = J;,. J; r 2 J 1 + 4r 2 rdrdO = J;,. dB J; r 2 v'1 + 4r2 r dr
ffs ydS =
x2 +z2~ ~ · .
[let u = 1 + 47' 2 =>
7' 2 = t(u- 1) and idu = 1' dr]
= 27T J:7 t(u- l )JU. id·u = ft1T ft17 (u3/2 ...:. ul/2) du
= ...!.. 7T (1u5/2 - 1u3/2)17 = ...!..7T(1(17)5 /2- 1(17? /2 - 1 + 1) = 2:.. (391 Jl7 + 1)
16. 5 3 1 16 :; 3 5 3 60
17. Using spherical coordinates and Example 16.6.1 0 we have r ( ¢,B) = 2 sin 4> cos B i + 2 sin 4> sin B j + 2 cos 4> k and
Jrq, x r~l = 4 sin¢. Then ffs(x 2 z + y 2 z) dS = .{ 0
2 " J; 12 (4sin 2 1p)(2 cos¢)( 4 sin 4>) d¢ dB = 167T sin 4 ¢] ~ 1 2 = 167T.
19. S is given by r (u,v) = ui + co~vj + sinv k, 0 $ u $ 3, 0 $ v $ 1Tj2. Then
r u X rv = i X (- sin v j +cosvk ) = - cosvj-sinv kand Jr,. X rvl = ) cos 2 v +sin 2 v = 1, so
.[[ 5
(z + x 2 y) dS = J; 12 J;(sinv + u 2 co~v)(l) dudv = f~.,. 12 (3sin v + 9 cos v) dv
= [- 3cosv + 9sin v] ~/ 2 = 0 + 9 + 3-0 = 12
21. From Exercise 5, r (u, v) = (u + v) i + (u - ·u) j + (1 + 2u + v) k, 0 $ u $ 2, 0 $ v $ 1, and r u x r., = 3 i + j - 2 k.
Then
F (r (u, v)) = (1 + 2u + v)e