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

308 0 CHAPTER 16 VECTOR CALCULUS
5. If we choose x as the parameter, parametric equations for Care x = x, y = v'xfor 1:::; x:::; 4 and
fc ( x2y3- v'x) dy = J;' [xz. (Vx)3 - v'x] 2 ~ dx = 4 J14 (x3- 1) dx
= 1 [1 x ·1 - x] 4 - ! (64 - 4 - 1 + 1) - ill
2 4 1- 2 4 - 8
7.
y
(2, 1)
On cl: X = x, y = ~X => dy = ~ dx, 0 :::; X :::; 2.
On Cz: x = x, y = 3 - :v => dy = -dx, 2 :::; x:::; 3.
Then
f 0
(x + 2y) dx + x 2 dy = fc
1
(x + 2y} dx + x 2 dy +.J 02
(x + 2y} dx + x 2 dy
=J: [x+2(~x)+x 2
(~)]dx+J: [x+2(3-x}+x 2 (-1}]dx
= f 0
2
(2x + ~x 2 ) dx + J 2
3
(6 - x- x 2 ) dx
9. x = 2sint, y = t , z = -2cost, 0:::; t:::; 7T. Then by Formula 9,
fc xyz ds = J 0
7T (2 sin t)(t)( - 2 cos th/ (!!fd + { !fltf + C¥tf dt
= J; -4t sin t cost .J(2 cos t)2 + 11)2 + (2 sin t) 2 dt = J 0
7T - 2tsin 2t .J 4(cos 2 t + sin 2 t) + 1 dt
= -2 y'5 j~'ff t sin2tdt = -2 y'5 [-~ t cos 2t +'i sin2t] ~
= -2¥'5(-~ - 0) = J57T
integrate by parts with ]
[
u = t , dv = s in 2t dt
11 . Parametric equations for Care x = t, y = 2t, z = 3t, 0 :::; t :::; l. Then
15. Parametric equations for Care x = 1 + 3t, y = t, z = 2t, 0 :::; t :5. 1. Then
fc z 2 dx + x 2 dy + y 2 dz = J;(2t? · 3 dt + (1 + 3t}~ dt + t 2 · 2 dt = J; (23t 2 + 6t + 1) dt
= [¥tl + 3t 2 + t]~ = ¥ + 3 + 1 = ¥
17. (a) Along the line x = - 3, the vectors ofF have positive y-components, so since the path goes upward, the integrand F ·Tis
always positive. Therefore f.e F · dr = f.e F · T ds is positive.
l 1 .
© 2012 Cengoge L