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

SECTION 16.2 LINE INTEGRALS D 309
37. From Example 3, p(x,y) = k(1- y), x = cost, y = sint, and ds = dt, 0 :S t :S 1r ::::}
1:. = Icy 2 p(x,y) ds = IoT< sin 2 t [k(1- sint)]dt = ki 0
" (sin 2 t- sin 3 t) dt
= ~k I 0
.,.(1- cos2t) dt- k Io"(1- cos 2 t)sintdt
Let u = cost, du = - sin t dt ]
[ in the second integral
= k[~ + I1- 1 (1-u 2 )du] =k (~ - ~)
lv = Icx 2 p(x, y)ds = ki 0
.,. cos 2 t(1 - sint)dt = ~ I 0
.,.(1 +cos2t)dt- kio" cos 2 tsintdt
= k(~- i), using t?e same substitution as above. ·
39. W = Ic F · dr = I;.,. (t-sint,3 - cost)· (1 - cost,sint) dt
= I; .,.(
t - t cost - sin t + sin t cos t + 3 sin t - sin t cos t) dt
=I;,. (t- tcost + 2sin t) dt = [~t 2 - (tsin t + cost) - 2 cos t)~ ,..
integrate by .parts ]
[ in the second term
41. r (t) = (2t, t, 1 - t), 0 :::; t :::; 1.
W = Ic F · dr = Io 1 (2t- t 2 , t- (1 - t?, 1 - t- (2t) 2 ) · (2,1, - 1) dt
= Io 1 (4t- 2t2 + t- 1 + 2t - t 2 - 1 + t + 4t 2 ) dt =I; (t 2 + 8t - 2) dt = [lt 3 + 4t2 - 2t)~ = ~
43. (a) r (t) = at2 i + bt 3 j ::::} v (t) = r' (t) = 2at i + 3bt 2 j ::::} a(t) = v' (t) = 2a i + 6btj, and force is mass times
acceleration: F(t) = ma(t) = 2mai + 6mbtj.
(b) W = fc F · dr = J 0 1 (2mai + 6mbtj) · (2at i + 3bt 2 j) dt = I 0
1
(4ma 2 t + 18mb 2 t 3 ) dt
= [2ma 2 t 2 + ~mb 2 t 4 )~ = 2ma 2 + ~mb 2
45. Let F = 185 k. To parametrize the staircase, let x = 20 cost, y = 20 sin t, z = ~~ t = 1!-t, 0 :::; t :::; 61r ::::}
W = fc F · dr =I:.,. (0, 0, 185) · ( -20 sin t, 20 cost, ~) dt = (185) ~ J:.,.. dt = (185)(90) :::::: 1.67 x 10 4 ft-lb
47. (a) r (t) = (cost, sin t}, 0 :S t:::; 271", and let F =(a, b). Then
W = Ic F · dr = I;" (a, b) · (- sin t,cost) dt = I~'lf(-asint + bcost) dt = [a cost + bsin t)~ .,.
(b) Yes. F (x, y) = k x = (kx, ky) and
=a+O - a+ O=O
W = Ic F · d r =I;,. (k cost, k s~ t} · (-sin t , cost} dt = J;" ( - k sin t cost+ k sin t cost) dt = I~ .,.. 0 dt = 0.
49. Let r(t) = (x(t), y(t), z(t)) and v = (v1, v2, va). Then
Ic v · dr = I: (v1, v2, va) · (x' (t), y' (t), z' (t)) dt = I: [v1 x' (t} + v2 y' (t) + va z' (t)] dt
= [v1 x(t) + v2 y(t) + va z(t)): = [v1 x (?) + V2 y(b) + va z(b)J - [v1 x(a) + V2 y(a) + vs z(a)]
= v 1 [x(b)- x(a)J + v2 [y(b)- y(a)] + va [z(b) - z(a)]
=: (v1, v2, va) · (x(b)- x(a), y(b) - y(a), z(b)- z(a))
= (v1, v2, va) · [(x(b), y(b), z(b)}- (x(a), y(a), z(a))] = v · (r(b)- r (a)]
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