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

CHAPTER 16 REVIEW 0 339
TRUE-FALSE QUIZ
1. False; div F is a scalar field.
3. True, by Theorem 16.5.3 and the fact that div 0 = 0.
5. False. See Exercise 16.3.35. (But the assertion is true if D is simply-connected; see Theorem 16.3.6.)
7. False. For example, div(y i) = 0 = div(xj) but y i =I x j.
9. True. See Exercise 16.5.24.
11. True. Apply the Divergence Theorem and use the fact that div F = 0.
. . .
EXERCISES
1. (a) Vectors starting on C point in roughly the direction opposite to C, so the tangential component F · T is negative.
Thus J c F · dr = j~ F · T ds is negative.
(b) The vectors that end near Pare shorter than the vectors that start near P, so the net flow is outward near P and·
div F(P) is positive.
3. fc y zcosx ds = J 0
71' (3cos t) (3sint) cost .j(1) 2 + ( -3sint)2 + (3cos t)2 dt = J 0
rr(9cos 2 t sin t)v'IO dt
= 9 JlO ( - t cos 3 t)]~ = - 3 VfO ( -2) = 6 VfO
5. fc y3 dx + x2 dy = J~1 [y3( -2y) + (1 - y2)2] dy = J~1 (-y4 - 2!/+ 1) dy
= [- !y5 - l y3 + y] 1 = _ ! - 1 + 1-! - 1 + 1 = ~
5 3 - 1 5 3 5 3 15
7.C:x=1+2t => dx=2dt,y = 4t => dy = 4dt,z=-1+3t => dz = 3dt,O~t~l.
fc xydx + y 2 dy + yzdz = f 0
1
[(1 + 2t)(4t)(2) + (4t) 2 (4) + (4t)(-1 + 3t)(3)] dt
9. F (r(t)) = e- t i + t 2 ( -t)j + (t 2 + t 3 ) k, r'(t) = 2t i + 3e j - k and
f. F · dr = rt (2te-t - 3t5 - (t 2 + t 3 ) ) dt = [-2te-t - 2e-t - !t 6 - !t 3 - !t 4 ] 1 = ll - 1
c Jo 2 3 4 o 12 c ·
11. : 11
[(1 + xy)e"'"] = 2xe"' 11 + x 2 ye"' 11 = :, [e" + x 2 e"' 11 ] and the domain ofF is JR?, so F is conservative. Thus there
exists a function f sue~ that F = 'V f. Then fu(x, y) = e 11 + x 2 e"' 11 implies f(x , y) = e 11 + xe"'ll + g(x) and then
f,(x, y) = xye"' 11 + e"' 11 + g'(x) = (1 + xy)e"'ll + g'(x). But J,(x, y) = (1 + xy)e"' 11 , so g'(x) = 0 => g(x) = K.
Thus f (x, y) = e 11 + xe"'" + [(is a potential function for F .
13. Since tu (4x 3 y 2 - 2xy 3 ) = Bx 3 y- 6xy 2 = :, (2x 4 y-3x 2 y 2 + 4y 3 ) and the domain ofF is IR?, F is conse~ative.
Furthermore f(x, y) = x 4 y 2 - x 2 y 3 + y 4 is a potential function for F . t = 0 corresponds to the point (0, 1) and t = 1
corresponds to (1, 1), so fc F · dr = /(1, 1) - /(0,.1) = 1 - 1 = 0.
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