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

340 0 CHAPTER 16 VECTOR CALCULUS
15. C1: r (t)=t i +t2 j , -l ~t~ l;
C2: r (t) = :-ti + j, - 1 ~ t ~ 1.
y
cl
. (- l , l ) ot--~~+---..., ( 1 . 1 )
Then
Ic xl dx- x 2 ydy = f 1
(t 5 - 2t 5 )dt + J~ 1 tdt
Using Green's Theorem, we have
j~ xy 2 dx - x 2 ydy = Ji [
~ [-.!t6)1 + [.!t2]1 = 0
6 -1 2 -1 '
:x (-x 2 y) - :y (xy 2 1
) ] dA = Ji (-2xy- 2xy) dA = /_ - 4xydydx
11:
= I~ 1 (-2xy 2 J~::2 dx = I~ 1 (2x 5 - 2x)dx = [ ~x 6 - x 2 t 1
= 0
X
17. Ic x 2 ydx- xy 2 dy. = II [ tfx ( -xy 2 )- %v (x 2 y)] dA = .f[ ( - y 2 - x 2 ) dA = - I;" I; r 3 dr dB= -81r
:r2 + y2 ~ 4 :z:2 + y2 ~ 4
19. If we assume there is such a vector field G, then div( curl G) = 2 + 3z - 2xz. But div( curl F) = 0 for all vector fields F.
Thus such a G cannot exist.
.21. For any piecewise-smooth simple closed plane curve C bounding a region D, we can apply Green's Theorem to
F(x, y) = f(x) i + g~y) j to get Ic f(x) dx + g(y) dy = IIo [ /!"' g(y) - /! 11
f(x)] dA = Jj 0 0 dA = 0.
. '
23. \7 2 f = 0 means that~:{ + ~:; = 0. Now ifF= fv i- f x j and C is any closed path in D , then applying Green's
Theorem, we get
Ic F · dr= Ic fvdx- f x dY = IIo [:x (-fx)- /! 11
Uv)] dA
= -.f[ 0 (fxx + f.uv) dA = - IJ'v 0 dA = 0
Therefore the line integral is independent of path, by Theorem 16.3.3.
25. z = f (x, y) = x 2 + 2y with 0 ~ x ~ 1, 0 ~ y ~ 2x. Thus
27. z = f(x, y) = x 2 + y 2 with 0 ~ x 2 + y 2 ~ 4 so r., x r 11 = -2x i - 2y j + k (using upward orientation). Then
II 5
z dS= II (x 2 +y 2 )J4x 2 +4y 2 +ldA
;r2 + y2 ~ 4
(Substitute u = 1 + 4r 2 and use tables.)
29. Since the sphere bounds a simple solid region, the Divergence Theorem applies and
IIsF · dS =III E divFdV = f.[f E(z- 2) dV = I.ff EzdV- 2III EdV
= O [ odd
fu~ction in =. ]
_ 2 . V(E) = _ 2 . h( 2 ? = _Q17r'
and E os symmctnc 3 3
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