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

316 0 CHAPTER 16 VECTOR CALCULUS
{)Q -2x(x 2 + y 2 ? - (y 2 - .x 2 ) • 2(x 2 + y 2 ) · 2x 2:~; 3 - 6xy 2 .
'* -{) = ( 2 2 ) 4
= a . Thus, as m the example,
.. X X + y (x2 + y2)
L P dx + Q dy + j_c, P dx + Q dy = I L ( ~~ -~:) dA = I L
and J c F · dr = .fc, F · dr. We parametrize C' as r (t) =a cost i +a :sin t j , 0:::; t:::; 21r. Then
1 1 d 1
0 dA = 0
2 " 2(acost)(asint)i+ (a 2 sin. 2 t - a 2 cos 2 t) j ( . . ·)
F · dr = F · r =
2
· - a sm t 1 + a cost J dt
c C' o (a 2 cos 2 t + a 2 sin 2 t)
. 1 l2~ 1 l 2"
= - (-cost sin 2 t - cos 3 t) dt = - (-cost sin 2 t- cost (1 - sin 2 t)) dt
a o
2
"
11 2 = -- ~ costdt = - -1 sint ] = 0
a o a o
.29. Since C is a simple closed path which doesn' t pass through or enclose the origin, there exists an open region that doesn't
a o
contain the origin but does contain D. Thus P = -y/(x 2 + y 2 ) and Q = xj(x 2 + y 2 ) have continuous partial derivatives on
this open region containing D and we can apply Green's Theorem. But by Exercise 16.3.35(a), aPjay = aQj8x, so
fcF · dr = ffv OdA ~ 0.
. . & M
31. Usmg the first part of(S), we have that I In dx dy = A(R) = Ion X dy. But X = g(u, v), and dy = au du + 8v dv,
and we orient as by taking the positive direction to be that which corresponds, under the mapping, to the positive direction
alongaR, so
{ X dy = { g(u, ·u) ( 8 ah du + {)ah dv) = r g(u, v) {)ah du + g(u, v) Bah dv
i on l as u v .los u v
= ± .ffs [ :v. (g(u, v) g~) - :v (g(u, v) ~~) ) dA [using Green's Theorem in the uv-plane]
= ± ff (!L9.. !ll! + g(u v) 0 2 " - !l.!l 81 ' - g(u v) 82 S Uu Dv > l:lul:lv l:l·v Vv. > 8vUu " ) dA [using the Chain Rule]
= ± .ffs (~~ ~- Z~ ~) dA _[by the equality of mixed partials] = ± Jj~ ~~::~\ dudv
The sign is chosen to be positive if the orientation that we gave to 8S corresponds to-the usual positive orientation, and it is
negative othe~ i se. In either case, since A(R) is positive, the sign chosen must be the same as the sign of ~i:: ~?.
ThereforeA(R) = ll dxdy= lfsl~~::~~ ~dudv .
16.5 Curl and Divergence
j k
1. (a) curlF = 'il x F = 8j 8x 8 / 8y 8/8z
x + yz y + xz z + xy
= [:Y(z + xy) ...:. :z(y + xz)] i - [:x(z + xy)- :z(x + yz)] j + [:x (y ~ xz) - :y (x + yz)] k
= (x- x) i- (y - y) j + (z- z) k = 0
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