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

312 0 CHAPTER 16 VECTOR CALCULUS
25. We know that if the vector field (call it F) is conservative, then around any closed path C, fc F · dr = 0. But take G to be a
circle centered at the origin, oriented counterclockwise. All of the field vectors that start on G are roughly in the direction of
motion along C, so the integral around G will be positive. Therefore the field is not conservative.
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From the graph, it appears that F is conservative, since around all closed
paths, the number and size of the field vectors pointing in directions similar
to that of the path seem to be roughly the same as the number and size of the
vectors pointing in the opposite direction. To check, we calculate
. .
~ + x cos y). Thus F is conservative, by
y
X
a o (sin y) = cos y = 0° (
Theorem 6.
29. Since F is conservative, there exists a function f such that F = \1 f, that is, P = /z, Q = ! 11 , and R = f= · Since P ,
Q, and R have continuous first order partial derivatives, Clairaut's Theorem says that 8P / EJy = fx 11 = fvx = EJQf EJx,
EJP/8z = fxz = /zx = 8R /.8x, and 8Q/8z = fyz = f zv = 8Rf8y.
31. D = {(x, y) I 0 < y < 3} consists of those points between, but not
on, the horizontal lines y = 0 andy = 3.
(a) Since D does not include any of its boundary points, it is open. More
formally, at any point in D there is a disk centered at that point that
lies entirely in D.
'(b) Any two points chosen in D can always be joined by a path that lies
y
3
----------- -----------
---------·--.:
0 X
entirely in D , so D is com1ected. (D consists of just one "piece.")
(c) Dis connected and it has no holes, so it's simply-connected. (Every simple closed curve in D encloses only points that are
in D.)
33. D = { (x, y) II ::; x 2 + y 2 ::; 4, y 2: 0} is the semiannular region
in the upper half-plane between circles centered at the origin of radii
l and 2 (including all boundary points).
(a) D includes boundary points, so it is not open. [Note that at any
boundary point, (1, 0) for instance, any disk centered there cannot lie
entirely in D .]
(b) .The region consists of one piece, so it's connected.
(c) D is connected and has no holes, so it's simply-connected.
y 8P y 2 - x 2 x 8Q y 2 - x 2 8P 8Q
35. (a)P= - x 2 +y 2 , .A..=
2 andQ=--- -- = 2 .Thus - VIJ (x2 + y2) x2 +y2' OX (x2 +y2) 8 y UX
(b) G 1: X = cost, y = sin t, 0 ::; t ::; ;rr' c 2: X = cost, y = sin t, t = 21!' to t = 1l'. Then
y
=,.-.
{ F ·dr = ("'(- sin t)(-s~n t)+.(~ost)(cost)dt= ("' dt = 7rand { F· dr = ("' dt= - 1!'
Jc, } 0 cos t + sm t ./ 0 lc 2 J 21r
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