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

314 0 CHAPTER 16 VECTOR CALCULUS
9. fc y 3 dx- x 3 dy = ffv [:., (-x 3 )- :U (y 3 ) ] dA = ff 0 (- 3x 2 - 3y 2 ) dA = J:"" J:(-3r 2 ) rdrdO
= - 3 J:" dO ..r: r 3 dr = -3(27!)(4) = - 2471"
11. F (x, y) = (y cosx - xysinx, xy + xcos x) and the region D enclosed by Cis given by
{(x, y) I 0 ~ x ~ 2, 0 ~ y ~ 4- 2x}. Cis traversed clockwise, so-C gives the positive orientation.
fc F · dr = - J_ 0 (ycosx- xys.inx)dx + (x~ + xcosx) dy = -
ffv [ :., (xy + xcosx) - :Y (ycosx- xysinx)] dA
=- ffv(Y - xsin x + cosx -: cosx + xsin x)dA = - J 0
2
.{0
4
-
2
:r ydydx
=- f 2 [ ly 2 ] v= 4 - 2 "' dx =- f 2 1(4 - 2x) 2 dx - - f 2 (B - Bx + 2x 2 ) dx- - (Bx- 4x 2 + ~x 3 ) 2
Jo 2 v=O Jo 2 - Jo - a o
= - (16 - 16 + 1 : - o) = - 1 6
3
13. F (x, y) = (y- cosy, x sin y) and the region D enclosed by Cis the disk with radius 2 centered at (3, -4).
C is traversed clockwise, so -C gives the positive orientation .
.f 0
F · dr = - J_ 0
(y - cosy)dx + (xsiny) dy = - JJ 0
[ :., (xsiny)- :u (y- cosy)] dA
= - ffv(siny - 1 - sin y) dA = ffv dA = area of D .= 7r(2? = 47r
15. Here C = C1 + C2 where
01 can be parametrized as X = t, y = 1, - 1 ~ t ~ 1, and
y
C2 isgivenbyx=-t, y=2 - t 2 , - 1 ~t~ l.
Then the line integral is
. 1
f y 2 e"' dx + x 2 ev dy = J_ 1
[1· et + t 2 e · 0] dt
~~ .
+ J~ 1 [(2- t2 ? e-t( - 1) + (- t) 2 e 2 - t· (- 2t)] dt
= J~ 1 [et - (2 - t 2 ) 2 e- t - 2t 3 e 2 2
-t ] dt =-Be+ 4Be- 1
-I 0
X
according to a CAS. The double integral is
!
. r (aQ aP) ; ·1 r2_,2 .
. J D ax - 8y dA = - 111 (2xeY - 2ye"') dy dx = - Be + 4Be- 1 ' verifying Green's Theorem in this case.
17. By Green's Theorem, W = fc F · dr = fc x(x + y) dx + xy 2 dy = J( 0
(y 2 - x) dA where C is the path described in the
question and D is the triangle hounded by C . So
W = J 01
J;-"'(y 2 - x) dy dx = f 0
1
[ty 3 - xy]~ = ~-"' dx = J; (i(1 - x) 3 - x(1 - x)) dx
• = [- f.I ( 1 - x) 4 - ~ x 2 + ~ x 3 ] ~ = (- ~ + ft ) - (- f2) = - 1
1 2
19. Let C1 be the arch of the cycloid from (0, 0) to (27r, 0), which corresponds to 0 ~ t ~ 271", and let C2 be the segment from
(271", 0) to (0, 0), so 02 is given by X = 271"..:. t, y = 0, 0 ~ t :::; 271". Then c = cl u 02 is traversed clockwise, so - C is
oriented positively. Thus -C encloses the area under one arch of the cycloid and from (5) we have
A = - f_ 0
y dx = .{ 0 1
y dx + f . 0 2
y dx = f~" (1 -cos t)(l - cost) dt + f~" 0 (- dt)
= J~" (1 - 2 cost+ cos 2 t) dt + 0 = (t- 2 sin t + ~t + ~sin 2t] ~" = 371"
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