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

254 0 CHAPTER 15 MULTIPLE INTEGRALS
27.
y
(0,3)
V = f 0
2
f~ - ~ "'
(6- 3x - 2y)dydx
r2( 2] v= 3- 1x
= Jo 6y - 3xy - y v = o 2 dx
= J; (6(3 - ~x)- 3x(3 - ~x) - (3 - ~ x) 2 ] dx
= J; (~x 2 - 9x + 9) dx = [~x 3 - ~ x 2 + 9x] ~ = 6 -0 = 6
29.
31.
y
1~ 11 [ V = 2 ] Y=~
y dy dx = 1L . dx
1 1
0 0 0 2 y= O
{ 1 1 - X 2 1 1 3 1 1
= Jo -2- dx = 2 [x - 3·-c Jo = 3
X
33.
From the graph, it appears that the two curves intersect at x = 0 and
at x ~ 1.213. Thus the desired integral is
Jf dA rl.213J3x- x 2 d .J. fl.213 [ ] y = 3 2
"'- "'
D X ~ Jo x•l X y u.'C - Jo xy d
X
y = x4.
r
= Jo L213( 3 X -X - X X= X - 4X - aX O
~ 0.713
2 3 6) d [ 3 1 4 1 G] 1.213
35. The two bounding curves y = 1 - x 2 and y = x 2 - 1 intersect at (±1, 0) with 1 - x 2 ~ x 2 - 1 on (-1, 1]. Within this
region, the p la n~ z = 2x·+ 2y + 10 is above the plane z = 2 - x - y, so
V = J_ •r1-x2 ( ) rl 1 Ji-~ 2 ( )
1
J,. 2 _ 1
2x+2y + 10 dydx -, _ 1
.,2 _ 1
2-x-y dydx
J •l J1-x2 ( ' ( )) •
= _ .,2_ 1 1
2x + 2y + 10- 2- x - y dy dx
= j _ 1
, 2 _ 1
(3x + 3y + 8) dy dx = f_ 1
3xy + 2Y + 8y y=z _ 2
dx
1
•l fl-:c2 1 [ 3 2 J y = l - :c2
= J~ 1 [3x(l- x 2 ) + ~(1 - x 2 ) 2 + 8(1- x 2 ) - 3x(x 2 - 1) - ~(x 2 - 1) 2 - 8(x 2 - 1)] dx
1 "3 1
= J (-6x - 16x 2 + 6x + 16) dx = [-.:!x 4 - .l&x 3 + 3x 2 + 16x]
- 1 2 3 -1
= -~ - ~ 6 + 3 + 16 + ~ - ¥ - 3 + 16 = 6 3 4
@) 2012 C