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

SECTION 15.3 DOUBLE INTEGRALS OVER GENERAL REGIONS D 255
37. The solid lies below the plane z = 1 - x - y
or x + y + z = 1 and above the region
y
D={(x,y)IO~x~l,O~y~ 1 -x}
in the xy-plane. The solid is a tetrahedron.
X ,
X
39. The two bounding curves y = x 3 - x andy = x 2 + x intersect at the origin and at x = 2, with x 2 + x > x 3 - x on (0, 2).
· Using a CAS, we find that the volume is
V = Z y X= X y . XY y -X= "
1o x3- x 0 xa - :z: 14,549,35
21x2 +x d d '121:r:2 +:r:( 3 4 ,+ 2) d d 13,984,7:'!5,616
41 . The two surfaces intersect in the circle x 2 + y2 = 1, z = 0 and the region of integration is the disk D : x 2 + y
2
~ 1.
43.
Using a CAS, the volume is (1- x
!In 2 1 11~
- y 2 ) dA = . (1 - x 2 - y 2 ) dydx = ~.
' D - 1 -Vl-x2 2
y
Because the region of integration is
D = {(x,y) I 0 ~ x ~ y, 0 ~ y ~ 1} = {(x, y) I x ~ y ~ 1, 0 ~ x ~ 1}
we have f 0
1
J~ f(x, y) dx dy = ffv f(x, y) dA = f 0
1
J: f(x, y) dy dx.
45.
Because the region of integration is
D = {(x,y) I 0 ~ y ~ cosx,O ~ x ~ 1rj2}
= {(x,y) I 0 ~ x ~ cos- 1 y, O ~ y ~ 1}
we have
foTr/2 focou f(x,y)dydx = ffv f(x, y)dA = fol focos-lv f(x,y)dx dy.
47.
49.
In 2
y
Because tlie reg_ion of integration is -
D = {(x, y) I 0 ~ y ~' Inx, 1 ~ x ~ 2} = {(x,y) J e 11 ~ x ~ 2, 0 ~ y ~ ln2}
we have
/,
211" x ~~ 11n 2[2
f(x,y).dydx = f(x,y)dA = f(x,y) dxdy
l 0 D 0 cV
113
2
2 131:r./3 2 13 [ ]u=x/3
ex dxdy = ex dydx = ex y dx
o 3u o o o u=O
(3, 1)
X
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