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

SECTION 15.3 DOUBLE INTEGRALS OVER GENERAL REGIONS 0 253
J J 0
y dA = f~ 1
J:/ 2 y dx dy. In either case, the resulting iterated integrals are not difficult to evaluate but the region D is
more simply described as a type II region, giving one iterated integral rather than a sum of two, so we evaluate the latter
integral:
ffo ydA = f~tf:/2 ydxdy = f~t (xyJ ::~~2 dy = J~1 (y + 2- y2)ydy = f~ t (y2 + 2y - ya) dy
= [jy3+y2- h4]~1 = (! +4-4) - (-~ +1 - i) = ~
19.
0 X
= [§.y 3 - y 4 ] 2 = .§1 - 16- ~ + 1- ll
3 1 3 3 - 3
21.
-2
y
! 2 !~ (2x - y) dydx
- 2 -~
2
- /
- 2 y=-y4- x 2
- 2xy - 2Y r.--;; dx
[ 12] !1=~
= }~ 2 [2x v'4 - x 2 :- H4 - x 2 ) + 2x v'4- x2 + H4 - x 2 )] dx
=J~24xv'4-x2dx= -H4-x2)3/2]2 = 0
-2
[Or, note that 4x v'4 - x 2 is an odd function, so J~
2 4x v'4- x2 dx = 0.]
23.
y
f1 J1-a: 2 ( ) f1 [ 2] y-l-x 2
V= 10 1 _., 1- x+2y dydx= 10 y-xy+ y !1;: 1
_., dx
=
1
[ ((1- x 2 ) - x(1- x 2 ) + (1 - x 2 ?)
- ((1- x)- x(1- x) + (1- x) 2 )] dx
X
= J 0
1
[(x 4 + x 3 - 3x 2 - x + 2)- (2x 2 - 4x + 2)] dx
25.
0 y
J2J7- 3y J2 [ 1 2 ] "' 7- 3y
V =
1 1
xy dx dy = 1 2 x y ., = dy
1
0
X
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