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

SECTION 15.4 DOUBLE INTEGRALS IN POLAR COORDINATES 0 259
17. In polar coordinates the circle (x- 1} 2 + y 2 = 1 ¢} x 2 + y 2 = 2x is r 2 = 2r cos 0 ~ 1' = 2 cos 0,
and the circle x 2 + y 2 = 1 is r = 1. The curves intersect in the first quadrant when
2 cos 0 = 1 ~ cos 0 = ~ · .~ 0 = 7r / 3, so the portion of the region in the first quadrant is given by
D = {(r, 0} 11 ~ r ~ 2 cos 0, 0 ~ 0 ~ 7r /2}. By symmetry, the total area
is twice the area of D:
2A(D} = 2ffvdA = 2Ja"/3 j l2coao rdrdO = 2fo"/3 ar2)~~~coao dO
O= Tr/3
,/ r=2oos(J
= J;/ 3 (4cos 2 0- 1) d(} = J;/ 3 [4 · ~( 1 +cos 20}- 1] dO
= J 0
" 13 (1 + 2 cos 20} dO= [0 + s~n20] ~;a = f + ~
21. The hyperboloid of two sheets - x 2 - y 2 + z 2 = 1 intersects the plane z = 2 when -x 2 - y 2 + 4 = 1 or x 2 + y 2 = 3. So tlie
solid region lies above the surface z = )1 + x 2 + y 2 and below the plane z = 2 for x 2 + y 2 :::; 3, and its volume is
23. By symmetry,
V = I I ( 2 - ) 1 + x 2 + y 2 ) dA = la 2 " 1av"J ( 2 - ) 1 + r2) r dr dO
:r:2 +y2 ~3
= f~" dO f ov"J (2r- rv'f+'Ti) dr = [ 0] ~.,.. [r 2 - t {1 + r 2 ?1 2 ] :
= 27r (3-i- 0 + 1) = ~7r
2
V = 2 // Ja 2 - x 2 - y 2 dA = 21a
"1aa )a 2 - r 2 rdrdO = 21az... dO loa
~+~~~ .
r Ja2 - r2 dr
26. The cone z = Jx 2 + y 2 intersects the sphere x 2 + y 2 + z 2 = 1 when x 2 + y 2 2
+ ( Jx 2 + y 2 ) = 1 or x
2
+ y 2 = ~· ~0
V = /1 ()1- x2- y2- )x2 +y2) dA = la 2 7r lal/-/2 () 1- r2 - r)rdrdO
.,2 + y2 ~ 1/ 2
= f~1r dO f 0
11 V2 (r Jl-T2- r 2 ) dr = [ 0]~ 7r [ - ~( 1 - r 2 ) 3 1 2 - ~r 3 ) :/-.12 = 271'( -~) ( ?z - 1) = f(2 - V2)
27. The given solid is the region inside the cylinder x 2 + y 2 = 4 between the surfaces z = J64 - 4x2 - 4y2
and z = - )64- 4x 2 - 4y 2 • So
J 2 J 64 - 4x2 - 4y2 dA
x2 + 11 2 s 4 +2 + 11 2J:
2 ~ 4
= 4J~ .,.. J~ ~ rdrdO = 4J:.,.. dO J: r v'16 - r 2 dr = 4 [ OJ~". [-1(16- r 2 ) 3 1
V = I I [ J 64 - 4x 2 - 4y 2 - ( --: J 64 - 4x 2 - 4y 2 ) ] dA = J
= 8Tr( - ~)( 12 3 1 2 -16 2 '
3 ) = 8
; .(64 - 24 v'3)
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