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

258 D CHAPTER 15 MULTIPLE INTEGRALS
15.4 Double Integrals in Polar Coordinates
1. The'region R is more easily described by polar coordinates: R = { (r, B) I 0 $ r::; 4, 0::; B::; 3 ; }.
Thus JJR f~x, y) dA = J:'~~' 12 4
f 0
f(r cos fJ, r sinO) r dr dB.
3. The region R is more easily described by rectangular coordinates: R = { ( x, y) I -1 :=:; x :=:; 1, 0 :=:; y :=:; ~ x + ~}.
Thus JJRf(x,y) dA = J~ 1 J 0
("'+ 1 )/ 2 f(x,y) dydx.
5. The integral J:;t J 1
2
r dr dB represents the area of the region .
R = {(r,B) 11::; r::; 2, 7r/4 $ '0::; 37r/4}, the top quarter portion of a
ring (annulus):
0=31T
4
y
J:; t j 1
2 r dr dO = ( J:; t d()) ( f 1
2 r dr)
= [B]3'11'/4 [!r2]2 = {3.,. _ .!!.) . 1 (4 _ 1) = 1!.. i! = 3'11'
'II' /4 2 - 1 4 4 2 2 2 4
X
7. The half disk D can be described in polar coordinates as D = { ( r, B) I 0 ::; r ::; 5, 0 ::; B ::; 1r}. Then
ffv· x 2 y dA = J 0
" J; (r cos B) 2 (r sin B) r dr dB= (J 0
.,. cos 2 Osin B dB) ( J~ r 4 dr)
= [-% cos 3 OJ~ [ir 5 ]~ = -%(-1-1) · 625 = 12 i 0
9. JJR sin(x 2 + y 2 ) dA = JJ:/ 2 J{ sin(r 2 ) r dr dB= (! 0
.,. 12 dO) (J 1
3
r sin(r 2 ) dr)
= [0]~ 12 [-~ cos(r 2 )]~
= (:~) [-~(cos9- cos 1)] = -;f(cos 1 - cos9)
_ ,2_ u2
11 dA- J'll'/ 2 f2 -r2 d dB- J"'/2 dB r2 -r2 d
. f·r
Jn e - -'11'/2 Jo e r r - -'11'/2 Jo re r
, = [B]"'/ 2 [-le-r 2 )
2
= 1r{-l)(e-
4 - e 0 ) = 1!.(1- e- 4 )
-'11'/2 2 0 2 2
13. R is the region shown in the figure, and can be described
by R = {(r,fJ) I 0::; B ::; rr/4, 1$ r::; 2}. Thus
JJR ~ctan(y/x) dA = f0 '~~' 14 f 1
2 arctan( tan B) r dr dO since yjx =tan 0.
Also, arctan( tan B) = B for 0 ::; B ::; 1r /4, so the integral becomes
. f":r/4 f20 d -'"- f'11'/ 4"d(} f2 d - [!02)"'/4 [l 2]2 - .,-2. ~- _1_ 2
.lo 1 r r uu - Jo u 1 r r - 2 o 2 r 1 - 32 2 - a.t7r ·
15. One loop is given by the region
D = {(r,O) 1-rr/6 S 0 $ rr/6, 0 S r $ cos30}, so the area is
11 !
7r /61cos SO . ~.,. /6 [ 1 2
] r=coo 30 _
dA = rdrdO =
2 r dO
D -'11'/6 0 -'11'/6 r=O
= {"' 16 _! cos 2 30 dfJ = 2 f" 16 _! ( 1 +cos 60 ) dO
}_71: 16 2 } 0 2 2
y
y=x
1[ 1 ]"'/ 6
sin 60
0
= 2
0 + G
= ;;
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