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

SECTION 10.4 AREAS AND LENGTHS IN POLAR COORDINATES 0 23
25. To find the area inside the l ~miniscate r 2 = 8 cos 29 and outside the circle r = 2,
we first note that the two curves intersect when r 2 = 8 cos 29 and ~ = 2,
. I
that is, when cos2B = ~-For -1r < B ::; 1r, cos2B = ~ {::} 2B = ±7r/ 3
r=2
or ± 511' /3 {::}
B = ±1r / 6 or ±57r / 6. The figure shows that the desired area is
4 times the area between the curves from 0 to 1r /6. Thus,
A = 4 J 0
" /. 6 [~(8 cos2B)- ~(2) 2 ] dB = 8 J" 16 0
(2 cos 2B - 1) dB
[ ] rr~ .
. =8 sin2B - B .=8(VS/2-7r/6) =4\1'3-411'/3
0
27. 3cosB=1+cosB {::} cosB=~ => B=tor-i.
A= 2 f 0
" 13 ~ [(3 cos 8?- (1 +cos 0) 2 ) dB
= J 0
"' 13 (8 cos 2 8-'- 2 cos B- 1) dB = f 0
" 13 [4(1 +cos 2B) - 2 cos B - 1) dB
8 =1!.
I 3
,l'
= J 0
.,. 13 (3 + 4cos28- 2cosB) dB= [3B + 2sin2B- 2sinf!] ~ 1 3
= 1T +v'3-.J3=1T
29. v'3cosB = sinB => J3 = sinBB => tanB = J3 => B = i·
. cos
A -
f "' 1 3 1 (sin B) 2 dB + 1'-"' 12 1 (J3 cos B) 2 d()
-Jo 2 hr/ 3 . 2
- f"'/ 3 1. 1(1 - cos28) dB+ f "/ 2 1 · 3 · 1(1 + cos2B) dB
- Jo 2 2 rr/3 2 2
- 1 [B - 1 · 28]" 13 +!! [B +.! · 28]1f 12
- 4 2 sm o 4 2 sm 7f/3
= t [ ( i - 4) - o] + ~ [ (~ + o)- ( f + 4)]
_ .l!.. _ _il + .!!: _ 1fl _ 5rr _ fl
- 12 16 . 8 16 - 24 4
r = J3cos8
31. sin 2B = cos 2B => sin 2 B = 1 => tan 28 = 1 => 28 = ~ =>
cos2B
B=. f =>
A= 8 · 2 f0"/8'~ sin 2 2B dB= 8 f 0
"' 18 H1-cos 48) dB
= 4[9- t sin48]~ 18 = 4(f - i ·1) = ~ - 1
33. sin2B = cos2B => tan2B = 1 => 2B = ~ => (} = f
A= 4 f 0 "' 18 ~sin 2B dB [since r 2 = sin 2B)
= f 0 "' 18 2 sin2B dB= [-;- cos 28] ; 18
= - ~ v'2- (- 1) = 1- ~ Y2.
.... ···B=i
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