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

SECTION 10.4 A~~S AND LENGTHS IN POLAR COORDINATES 0 21
3. r 2 = 9sin28, r ~ 0, ·a~ 8::; 7rj2.
1
~/2 1~ /2 / 2
A - lr
2 2 d8
- 1(9sin2B 2 )d8 -l!. -2 [- lcos2B)~ 2 0 ---l!.(- 4
1 - 1)=l!.
2
0 . 0
7. 1' = 4 + 3sin8, -%::; () ::; %·
1
~/2
= ~ (16 + 9sin 2 8) dO
-rr/2
= ~·2} 0
rr/ 2
r /2
[by Theorem 4.5.6(b) [ET 5.5.7(b)]]
[16 + 9 ·~( 1 -cos28))d~ [by Theorem 4.5.6(a) [ET 5.5.7(a)]] .
= 1 ("; - ~cos28) d8 = [~ 1 8 - tsin2BJ; 12 = ( 4 ~" - O)- (0 _ 0) = 4~"
'
9. The area is bounded by r = 2 sin 8 for 8 = 0 to 8 = 7r.
(2, 'TT/ 2)
A= 1"' ~r 2 d8 := ~ 1 -rr (2 sinfJ) 2 dB= ~ 1"' 4sin 2 Bd8
=21" ~( 1 - cos2fJ)dB ~ [e - ~sin2eJ: =1r
Also, note that th is is a circle with radius 1, so its area is 1r(1) 2 = 1r.
" ~r 2 2
d8 = l
11. A = 12
rr H3 + 2cos8) 2 d8=! fo 2 rr(9+12.cos8+4cos 2 8)d8
= ~ 12.,.. [9 + 12cos8 + 4 · t(1.+ cos 28)] dB _(_l ,_'TT--:) H-=-- - --'-:..._:..1----+
{2-rr
= ~ lo (11 + 12 cos 8 + 2 cos 28) d8 = Hne + 12 sin O+ sin 28) ~71'
= H227r) = 111r
13. A = r 71' ~r 2 d8= r rr t{2 + sin4fJ) 2 dfJ = ~ r"(4+4si~4fJ+sin 2 4fJ)d8
Jo lo · Jo ·
3
= ~ fo 2 " I4 + 4sin48 + ~(1- cos88)] dB
2"
= ~ (~ + 4 sin48 - ~ cos88) dB=~ [~B- cos4fJ-
1
fs- sin8B] ~"'
0 .
= ~[(97r- 1) - (- 1)) = ~7('
® 2012 Ccngagc Learning. All RighiS R=:rvod. May not be sconncd, copied, or duplicotcd, or posted to a publicly accessible website, in whole or in pan.