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

SECTION 10.4 AREAS AND LENGTHS IN POLAR COORDINATES 0 25
43.
. 3.4 3 y = 2x
r= 2(}
-3
r = 1 +sinO
-1.4 1.4
- 0.3 -3
From th~ first graph, we see that the pole .is one point of intersection. By zooming in or usin~ the cursor, we find the 0-values
of the intersection points to be a:~ 0.88786 ~ 0.89 and :r - a ~ 2.25. (The first of these values may be more easily
estimated by plotting y = 1 +sin x andy = 2x in rectangular coordinates; see the second graph.) By symmetry, the total
area contained is twice the area contained in the first quadrant; that is,
A= 21o. ~(20) 2 dO+ 2[ 1r/Z ~( 1 +sin 0.) 2 d8 = 1"' 40 2 .d8 + l1r/~ [1 + 2sinU + H1- cos 2B)] d8
= (~0 3 ] ~ + [8- 2cosB + (~0 - t sin2B)J: 12 = ~a 3 + [(~+~)-(a- 2cosa + ~a: - t sin2o:)) ~ 3.4645
45. L = 1b Jr2+(d7·jd0) 2 dB= fo7r }(2cos0) 2 +(-2sinB.)2d8
= 17r V 4(cos 2 B + sin 2 B) dB = 1"' ..f4d8 = [2B]; = 21r
As a check, note that the curve is a circle of radius 1, so its circumference is 27r(1) = 21r.
= 12
"' Je 2 (0 2 + 4)d0 = 12
1f BV0 2 +4d0
Now let u = 0 2 + 4, so that du = 20d8 [Ode= t du] and
49. The curve r = cos 4 (0 / 4) is completely traced with 0 :::; () :::; 47r.
r 2 + (drfd8) 2 = [cos 4 (B/4)] 2 + [4cos 3 (B/ 4) · (-sin(B/4)) · t ] 2
= cos 8 .(B/4) + cos 6 (8/4) sin 2 (B/ 4)
= cos 6 (8/4)[cos 2 (B/4) +sin 2 (0/4)] = cos 6 (B/4)
L = f 0
41 r Jcos 6 (0/4) dO = J 0
4
"' jcos 3 .(0/4) j rj,B .
= 2 f~" cos 8 (B/4)dB [sincecos 3 (B/4) ~ 0 for O :S B $ 27r] = 8J; 12 cos 3 udu [u = tB]
= 8 J; 12 (1 - sin 2 u) cosudu = 8 f 0 1 (1- x 2 ) dx
= 8[x- ~x 3 ) ~ = 8(1- ~) = lf
x sinu, ·]
[ dx = cosudu
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