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CHAPTER 35
35.69 Find the area inside the circle r = sin 0 and outside the cardioid r=l— cos 8.
Fig. 35-17
35.70
Find the area swept out by the Archimedean spiral r = 0 from 8 = 0 to 0 = 2ir.
35.71
Find the area swept out by the equiangular spiral r = e° from 9 = 0 to 0 = 2-rr.
35.72
35.73 Find the centroid of the right half of the cardioid r = 1 + sin 0. (See Problem 35.36.)
By Problem 35.55 (and a rotation), the area A is 37r/4. The region in question is swept out from 0 = — -n-12
to
0=77/2. Hence,
je = 16/977. In a similar manner, one finds y = |.
35.74
Sketch the graph of r
2 = 1 + sin 6.
See Fig. 35-18; there are two loops inside a larger oval.
For 0<0<7r/2, sin
2 0 = 1 -cos
2 Sal -cos0>(l- cos 0)
2
, so that sin0al-cos0 and the desired area is as shown in Fig. 35-17.
The area A is
The area A is
\ J0"'2 sin2 28 d0 = \ J0"'2 |(1 - cos 40) dO=$(6- $ sin 40) ]^'2 = 77/8. The general
formulas for the centroid (x, y) are Ax = | Je^ r3 cos 0 ^0 and Ay = 5 Js*2 r3 sin 0 d0, where yl is the
area. In this case, we have (7r/8)f = j J 0 "'
2 sin
3 20 cos 0 d0 = § J n "'
2 sin
3 0 cos
4 0 d0 = § J 0 "'
2 (1 - cos
2 0) -
cos"0sin0d0 = § So'
2 (cos
4 0 sin 0 - cos
6 0 sin 0) d0 = |(-± cos
5 0 + $ cos
7 0) ]£'
2 = -|(-^ + ^) = ^.
Hence, jc = 128/105u-. By symmetry, y = 128/10577.
Find the centroid of the region inside the first-quadrant loop of the rose r = sin 26 (Fig. 35-6).
Thus,