Answer to Question #336502 in Calculus for PEAC_BOY

Question

1.Find the area of the region that lies inside the circle r =3sinx and outside the cardioid

r = 1+sinx

Answer 1

I will refer to the curves as r(t) (where t - polar angle) to avoid confusion with Cartesian coordinates. I.e.

circle: r=3sin(t), cardioid r=1+sin(t)

Area in polar coordinates

$1/2\int r^2(t)dt$

Draw simple chart

Region area

$A = 1/2\int_{t_1}^{t_2} (3\sin(t))^2 - 1/2\int_{t_1}^{t_2} (1+\sin(t))^2dt\\ = \int_{t_1}^{t_2} 4\sin^2(t) - \sin(t) -1/2 dt$

Find integration bounds as intersections of curves

$3\sin(t) = 1+ \sin(t), 0 \le t \le 2\pi\\ \sin(t) = 1/2\\ t_1 = \pi/6, t_2=5\pi/6$

Integrate

$A= \int_{\pi/6}^{5\pi/6}4\sin^2(t) - sin(t)dt - 1/2dt\\ = \Big[2t -\sin(2t) +\cos(t) - 1/2t\Big]_{\pi/6}^{5\pi/6}\\ = 3/2(5\pi/6 -\pi/6) - (\sin(5\pi/3) - \sin(\pi/3)) + (\cos(5\pi/6) - \cos(\pi/6))\\ = \pi +2\sin(\pi/3) - 2\cos(\pi/6)\\ = \pi + 2\sqrt{3}/2 - 2\sqrt{3}/2\\ = \pi \approx 3.1416$

Answer:  3.1416