Answer to Question #349425 in Calculus for Esh

Question

Find the area that us inside r=4 and outside r=4-2sinθ

Answer 1

$4=4-2\sin \theta$

$\theta_1=0, \theta_2=\pi$

$A=\dfrac{1}{2}\displaystyle\int_{0}^{\pi}\bigg((4)^2-(4-2\sin \theta)^2\bigg)d\theta$

$=\dfrac{1}{2}\displaystyle\int_{0}^{\pi}\bigg(16\sin \theta-4\sin ^2\theta\bigg)d\theta$

$=\dfrac{1}{2}\displaystyle\int_{0}^{\pi}\bigg(16\sin \theta-2+2\cos 2\theta\bigg)d\theta$

$=[-8\cos \theta-\theta+\dfrac{1}{2}\sin 2\theta]\begin{matrix} \pi\\ 0 \end{matrix}$

$=8-\pi+0-(-8-0+0)$

$=16-\pi ({units}^2)$