## Calculus: Early Transcendentals 8th Edition

$\frac{11 \pi}{4}-\frac{11sin^{-1}({1/3})}{2}-3\sqrt 2$
Given: $r=1-3 sin \theta$ or, $sin \theta= \frac {1}{3}$ Area inside inner loop$,A=\frac{1}{2}\int_{sin^{-1}({1/3})}^{\pi-sin^{-1}({1/3})}(1-3sin \theta)^2d\theta$ $=\frac{1}{2}\int_{sin^{-1}({1/3})}^{\pi-sin^{-1}({1/3})}1+\frac{9}{2}-\dfrac{9 cos 2 \theta}{2}-6 sin \theta d\theta$ $=[\frac{11 \theta}{4}-\dfrac{9 sin \theta cos \theta}{4}+3 cos \theta]_{sin^{-1}({1/3})}^{\pi-sin^{-1}({1/3})}$ $=\frac{11 \pi}{4}-\frac{11sin^{-1}({1/3})}{2}-3\sqrt 2$