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