Find the area of the region inside: r = 9 sinθ but outside: r = 1
$\displaystyle r= 9 sin(\theta)$ is a circle with center at (0, 4/2) and radius 4/2 while r= 1 is a circle with center at (0, 0) and radius 1.
The two curves intersect where [/tex]r= 1= 9 sin(\theta)[/tex] or $\displaystyle sin(\theta)= 1/9$. For $\displaystyle 0\le\theta\le\pi$, that is satified by $\displaystyle \theta= sin^{-1}(1/9)$ and $\displaystyle \theta= \pi- sin^{-1}(1/9)$.
For $\displaystyle \theta$ between those two limits, r ranges from 1 up to $\displaystyle 9 sin(\theta)$. The "differential of area" in polar coordinates is $\displaystyle r drd\theta$ so the area is given by $\displaystyle \int_{\theta= sin^{-1}(1/9)}^{\pi- sin^{-1}(1/9)}\int_{r= 1}^{9 sin(\theta)} r dr d\theta$.