polar coordinates - Find the area of the region inside: $r= 6\sin(\theta)$ but outside of $r
You want the region where $r>1$. That's the same as $\sin\theta > 1/6$. The boundary of that region is where $\sin\theta=1/6$. That happens when $\theta = \arcsin(1/6)$ and when $\theta = \frac\pi 2 - \arcsin(1/6)$....