# Thread: Polar Coordinates problem area of region

1. ## Polar Coordinates problem area of region

Find the area of the region inside: r = 9 sinθ but outside: r = 1

2. Originally Posted by muddyjch
Find the area of the region inside: r = 9 sinθ but outside: r = 1
$r= 9 sin(\theta)$ 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 $sin(\theta)= 1/9$. For $0\le\theta\le\pi$, that is satified by $\theta= sin^{-1}(1/9)$ and $\theta= \pi- sin^{-1}(1/9)$.

For $\theta$ between those two limits, r ranges from 1 up to $9 sin(\theta)$. The "differential of area" in polar coordinates is $r drd\theta$ so the area is given by $\int_{\theta= sin^{-1}(1/9)}^{\pi- sin^{-1}(1/9)}\int_{r= 1}^{9 sin(\theta)} r dr d\theta$.