Hello, casesam!
Find the area inside $\displaystyle r\:=\:18\sin\theta$, but outside $\displaystyle r\:=\:2$ A sketch might help . . . Code:
18
* * *
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:::::::ooo:::::::
*:::o  o:::*
*:o  o:*
*  *
o***o
 2
o  o
o  o
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The graphs intersect where: .$\displaystyle 18\sin\theta \:=\:2 \quad\Rightarrow\quad \sin\theta \:=\:\tfrac{1}{9}$
Hence: .$\displaystyle \theta \;=\;\sin^{1}(\tfrac{1}{9}) \;\approx\;\begin{Bmatrix}0.11134 \\ 3.03025 \end{Bmatrix} $
Due to the symmetry, we can integrate from 0.11134 to $\displaystyle \tfrac{\pi}{2}$ . . . and double.