Find the area between the outer and inner loops of $=2+3\sin(\theta)$
So, firstly I integrated from $\arcsin(-2/3)$ to $\pi/2$ to find the half area of the outer loop. Then, I integrated from $\arcsin(-2/3)$ to $3\pi/2$ to find the half area of the inner loop.
Area = (Half outer area - half inner area)*2
I got $17\pi/4$, but I'm not sure if it's right