It is a known deduction of Green's Theorem that, for an area of a disc in a plane,
$\displaystyle \displaystyle A = \frac{1}{2}\oint_D xdy-ydx$
Find an expression of x and y in terms of $\displaystyle r=f(\theta)$. and substitute into this equation. The only problem with this method is that this does not use Green's theorem directly, only a major consequence of it. but is probably still valid.
For $\displaystyle r=3\sin{2\theta}$,
$\displaystyle \displaystyle A = \frac{1}{2}\int_0^{\pi/2} r^2 d\theta = \frac{1}{2}\int_0^{\pi/2} (3\sin{2\theta})^2 d\theta = \frac{9}{2}\int_0^{\pi/2}\sin^2 2\theta d\theta = \frac{9}{2}\int_0^{\pi/2}\frac{1-cos4\theta}{2}d\theta =$