Find the area of the Rose's petal. 
If a Rose leaf is described by the equation $r = \sin 3\theta$, find the area of one petal.


A sketch is useful here, but the only important observation is that $r=0$ when $\theta=0$, and again at $\frac{\pi}{3}$. These are your limits for one petal.  
Since the area of a polar curve between the rays $\theta=a$ and $\theta=b$ is given by $\int_{a}^{b}\frac{1}{2}r^{2}d\theta$, we have
$$A=\int_{0}^{\pi/3}\frac{1}{2}\sin^{2}(3\theta)d\theta=\frac{1}{2}\int_{0}^{\pi/3}\frac{1-\cos(6\theta)}{2}d\theta$$
$$=\frac{1}{4}\left[\theta-\frac{\sin(6\theta)}{2}\right]^{\pi/3}_{0}=\frac{1}{4}\left(\frac{\pi}{3}-\frac{1}{2}\sin\left(\frac{6\pi}{3}\right)\right)=\frac{\pi}{12}$$
