Area boundes by two circles 
Find the area bounded by $r \le 3\sin(t)$ and $r \le -3 \sqrt3 \cos(t)$. 

Well, I've found the intersection point, which is $t=\frac{2 \pi}{3}$ . Then, I set up two integrals
$$
A=\frac{1}{2} \int_{\pi/2}^{2\pi/3} (3\sin(t))^2 dt
$$
and
$$
A=\frac{1}{2} \int_{2\pi/3}^{\pi}(-3 \sqrt3 \cos(t))^2 dt
$$
Is that correct?


$$
\text{Area}_1=\frac12\,\frac\pi3\left(\frac{3\sqrt3}2\right)^2=\frac{9\pi}8
$$

$$
\text{Area}_2=\frac12\,\frac{2\pi}3\left(\frac32\right)^2=\frac{3\pi}4
$$

$$
\text{Area}_3=\frac32\,\frac{3\sqrt3}2=\frac{9\sqrt3}4
$$

$$
\begin{align}
\text{Area}
&=\text{Area}_1+\text{Area}_2-\text{Area}_3\\[3pt]
&=\frac{9\pi}8+\frac{3\pi}4-\frac{9\sqrt3}4\\
&=\frac{15\pi}8-\frac{9\sqrt3}4
\end{align}
$$
