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?
 A: The correct integral expressions are
$$
A_s=\frac{1}{2} \int_{2\pi/3}^{\pi} [3\sin(t)]^2 dt=\frac{3\pi}{4}-\frac{9\sqrt{3}}{16}
$$
$$
A_l=\frac{1}{2} \int_{\pi/2}^{2\pi/3}[-3\sqrt3 \cos(t)]^2 dt =\frac{9\pi}{8}-\frac{27\sqrt{3}}{16}
$$
$$Area=A_s+A_l =\frac{15}{8}\pi-\frac{9}{4}\sqrt{3}.$$
The integration ranges are specified incorrectly in the post.
———————
For the bounded area, it is often easier just to visualize and quantify how the two circles overlap, as opposed to performing integration. 
The bounded area is enclosed by two arcs, one spanning $\pi/3$ on the larger circle and $2\pi/3$ on the smaller one. The area is made up of two parts, separated by the line joining the two intersecting points.
The contribution from the larger circle is the difference between 1/6 of the full circle area and an overlaid 60-60-60 equilateral triangle, i.e.
$$A_l=\frac{1}{6}\pi r_l^2 - \frac{\sqrt{3}}{4}r_l^2$$
Similar, the contribution from the smaller circle is 1/3 of the full circle area and an overlaid 30-30-120 isosceles triangle
$$A_s=\frac{1}{3}\pi r_s^2 - \frac{\sqrt{3}}{4}r_s^2$$
$r_l=3\sqrt{3}/2$ and $r_s=3/2$ are the radii of the two circles, respectively. 
Adding the two contributions together, 
$$Area=\frac{\pi}{6} (r_l^2+2r_s^2)-\frac{\sqrt{3}}{4} (r_l^2+r_s^2)=\frac{15}{8}\pi-\frac{9}{4}\sqrt{3}.$$
A: 
$$
\text{Area}_1=\frac12\,\frac\pi3\left(\frac{3\sqrt3}2\right)^2=\frac{9\pi}8
$$

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

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

$$
\begin{align}
\text{Area}
&=\text{Area}_1+\text{Area}_2-\text{Area}_3\\[3pt]
&=\frac{9\pi}8+\frac{3\pi}4-\frac{9\sqrt3}4\\
&=\frac{15\pi}8-\frac{9\sqrt3}4
\end{align}
$$
A: Your question is unclear. Do you have two problems? In that case, the answers for the disks are trivial: $\pi (3/2)^2$ and $\pi(3\sqrt{2}/2)^2$. 

If instead you have a single problem and need to find the area between the two curves, then you must recast the functions so they intersect at the same value  and  and set up a single integral for the area between them. 
