Find the area of the region determined by two curves 
Find the area of the region $R$ given by two curves.

So the region $R$ describes the area that is common between the two curves:
$$\begin{align*}
\text{Function 1: } r&= 2\sin(\theta)\\
\text{Function 2: }r&= \frac{3}{2} - \sin(\theta).
\end{align*}$$
I have to find the area that is common between the two graphs..
What I did was just take the integral of function 1 (from 0 to $2\pi$) then subtract it from the integral of function 2 (from 0 to $2\pi$)... which gave me ($2-3\pi$) as my final answer... is this correct?
 A: The graph of $r=2\sin(\theta)$ is a circle with radius $1$ centered at $(0,1)$. It lies entirely in qu
The graph of $r = \frac{3}{2}-\sin(\theta)$ is a cardiod; call it $\mathcal{C}_2$
From $\theta=0$ to $\theta=\frac{\pi}{6}$, $\mathcal{C}_2$ is "outside" of $\mathcal{C}_1$. They intersect at $\theta=\frac{\pi}{6}$. 
From $\theta=\frac{\pi}{6}$ to $\theta=\frac{5\pi}{6}$, $\mathcal{C}_1$ is "outside" $\mathcal{C}_2$. They intersect when $\theta=\frac{5\pi}{6}$. 
From $\theta=\frac{5\pi}{6}$ to $\theta=\pi$, $\mathcal{C}_2$ is "outside $\mathcal{C}_1$.
From $\theta=\pi$ to $\theta=2\pi$, $\mathcal{C}_1$ retraces the circle above the $y$-axis, whereas $\mathcal{C}_2$ lies in the 3rd and 4th quadrant. 
So the region common to both graphs lies inside $\mathcal{C}_1$ from $\theta=0$ to $\theta=\frac{\pi}{6}$, inside $\mathcal{C}_2$ from $\theta=\frac{\pi}{6}$ to $\theta=\frac{5\pi}{6}$, and inside $\mathcal{C}_1$ from $\theta=\frac{5\pi}{6}$ to $\theta=\pi$.
This analysis tells you what integrals you need to evaluate and add 
