Finding area between polar curves using double integral The question asks to find the area inside $r = 1 + \sin\theta$ and outside $r = 2 \sin\theta$ using double integrals.

In my attempt, I found the intersection to be $\theta = \frac{\pi}{2}$.

I bounded $r$ as $[2 \sin \theta, 1 + \sin \theta]$, since I couldn't find another intersection for the curves, I bounded $\theta$ as $[-\frac{\pi}{2}, \frac{\pi}{2}]$ to solve for half of the area required. 

Then multiplied everything by $2$ to get the full area. I ended up with $\frac{-\pi}{2}$ while the correct answer is $\frac{\pi}{2}$.

What am I doing incorrectly?
The polar equation $$r(\theta):=2\sin\theta\qquad(0\leq\theta\leq 2\pi)$$ describes a circle of radius $1$ centered at $(0,1)$, covered twice. On the other hand, the area inside the curve $$r(\theta):=1+\sin\theta\geq0\qquad(0\leq\theta\leq 2\pi)$$ is given by
$${1\over2}\int_0^{2\pi}(1+2\sin\theta+\sin^2\theta)\&gt;d\theta={1\over2}(2\pi+0+\pi)={3\pi\over2}\ .$$
The area $A$ in question therefore has the value ${\displaystyle{\ 3\pi\over2}-\pi={\pi\over2}}$.

If you want to compute $A$ in terms of a double integral you should stay with polar coordinates. You then obtain
$$A=2\left(\int_{-\pi/2}^0\int_0^{1+\sin\theta} r\&gt;dr\ d\theta+\int_0^{\pi/2}\int_{2\sin\theta}^{1+\sin\theta} r\&gt;dr\ d\theta\right)\ .$$
$$\frac12\int_\frac{-\pi}{2}^\frac{\pi}{2}((1+sin\theta)^2-(2\sin\theta)^2)\,d\theta$$

$$= \frac{\pi}{4}$$

$$=\frac{\pi}{4} \cdot 2$$

$$=\frac{\pi}{2}$$