Evaluate double integral: Evaluate:
$$\iint_{D} \arcsin(x^2+y^2)\, dx\,dy$$
where $D$ is defined by the following polar equation $\rho=\sqrt{\sin \theta}$ and $0\le\theta\le\pi$
 A: change it into polar, it changes into 
$$\int_0^\pi \int_0^{\sqrt{\sin\theta}} \arcsin(r^2) r dr d\theta$$
Just evaluate this integral first
$$\int_0^{\sqrt{\sin\theta}} \arcsin(r^2) r dr$$
you can evaluate the indefinite integral using substitution and by parts as 
$$\frac 1 2 \int \arcsin(r^2) d(r^2) = \frac 1 2  \left( \sqrt{1-r^4}+r^2 \sin ^{-1}(r^2) \right)$$
Putting bounds $r = \sqrt{\sin \theta}$ and $r=0$ gives you $\displaystyle \frac 1 2 \left( |\cos \theta| + \arcsin( \sin \left(\theta\right)) \sin\theta \right) - \frac 1 2$ which you will get 
$$\int_0^\pi \frac 1 2 \left( |\cos \theta| + \arcsin( \sin \left(\theta\right)) \sin\theta  - 1\right) d\theta  = \frac{4 - \pi}{2}$$
Note that $\cos \theta$ is negative on $\pi/2 \to \pi$ but $\sqrt{1 - \sin^2\theta}$ is always positive. I think you can easily integrate $\theta \sin \theta$ by parts. For $|\cos \theta|$ you can use that fact that 
$$\int_0^{\pi}|\cos\theta|d\theta = 2 \int_0^{\pi/2}\cos\theta d\theta = 2$$
Since $\cos $ is symmetric at middle for interval $(0, \pi)$.
For integral you can use the same argument
$$\int_0^{\pi} \arcsin( \sin \left(\theta\right)) d\theta = 2 \int_0^{\pi/2} \theta \sin \theta d\theta = 2$$
