integrate $\iint\limits_D \frac{2a}{\sqrt{4a^2-x^2-y^2}}dxdy, D: (x-a)^2+y^2 \le a^2$ $$I=\iint\limits_D \frac{2a}{\sqrt{4a^2-x^2-y^2}}dxdy, D: (x-a)^2+y^2 \le a^2(a>0)$$
The difficulty is to find a proper and easy way to solve this double integrals.
If do it like this, $ 0\le x \le 2a, -\sqrt{2ax-x^2} \le y \le \sqrt{2ax-x^2}$, wolfram calculation exceeds the standard time.
Maybe using polar coordinate is easier? $x-a=r\cos\theta, y=r\sin\theta(-\frac\pi2 \le \theta \le \frac\pi2, 0\le r \le 2a)$, then $I=\int_{-\pi/2}^{\pi/2} d\theta\int 2a/\sqrt{4a^2-(a+r\cos\theta)^2-(r\sin\theta)^2}rdr$ which looks rather complex too.
Am I doing it wrong? How to integrate this $I$ ?
 A: Let $x=r\cos(\theta),y=r\sin(\theta)$. We know that $0\le x\le 2a,-a\le y\le a$, hence $0\le r\le 2a$ and $\displaystyle -\frac{\pi}{2}\le \theta\le \frac{\pi}{2}$. The Jacobian is $r$, thus \begin{align*}\iint\limits_D \frac{2a}{\sqrt{4a^2-x^2-y^2}}\text{d}x\text{d}y&=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int\limits_{0}^{2a}\frac{2ar}{\sqrt{4a^2-r^2}}\text{d}r\text{d}\theta=\begin{bmatrix}4a^2-r^2=u\\-2r\text{d}r=\text{d}u\end{bmatrix}\\&=\pi a\int\limits_{0}^{4a^2}\frac{1}{\sqrt{u}}\text{d}u=\Big.2\pi a\sqrt{u}\Big\vert_{0}^{4a^2}=4\pi a^2\end{align*}
A: Believe it or not, it's actually easier without polar coordinates.
$\int_{0}^{2a}\int_{-\sqrt{2ax-x^2}}^{\sqrt{2ax-x^2}}\frac{2a}{\sqrt{4a^2-x^2-y^2}}dydx = \int_{0}^{2a}2a\arctan{\frac{y}{\sqrt{4a^2-x^2-y^2}}}\Big\vert_{-\sqrt{2ax-x^2}}^{\sqrt{2ax-x^2}}dx$ 
$= 4a\int_{0}^{2a}\arctan{\frac{\sqrt{-2x(x-2a)}}{2\sqrt{-a(x-2a)}}} = 4a\int_{0}^{2a}\arctan{\sqrt{\frac{x}{2a}}}dx = 4ax\arctan{\sqrt{\frac{x}{2a}}}+8a^2\arctan{\sqrt{\frac{x}{2a}}}-4a^2\sqrt{\frac{2x}{a}}\Big\vert_{0}^{2a} = 4a^2(\pi-2)$
