<s>[INST] 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 $I$ ?
 [/INST] Believe it or not, it's actually easier without polar coordinates.
$\int_{0}^{2a}\int_{-\sqrt{2ax-x^2}}^{\sqrt{2ax-x^2}}\frac{2a}{\sqrt{4a^2-x^2-y^2}}dydx = \int_{0}^{2a}2a\arctan{\frac{y}{\sqrt{4a^2-x^2-y^2}}}\Big\vert_{-\sqrt{2ax-x^2}}^{\sqrt{2ax-x^2}}dx$ 
$= 4a\int_{0}^{2a}\arctan{\frac{\sqrt{-2x(x-2a)}}{2\sqrt{-a(x-2a)}}} = 4a\int_{0}^{2a}\arctan{\sqrt{\frac{x}{2a}}}dx = 4ax\arctan{\sqrt{\frac{x}{2a}}}+8a^2\arctan{\sqrt{\frac{x}{2a}}}-4a^2\sqrt{\frac{2x}{a}}\Big\vert_{0}^{2a} = 4a^2(\pi-2)$
</s>