# Calculate double integral of …

I was doing a homework problem but now I'm stuck. The problem says:

Calculate $\iint_{S} \frac{dx dy}{\sqrt{2a - x}}$ where S is a circle of radius $a$ which is tangent to to both coordinate axes and is in the first quadrant

The cartesian equation for a circle of radius a, and center C(a,a) is:

$$(x-a)^{2}+(y-a)^{2} = a^{2}$$

Sketching this equation for different values of $a$ I obtained the limits of integration for the double integral:

$$4\int_{a}^{2a} \int_{a}^{a+\sqrt{2ay-y^{2}}} \frac{1}{\sqrt{2a-x}}dx dy$$

It is multiplied by 4 because that double integral only calculates a quarter of the desired area. Then evaluating the double integral I obtained this:

$$4 \left ( 2a^{3/2}+2\int_{a}^{2a} \sqrt{a+\sqrt{2ay-y^{2}}} dy \right )$$

But I was stuck on this (I've tried to put the indefinite integral$\int \sqrt{a+\sqrt{2ay - y^{2}}} dy$ into Wolfram Alpha and I obtained this ), then I said "why I don't use polar coordinates?" and I used them, and I obtained this double integral:

$$4\int_{0}^{\pi /2} \int_{0}^{a}\frac{r}{\sqrt{2a-rcos\theta}} dr d\theta$$

I evaluated it, and I obtained:

$$4\left ( \int_{0}^{\pi/2} -\frac{2}{3}\left ( \sec ^{2} \theta \sqrt{2a-\arccos {\theta}} \left ( 4a + acos\theta \right ) \right ) d\theta \right )$$

But Wolfram says this for the indefinite integral so... What can I do? Am I wrong in something? Is there another way to do it? I would appreciate any help.

\begin{align} \iint_S \frac{dx\,dy}{\sqrt{2a-x}} &= \int_0^{2a} \frac{1}{\sqrt{2a-x}}\left(\int_{a-\sqrt{a^2-(x-a)^2}}^{a+\sqrt{a^2-(x-a)^2}}dy\right) dx\\ &= \int_0^{2a}\frac{2\sqrt{2ax-x^2}}{\sqrt{2a-x}}\,dx\\ &= \int_0^{2a} 2\sqrt{x}\,dx. \end{align}