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.
