I am calculating:
$$ \int\int_R (2ax-x^2-y^2)^{\frac{1}{2}} \, dA$$
Where $R$ is the region determined by the inside of $x^2+y^2-2ax=0$
So far, I tried using polar coordinates, wich turns the problem into;
$$ \int\int_R (2ax-x^2-y^2)^{\frac{1}{2}} \, dA=\int_{-\pi/2}^{\pi/2}\int_{0}^{2acos\theta} r(r-r^2)^{\frac{1}{2}} \, dr ~d\theta$$ $$=-1/3\int_{-\pi/2}^{\pi/2} (2acos\theta-4a^2cos^2\theta)^{\frac{3}{2}} \, d\theta$$
The thing is I haven't figured out how to compute this last integral. So I want to ask if you if what I've done so far is right, or if you would suggest a simpler method.
The region is a disk centered at $(a,0)$ with radius $a$. You can express this in polar coördinates. You might try shifting coördinates first. Complete the square in the integrand as well. Can you translate the whole thing?