Evaluating a Double Integral $\iint \sqrt{x(2a-x)+y(2b-y)}$ Please help me in evaluating the double integral
$$\iint \sqrt{x(2a-x)+y(2b-y)}$$
over the region bounded by the circle $x^2+y^2-2ax-2by=0$.
My thought was to change the variable by substituting $x=r(1+cos \theta)$ and $y=r(1+sin \theta)$. so that $0\leq r\leq \sqrt{a^2+b^2}$ and $0\leq \theta \leq 2\pi$. But i could not evaluate the outcome of the substitution. Any help in this regard is highly appreciated. Thank you in advance.

Completing the square we have that the integrand is really
$$\sqrt{2ax-x^2+2by-y^2} = \sqrt{a^2+b^2-(x-a)^2-(y-b)^2}$$
which, when integrated over the circle, $$(x-a)^2+(y-b)^2=a^2+b^2$$ gives half the volume of the sphere $$(x-a)^2+(y-b)^2+z^2=a^2+b^2$$ Thus the answer is
$$I = \frac{2\pi}{3}(a^2+b^2)^{\frac{3}{2}}$$
