Answer
$\int_0^2\int_0^{\sqrt{2x-x^2}}\sqrt{x^2+y^2}dydx=\frac{16}{9}$
Work Step by Step
Notice that:
$y=\sqrt{2x-x^2}$
$y^2=2x-x^2$
$x^2+y^2-2x=0$
$(x-1)^2+y^2=1$
For $0\leq x\leq 2$, it means that $y=\sqrt{2x-x^2}$ describes an upper semicircle in quadrant I center at $(1,0)$ with the radius $1$.
Write this equation in polar coordinates:
$x^2+y^2=2x$
$r^2=2r\cos \theta$
$r=2\cos\theta$
Then, $R=\{(x,y)|0\leq x\leq 2,0\leq y\leq \sqrt{2x-x^2}\}$ can be converted to $R=\{(r,\theta)|0\leq r\leq 2\cos\theta,0\leq\theta\leq \pi/2\}$.
Convert the integral to polar coordinates and evaluate:
$\int_0^2\int_0^{\sqrt{2x-x^2}}\sqrt{x^2+y^2}dydx=\int_0^{\pi/2}\int_0^{2\cos\theta}\sqrt{r^2}\cdot rdrd\theta$
$=\int_0^{\pi/2}\int_0^{2\cos\theta}r^2drd\theta$
$=\int_0^{\pi/2}\frac{r^3}{3}]_0^{2\cos\theta}d\theta$
$=\int_0^{\pi/2}\frac{8\cos^3\theta}{3}d\theta$
$=\int_0^{\pi/2}\frac{8\cos\theta}{3}-\frac{8\sin^2\theta\cos\theta}{3}d\theta$
$=\frac{8\sin\theta}{3}-\frac{8\sin^3\theta}{9}]_0^{\pi/2}$
$=\frac{8}{3}-\frac{8}{9}$
$=\frac{16}{9}$
