Answer
$=\frac{16}{9}$
Work Step by Step
$\int_{0}^{2} \int_{0}^{\sqrt{2 x-x^{2}}} \sqrt{y^{2}+x^{2}} d y d x$
convert to polar coordinates $R=\left\{(r, \theta): 0 \leqslant \theta \leqslant \frac{\pi}{2}, 0 \leqslant r \leqslant 2 \cos \theta\right\}$
so
$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2 \cos \theta} \sqrt{r^{2}} r d r d \theta=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2 \cos \theta} r^{2} d r d$
integrate
$=\int_{0}^{\frac{\pi}{2}}\left[\frac{r^{3}}{3}\right]_{0}^{2 \cos \theta} d \theta$
$=\frac{1}{3} \int_{0}^{\frac{\pi}{2}} 8 \cos ^{3} \theta d \theta=\frac{8}{3} \int_{0}^{\frac{\pi}{3}} \cos ^{2} \theta \cos \theta d \theta$
$=\frac{8}{3} \int_{0}^{\frac{\pi}{2}}\left(-\sin ^{2} \theta+1\right) \cos \theta d \theta$
integrate and evaluate
$=\frac{8}{3}\left[\sin \theta-\frac{\sin ^{3} \theta}{3}\right]_{0}^{\frac{\pi}{2}}$
$=\left(\sin \frac{\pi}{2}-\frac{\sin ^{3} \frac{\pi}{2}}{3}\right)\frac{8}{3} $
$=\left(1-\frac{1}{3}\right)\frac{8}{3}$
$=\left(\frac{2}{3}\right)\frac{8}{3}$
$=\frac{16}{9}$
