Answer
$$\frac{2}{3}\pi $$
Work Step by Step
$$\eqalign{
& \int_0^2 {\int_y^{\sqrt {8 - {y^2}} } {\sqrt {{x^2} + {y^2}} } dxdy} \cr
& R{\text{ has the bounds}} \cr
& y \leqslant x \leqslant \sqrt {8 - {y^2}} ,{\text{ }}0 \leqslant x \leqslant 2 \cr
& {\text{The region }}R{\text{ in polar coordinates is}} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant r \leqslant 2,{\text{ }}\frac{\pi }{4} \leqslant \theta \leqslant \frac{\pi }{2}} \right\} \cr
& {\text{Where: }}x = r\cos \theta ,{\text{ }}y = r\sin \theta {\text{ and }}dydx = rdrd\theta \cr
& {\text{So}},{\text{ we have}} \cr
& \int_0^2 {\int_y^{\sqrt {8 - {y^2}} } {\sqrt {{x^2} + {y^2}} } dxdy} = \int_{\pi /4}^{\pi /2} {\int_0^2 {\sqrt {{{\left( {r\cos \theta } \right)}^2} + {{\left( {r\sin \theta } \right)}^2}} } rdrd\theta } \cr
& = \int_{\pi /4}^{\pi /2} {\int_0^2 {\sqrt {{r^2}{{\cos }^2}\theta + {r^2}{{\sin }^2}\theta } } rdrd\theta } \cr
& = \int_{\pi /4}^{\pi /2} {\int_0^2 {{r^2}} drd\theta } \cr
& {\text{Integrate with respect to }}r \cr
& = \int_{\pi /4}^{\pi /2} {\left[ {\frac{{{r^3}}}{3}} \right]_0^2d\theta } \cr
& = \int_{\pi /4}^{\pi /2} {\frac{8}{3}} d\theta \cr
& {\text{Integrate }} \cr
& = \frac{8}{3}\left( {\frac{\pi }{2} - \frac{\pi }{4}} \right) \cr
& = \frac{2}{3}\pi \cr} $$
