Answer
$$\frac{{\sin 2 - 2\cos 2}}{2}\pi $$
Work Step by Step
$$\eqalign{
& \int_0^2 {\int_0^{\sqrt {4 - {x^2}} } {\sin \sqrt {{x^2} + {y^2}} } dydx} \cr
& R{\text{ has the bounds}} \cr
& 0 \leqslant y \leqslant \sqrt {4 - {x^2}} ,{\text{ }}0 \leqslant x \leqslant 2 \cr
& {\text{Is quarter of a circle of radius 2}},{\text{ centered at the origin}} \cr
& {\text{The region }}R{\text{ in polar coordinates is}} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant r \leqslant 2,{\text{ }}0 \leqslant \theta \leqslant \frac{\pi }{2}} \right\} \cr
& {\text{Where: }}x = r\cos \theta ,{\text{ }}y = r\sin \theta {\text{ and }}dydx = rdrd\theta \cr
& {\text{So}},{\text{ we have}} \cr
& \int_0^2 {\int_0^{\sqrt {4 - {x^2}} } {\sin \sqrt {{x^2} + {y^2}} } dydx} \cr
& = \int_0^{\pi /2} {\int_0^2 {\sin \sqrt {{r^2}{{\cos }^2}\theta + {r^2}{{\sin }^2}\theta } rdrd\theta } } \cr
& = \int_0^{\pi /2} {\int_0^2 {\left( {\sin r} \right)rdrd\theta } } \cr
& {\text{Integrating by parts with respect to }}r,{\text{ we obtain}} \cr
& x\sin x \cr
& = \int_0^{\pi /2} {\left[ {\sin x - x\cos x} \right]_0^2} d\theta \cr
& = \int_0^{\pi /2} {\left[ {\left( {\sin 2 - 2\cos 2} \right) - \left( {\sin 0 - 0\cos 0} \right)} \right]} d\theta \cr
& = \left( {\sin 2 - 2\cos 2} \right)\int_0^{\pi /2} {d\theta } \cr
& = \left( {\sin 2 - 2\cos 2} \right)\left( {\frac{\pi }{2}} \right) \cr
& = \frac{{\sin 2 - 2\cos 2}}{2}\pi \cr} $$
