Answer
$\int_{-3}^3\int_0^{\sqrt{9-x^2}}\sin (x^2+y^2)dydx=\frac{\pi}{2}-\frac{\pi\cos 9}{2}$
Work Step by Step
The set of points $R=\{(x,y)|-3\leq x\leq 3,\leq y\leq \sqrt{9-x^2}\}$ describes the region of an upper semicircle center at the origin with the radius of $3$.
Then, in polar coordinates $R=\{(r,\theta)|0\leq r\leq 3,0\leq\theta \leq\pi\}$.
Convert the integral to polar coordinates and evaluate:
$\int_{-3}^3\int_0^{\sqrt{9-x^2}}\sin(x^2+y^2)dydx=\int_0^3\int_0^\pi \sin(r^2)\cdot rd\theta dr$
$=\int_0^3\theta r\sin(r^2)]_0^{\pi}dr$
$=\int_0^3\pi r\sin(r^2)dr$
$=-\frac{\pi\cos(r^2)}{2}]_{0}^3$
$=\frac{\pi}{2}-\frac{\pi\cos 9}{2}$
