Looks like the integral is
[tex]\displaystyle\int_{-8}^8\int_0^{\sqrt{64-x^2}}\sin(x^2+y^2)\,\mathrm dy\,\mathrm dx[/tex]
The integration region is the upper half of a circle in the x-y plane centered at the origin with radius 8. So, in polar coordinates, the integral is
[tex]\displaystyle\int_0^{2\pi}\int_0^8r\sin(r^2)\,\mathrm dr\,\mathrm d\theta=2\pi\int_0^8r\sin(r^2)\,\mathrm dr[/tex]
[tex]=\pi\displaystyle\int_0^{64}\sin s\,\mathrm ds[/tex]
where we substituted [tex]s=r^2[/tex] and [tex]\mathrm ds=2r\,\mathrm dr[/tex], which gives
[tex]=\boxed{\pi(1-\cos(64))}[/tex]
