Evaluate the double integral $$\int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} (x^2+y^2+\sin(\pi(x^2+y^2)))\,dy\,dx$$
*sorry if the mathjax is off, I'm new at it.
Anyways, I can use the properties of double integrals to make it
$$\int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} x^2+y^2 \,dy\,dx + \int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \sin(\pi(x^2+y^2))\,dy\,dx$$
From there I can solve the first double integral expression but I'm not sure on the second double integral expression:
$$\int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \sin(\pi(x^2+y^2))\,dy\,dx$$
A nudge in the right direction would be appreciated on how to solve this second expression.

Convert the integral into polar coordinates. Note that the region you are integrating is the semi-circle of radius lying on the right half plane. Hence, the integral becomes
$$\int_{-\pi/2}^{\pi/2} \int_0^3 (r^2 + \sin(\pi r^2)) \,r\, dr \,d \theta = \pi \cdot \int_0^3 (r^2 + \sin(\pi r^2)) \,r\, dr$$
All we need to evaluate now is $\displaystyle \int_0^3 (r^2 + \sin(\pi r^2)) \,r\, dr$.
\begin{align}
\int_0^3 (r^2 + \sin(\pi r^2)) \,r\, dr & = \int_0^3 r^3 dr + \int_0^3 r \sin( \pi r^2) \,dr = \left. \dfrac{r^4}4 \right \vert_{r=0}^{r=3} + \int_0^3 \dfrac{d( \cos(\pi r^2))}{-2 \pi}\\
& = \dfrac{81}4 - \dfrac{\cos(9 \pi) - \cos(0)}{2 \pi} = \dfrac{81}4 + \dfrac1{\pi}
\end{align}
Hence, the integral is $$\pi \left(\dfrac{81}4 + \dfrac1{\pi}\right) = 1 + \dfrac{{81}\pi}4$$
