<s>[INST] 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.
 [/INST] 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 $$\pi \left(\dfrac{81}4 + \dfrac1{\pi}\right) = 1 + \dfrac{{81}\pi}4$$
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