<s>[INST] Using change of coordinates to find the exact value of an integral Use an appropriate change of coordinates to find the exact value of the integral
$$\int_{-\sqrt{3}}^{\sqrt{3}}\int_{-\sqrt{3-x^2}}^{\sqrt{3-x^2}}\int_{-3+x^2+y^2}^{3-x^2-y^2}x^2dzdydx$$
My work so far:
$-3+x^2+y^2\leq z\leq 3-x^2-y^2$
$-\sqrt{3-x^2}\leq y\leq \sqrt{3-x^2}$
$-\sqrt{3}\leq x \leq \sqrt{3}$
I can see that $y$ includes the entire circle, so that means
$0 \leq \theta \leq 2\pi$
Similarly, graphing the circle centered at the origin with radius $\sqrt{3}$, I can see that $0\leq r \leq \sqrt{3}$
I just need to change $z$ to polar now:
$3-x^2-y^2=3-r^2$
$-3+x^2+y^2=-3+r^2$
So the integral becomes:
$$\int_{0}^{2\pi}\int_{0}^{\sqrt{3}}\int_{-3+r^2}^{3-r^2} r(r\sin\theta)^2 dz dr d\theta$$
$$=\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} r^3\sin^2\theta(3-r^2-(-3+r^2))drd\theta$$
$$=\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} r^3\sin^2\theta(6-2r^2)drd\theta$$
$$=\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} \sin^2\theta(6r^3-2r^5) drd\theta$$
$$=\int_{0}^{2\pi} \frac{9}{2}\sin^2\theta d\theta$$
$$=\frac{9\pi}{2}$$
Is my solution correct?
 [/INST] Yes, I confirm that your calculation is correct.

By symmetries, one can either use the cylindrical coordinates for the triple integral or the polar coordinates after integrating with the variable $z$.
The iterated triple integral can be written as
$$
J=\iint_D (6-2x^2-2y^2)x^2\;dxdy
$$
where $D=\{(x,y)\mid x^2+y^2\le 3\}$. Applying the polar coordinates, one has
$$
\begin{align}
J&=\int_{0}^{2\pi}\int_0^{\sqrt3} (6-2r^2)r^3\cos^2(\theta) \;drd\theta\\
&=\left(\int_0^{\sqrt3} (6-2r^2)r^3\;dr\right)\cdot
\left(\int_{0}^{2\pi} \cos^2\theta\;d\theta\right)\\
&=\frac{9}{2}\cdot \int_0^{2\pi}\frac{1+\cos(2\theta)}{2}\;d\theta\\
&=\frac{9\pi}{2}
\end{align}
$$
</s>