Calculate $\displaystyle \iiint_V x^2 \,dx\, dy\, dz$ where $V=\{(x, y, z)|z^2\le \frac{x^2}{4}+\frac{y^2}{9}\le 2z\}$ Calculate $\displaystyle \iiint_V x^2 \,dx\, dy \,dz$, where $V=\{(x, y, z)\mid z^2\le \frac{x^2} {4} + \frac{y^2}{9}\le 2z\}$.
I made the substitution $x=2r\cos\theta$, $y=3r\sin \theta$ and $z=z$. From here I got that $z^2\le r^2\le 2z$, which means that $z\le r \le \sqrt{2z}$ and also implies that $0\le z \le 2$. Furthermore, $0\le \theta \le 2\pi$ (this may also be seen if we draw a picture), so
$$\iiint_V x^2 \, dx \, dy \, dz = \int_0^2 \left(\int_0^{2\pi}\left(\int_z^{\sqrt{2z}}24r^3 \cos^2 \theta \,dr\right)d\theta\right)dz$$and this is not hard to compute. Is my substitution correct? Did I get this one right?

Your working is absolutely correct. The region is bound below by the paraboloid and above by the cone and we could have alternatively set up the integral wrt $z$ first. The surfaces intersect at $z = 2$ and at intersection, $r \leq  2$.
We have $z^2 \leq r^2 \leq 2z \implies \dfrac{r^2}{2} \leq z \leq r$. So the integral becomes,
$\displaystyle \int_0^{2\pi} \int_0^2 \int_{r^2/2}^r 24r^3 \cos^2 \theta \ dz \ dr \ d\theta = \dfrac{128 \pi}{5}$
