<s>[INST] 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?
 [/INST] 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,
$\displaystyle \int_0^{2\pi} \int_0^2 \int_{r^2/2}^r 24r^3 \cos^2 \theta \ dz \ dr \ d\theta = \dfrac{128 \pi}{5}$
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