Lebesgue measure of $\{(x, y, z) \in \mathbb{R^3}: 2x^2 - y^2 + z^2 \le 2z, |y| \le 1\}$

Question

I would like to calculate Lebesgue measure of set $M := \{(x, y, z) \in \mathbb{R^3}: 2x^2 - y^2 + z^2 \le 2z, |y| \le 1\}$

First thing that came to my mind was using polar coordinates in form: $(x, y, z) = (r\cos t , \sqrt2r\sin t, s)$, but problem I see is minus next to y, so $\sin$ and $\cos$ don't cancel out in equation after applying substitution.

What is the correct approach here?

Hints are appreciated, thanks.

Answer 1

Consider this change of variables: $x=\rho\cos\theta$ and $z=1+\sqrt2\rho\sin\theta$ ($y$ is left unganged). Then\begin{align}2x^2-y^2+z^2\leqslant2z&\iff2x^2-y^2+(z-1)^2\leqslant1\\&\iff2\rho^2-y^2\leqslant1\\&\iff\rho\leqslant\sqrt{\frac{1+y^2}2}.\end{align}The Jacobian corresponding to this change of variables is $\sqrt2\rho$. So, the volume of the region that you are interested in is\begin{align}\int_{-1}^1\int_0^{2\pi}\int_0^{\sqrt{(1+y^2)/2}}\sqrt2\rho\,\mathrm d\rho\,\mathrm d\theta\,\mathrm dy&=2\sqrt2\pi\int_{-1}^1\frac{y^2+1}4\,\mathrm dy\\&=\frac{4\sqrt2\pi}3.\end{align}