I would like to calculate Lebesgue measure of set: $M := \{(x, y, z) \in \mathbb{R^3}: ax^2 + y^2 \le 1, |z| + |y| \le 2\}$, $a > 0$.
My approach, use substitution: $\rho(r, t, s) = (r\cos t, \sqrt{a}r\sin t, s)$
Where $(r, t, s) \in (0, \infty)\times(-\pi, \pi)\times\mathbb{R}$ and Jacobian of $\rho$ is $\sqrt{a}r$.
Let: $\tilde{M} := \{(r, t, s) \in \mathbb{R^3}: 0 < r \le \frac{1}{\sqrt{a}}, |s| + |r\sqrt{a}\sin t| \le 2, -\pi < t < \pi\}$.
Then: $\rho(\tilde{M}) = M$.
My question is, how to handle double absolute value in inequality: $|s| + |r\sqrt{a}\sin t| \le 2$ to properly specify bounds to form integral of variable s? Or should I use different substitution to avoid this?
Thanks.
You can observe that the region $M$ is symmetric respect to the values of $s$ (or $z$ as $z=s$), then you can consider to write $M=M_1\cup M_2$ where $\rho (M_1)=\rho (M_2)=\tfrac1{2}\rho (M)$ for $M_1:=M\cap \{(x,y,z):z\geqslant 0\}$ and $M_2:=M\cap \{(x,y,z):z<0\}$, noticing that $\rho (M\cap \{(x,y,z):z=0\})=0$.
Then, for $M_1$, you have the conditions
$$ s\in[0,\infty )\,\land\, r\in(0,1/\sqrt{a})\,\land\, t\in\left(- \arcsin \left(\frac{2-s}{r\sqrt{a}}\right),\arcsin \left(\frac{2-s}{r\sqrt{a}}\right)\right) $$
But in my opinion it will be easier to handle all of this with the change of variables $x=\frac1{\sqrt{a}}\tilde x$, this will have the conditions for $M_1$ as
$$ \tilde x^2+y^2\leqslant 1\,\land\, 0\leqslant z\leqslant 2-|y| $$
knowing that $dx=\frac1{\sqrt{a}}d\tilde x$.
Using again symmetry we can divide $M_1$ in two pieces $M_{1,1}$ and $M_{1,2}$ of the same volume, for $y\geqslant 0$ and $y<0$ respectively, giving the conditions for $M_{1,1}$ as
$$ \tilde x^2+y^2\leqslant 1\,\land\, 0\leqslant z\leqslant 2-y\,\land\, y\geqslant 0 $$
And now with cylindrical coordinates $(\tilde x, y,z)=(r\cos \theta ,r\sin \theta ,z)$ it gives
$$ \rho (M)=4 \rho (M_{1,1})=\frac{4}{\sqrt{a}}\int_{[0,1]}\int_{[0,\pi]}\int_{[0,2-r\sin \theta]}r\,d z\,d\theta \,d r $$
as $d \tilde xdy=r drd\theta $.