Volume of $\{(x,y,z)\colon z\in [0,2]\text{ and }\frac{1}{2}(1+z^2) \leqslant x^2+y^2 \leqslant 2(1+z^2)\}$

Question

Let

$$G=\left\{(x,y,z)\colon z\in [0,2]\text{ and }\frac{1}{2}(1+z^2) \leqslant x^2+y^2 \leqslant 2(1+z^2)\right\}.$$

This looks to me like the interior of two hyperboloids, is that right?

Now consider a map $F$

$$(r, \phi, s) \mapsto(sr\cos \phi, sr \sin \phi, \sqrt{r^2-1})$$

The Jacobian of that map is messy but computable. I would like to find the volume of $F^{-1}(G)$. Can one help me to find the limits of integration?

Answer 1

$$G=\left\{(r\cos\theta,r\sin\theta,z)\mid \theta\in[0,2\pi[, r\in[0,1],\sqrt{\frac{1}{2}(r^2-1)}\leq z\leq \sqrt{2(r^2-1)}\right\}$$

therefore,

$$Volume= \iiint_G r\mathrm{d}r\mathrm{d}\theta \mathrm{d}z=\int_0^{2\pi}\mathrm{d}\theta\int_0^1 r\mathrm{d}r\int_\sqrt{\frac{1}{2}(r^2-1)}^\sqrt{2(r^2-1)}\mathrm{d}z=...$$

Answer 2

$$0\le z=\sqrt{r^2-1}\le 2\implies 1\le r=\sqrt{1+z^2}\le\sqrt5$$ $$\frac12 r^2=\frac12(1+z^2)\le s^2r^2=x^2+y^2\le 2(1+z^2)=2r^2$$ $$\frac{\sqrt 2}{2} \le s\le\sqrt 2$$ And no restriction on $\phi$, i.e., $0\le\phi\le 2\pi$.