Answer
${I_z} = \frac{{14\pi }}{{15}}$
Work Step by Step
We have the solid region ${\cal W}$ inside the hyperboloid ${x^2} + {y^2} = {z^2} + 1$ between $z=0$ and $z=1$.
From the figure attached, we see that ${\cal W}$ is bounded below by $z=0$ and bounded above by $z=1$. So, $0 \le z \le 1$.
The hyperboloid in cylindrical coordinates is ${r^2} = {z^2} + 1$. Since ${\cal W}$ is bounded by the hyperboloid, $r$ varies from $0$ to $r = \sqrt {{z^2} + 1} $.
Using the cylindrical coordinates, the description of ${\cal W}$:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le \theta \le 2\pi ,0 \le z \le 1,0 \le r \le \sqrt {{z^2} + 1} } \right\}$
Assume that the mass density is $\delta \left( {x,y,z} \right) = 1$.
Using ${x^2} + {y^2} = {r^2}$, we evaluate the moment of inertia ${I_z}$ in cylindrical coordinates:
${I_z} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \left( {{x^2} + {y^2}} \right)\delta \left( {x,y,z} \right){\rm{d}}V$
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{z = 0}^1 \mathop \smallint \limits_{r = 0}^{\sqrt {{z^2} + 1} } \left( {{r^2}} \right)r{\rm{d}}r{\rm{d}}z{\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{z = 0}^1 \left( {{r^4}|_0^{\sqrt {{z^2} + 1} }} \right){\rm{d}}z{\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{z = 0}^1 \left( {{z^4} + 2{z^2} + 1} \right){\rm{d}}z{\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( {\frac{1}{5}{z^5} + \frac{2}{3}{z^3} + z} \right)|_0^1} \right){\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{1}{5} + \frac{2}{3} + 1} \right){\rm{d}}\theta $
$ = \frac{7}{{15}}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \frac{{14\pi }}{{15}}$
So, ${I_z} = \frac{{14\pi }}{{15}}$.
