Answer
The volume is $\frac{4\pi}{3}$.
Work Step by Step
Since the plane $z=2$ ia above the xy-plane, rewrite the hyperboloid as a function $z=f(x,y)$ when $z>0$.
$-x^2-y^2+z^2=1$
$z^2=1+x^2+y^2$
$z=\sqrt{1+x^2+y^2}$ for $z>0$
Find the region $R$ of this function under the plane z=2:
$z\leq 2$
$z^2\leq 2^2$
$1+x^2+y^2\leq 4$
$x^2+y^2\leq 3$
Then, $R$ is a disk center at the origin and radius $\sqrt{3}$ expressed in polar coordinates as the set of points $R=\{(r,\theta)|0\leq r\leq \sqrt{3},0\leq \theta\leq 2\pi\}$.
Evaluate the volume of the solid above the hyperboloid and above the plane z=2:
$V=\int_0^{2\pi}\int_0^{\sqrt{3}}(2-\sqrt{1+r^2})\cdot r dr d\theta$
$=\frac{4\pi}{3}$
