Answer
$(\dfrac{8\sqrt 2-7}{6}) \pi$
Work Step by Step
$V=\int_0^{2\pi} \int_{0}^{1}\int_{r^2}^{\sqrt{2-r^2}} r dz dr d\theta$
and $x=r \cos \theta; y=r \sin \theta, z=z$
This implies that
$\int_0^{2\pi} \int_{0}^{1}[rz]_{r^2}^{\sqrt{2-r^2}} \times (r dz dr d\theta) =\int_0^{2\pi} \int_{0}^{1}(r\sqrt{2-r^2}-r^3) dr d\theta$
Consider $p=2-r^2$ and $dp=-2rdr$
or, $=-\int_0^{2\pi} [\int_{2}^{1} (\dfrac{1}{2}) \sqrt p dp-\int_0^1 r^3 dr] d\theta$
or, $=\int_0^{2\pi} [\int_{1}^{2} (\dfrac{1}{2}) \sqrt p dp -\int_0^1 r^3 dr] d\theta$
or, $=\int_0^{2\pi} (\dfrac{8\sqrt 2-7}{12}) d\theta$
or, $=(\dfrac{8\sqrt 2-7}{12}) [\theta]_0^{2\pi}$
Hence, $V=(\dfrac{8\sqrt 2-7}{6}) \pi$
