Answer
$\dfrac{4\pi}{3}(\sqrt 2 -1)$
Work Step by Step
Here, we have $V=\int_0^{2\pi} \int_{0}^{1}\int_{r}^{\sqrt{2-r^2}} r dz dr d\theta$
and $x=r \cos \theta; y=r \sin \theta, z=z$
This implies that
$=\int_0^{2\pi} \int_{0}^{1}[rz]_{r}^{\sqrt{2-r^2}} r dz dr d\theta=\int_0^{2\pi} \int_{0}^{1}(r\sqrt{2-r^2}-r^2) dr d\theta$
Consider $p=2-r^2$ and $dp=-2rdr$
Then, we get $V=\int_0^{2\pi} [\int_{1}^{2} (\dfrac{1}{2}) \sqrt p dp -\int_0^1 r^2 dr] d\theta$
or, $=\int_0^{2\pi} [\dfrac{1}{3}(2 \sqrt 2 -1-1) d\theta$
or, $=\int_0^{2\pi} (\dfrac{2}{3}) (\sqrt 2 -1) d\theta$
or, $=\dfrac{2}{3}(\sqrt 2 -1)[\theta]_0^{2\pi}$
Hence, $V=\dfrac{4\pi}{3}(\sqrt 2 -1)$
