Answer
$\dfrac{4\pi}{3}(\sqrt 2 -1)$
Work Step by Step
Consider $V=\int_0^{2\pi} \int_{0}^{1}\int_{r}^{\sqrt{2-r^2}} r \times dz dr d\theta$
$=\int_0^{2\pi} \int_{0}^{1}[rz]_{r}^{\sqrt{2-r^2}} \times r dz dr d\theta$
$=\int_0^{2\pi} \int_{0}^{1}(r \times \sqrt{2-r^2}-r^2) dr d\theta$
Plug $a=2-r^2 \implies -2rdr=da $
$=\int_0^{2\pi} [\int_{1}^{2} (1/2) \times a^{1/2} dk -\int_0^1 r^2 dr] d\theta$
$=\int_0^{2\pi} [\dfrac{2}{3}(\sqrt 2 -1) d\theta$
$=\dfrac{2}{3}(\sqrt 2 -1) \times [\theta]_0^{2\pi}$
$=\dfrac{4\pi}{3}(\sqrt 2 -1)$
