Answer
$(\dfrac{8\sqrt 2-7}{6}) \pi$
Work Step by Step
Consider $V=\int_0^{2\pi} \int_{0}^{1}\int_{r^2}^{\sqrt{2-r^2}} r dz dr d\theta$
$=\int_0^{2\pi} \int_{0}^{1}[rz]_{r^2}^{\sqrt{2-r^2}} r \times dz dr d\theta$
$=\int_0^{2\pi} \int_{0}^{1}(r\sqrt{2-r^2}-r^3) \times dr d\theta$
Plug $a=2-r^2 \implies -2rdr=da$
$=-\int_0^{2\pi} [\int_{2}^{1} (1/2) \times a^{1/2} da -\int_0^1 r^3 dr] d\theta$
$=\int_0^{2\pi} [\dfrac{1}{3}(2 \sqrt 2 -1-\dfrac{1}{4}) d\theta$
$=\int_0^{2\pi} (\dfrac{8\sqrt 2-7}{12}) d\theta$
$=(\dfrac{8\sqrt 2-7}{12}) \times [\theta]_0^{2\pi}$
$=(\dfrac{8\sqrt 2-7}{6}) \pi$
