Answer
$(\dfrac{8\sqrt 2-7}{6}) \pi$
Work Step by Step
Conversion of rectangular to cylindrical coordinate system gives: $r^2=x^2+y^2 \\ \tan \theta=\dfrac{y}{x} \\z=z$
Here, $x=r \cos \theta; y=r \sin \theta, z=z$
Let $V=\int_0^{2\pi} \int_{0}^{1}\int_{r^2}^{\sqrt{2-r^2}} r dz dr d\theta$
or, $=\int_0^{2\pi} \int_{0}^{1}[rz]_{r^2}^{\sqrt{2-r^2}} r dz dr d\theta=\int_0^{2\pi} \int_{0}^{1}(r\sqrt{2-r^2}-r^3) dr d\theta$
Plug $k=2-r^2$ and $dk=-2rdr$
Thus, $V=-\int_0^{2\pi} [\int_{2}^{1} (1/2) k^{1/2} dk -\int_0^1 r^3 dr] d\theta=\int_0^{2\pi} [\int_{1}^{2} (1/2) k^{1/2} dk -\int_0^1 r^3 dr] d\theta$
or, $V=\int_0^{2\pi} [\dfrac{1}{3}(2 \sqrt 2 -1-\dfrac{1}{4}) d\theta=\int_0^{2\pi} (\dfrac{8\sqrt 2-7}{12}) d\theta$
Hence, $V=(\dfrac{8\sqrt 2-7}{12}) [\theta]_0^{2\pi}=(\dfrac{8\sqrt 2-7}{6}) \pi$
