[tex]\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=\zeta\end{cases}[/tex]
Let [tex]R[/tex] be the region bounded by the two surfaces. Then the volume of the region is given by
[tex]\displaystyle\iiint_R\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{\zeta=r^2}^{\zeta=\sqrt{2-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle2\pi\int_{r=0}^{r=1}r(\sqrt{2-r^2}-r^2)\,\mathrm dr[/tex]
[tex]=\displaystyle2\pi\int_0^1(r\sqrt{2-r^2}-r^3)\,\mathrm dr[/tex]
[tex]=\dfrac{(8\sqrt2-7)\pi}6[/tex]
