Let [tex]R[/tex] be the solid. Then the volume is
[tex]\displaystyle\iiint_R\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta[/tex]
which follows from the facts that
[tex]\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=\zeta\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta[/tex]
(by computing the Jacobian)
and
[tex]z=x^2+y^2=r^2\implies z+z^2=72\implies z=-9\text{ or }z=8[/tex]
(we take the positive solution, since it's clear that [tex]R[/tex] lies above the [tex]x[/tex]-[tex]y[/tex] plane)
[tex]r^2+z^2=72\implies z=\pm\sqrt{72-r^2}[/tex]
(again, taking the positive root for the same reason)
[tex]z=r^2\implies 8=r^2\implies r=\sqrt8[/tex]
[tex]\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr[/tex]
[tex]=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}r(\sqrt{72-r^2}-r^2)\,\mathrm dr[/tex]
[tex]=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}(r\sqrt{72-r^2}-r^3)\,\mathrm dr[/tex]
[tex]=\dfrac{32(27\sqrt2-35)\pi}3[/tex]