[tex]\begin{cases}z^2=64+x^2+y^2\\z^2=2x^2+2y^2\end{cases}[/tex]
It's clear enough that the upper half of the cone falls below the upper sheet of the paraboloid, so that [tex]\sqrt{2x^2+2y^2}\le z\le\sqrt{64+x^2+y^2}[/tex]. Right away you can see that converting to cylindrical coordinates will be quite advantageous.
The intersection of the two surfaces occurs as a circle:
[tex]64+x^2+y^2=2x^2+2y^2\implies64=x^2+y^2[/tex]
which is parallel to the x-y plane, has radius 8, and is centered at [tex](0,0,8\sqrt2)[/tex].
The volume of this space is given by the integral
[tex]\displaystyle\iiint_S\mathrm dV=\int_{x=-8}^{x=8}\int_{y=-\sqrt{64-x^2}}^{y=\sqrt{64-x^2}}\int_{z=\sqrt{2x^2+2y^2}}^{z=\sqrt{64+x^2+y^2}}\mathrm dz\,\mathrm dy\,\mathrm dz[/tex]
where [tex]S[/tex] denotes the bounded space between the surfaces. Converting to cylindrical coordinates, this can be expressed as
[tex]\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=8}\int_{z=\sqrt2r}^{z=\sqrt{64+r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta[/tex]
which evaluates to [tex]\dfrac{1024\pi(\sqrt2-1)}3[/tex].
