I assume you mean the paraboloid [tex]x=6y^2+6z^2[/tex]. Denote the space between this surface and the plane [tex]x=6[/tex] by the symbol [tex]D[/tex].
The volume is then
[tex]\displaystyle\iiint_D\mathrm dV=\int_{-1}^1\int_{-1}^1\int_{6y^2+6z^2}^6\mathrm dx\,\mathrm dy\,\mathrm dz=8[/tex]