The intersection of the two surfaces occurs along the cylinder [tex]x^2+y^2=4[/tex], so we have the [tex]x[/tex]- and [tex]y[/tex]-coordinates of the points in the bounded region contained within [tex]0\le x\le2[/tex] and [tex]0\le y\le2[/tex], while [tex]y^2\le z\le8-2x^2-y^2[/tex].
The triple integral is then given by
[tex]\displaystyle\iiint_V12xz\,\mathrm dV=\int_{x=0}^{x=2}\int_{y=0}^{y=2}\int_{z=y^2}^{z=8-2x^2-y^2}12xz\,\mathrm dz\,\mathrm dy\,\mathrm dx[/tex]
where [tex]V[/tex] denotes the bounded region in the first quadrant.
Integrating in Cartesian coordinates is easy enough to do in this order.
[tex]\displaystyle\int_{x=0}^{x=2}\int_{y=0}^{y=2}\int_{z=y^2}^{z=8-2x^2-y^2}12xz\,\mathrm dz\,\mathrm dy\,\mathrm dx[/tex]
[tex]=\displaystyle24\int_{x=0}^{x=2}\int_{y=0}^{y=2}x(x^2-4)(x^2+y^2-4)\,\mathrm dy\,\mathrm dx[/tex]
[tex]=\displaystyle\int_{x=0}^{x=2}(512x-320x^3+48x^5)\,\mathrm dx[/tex]
[tex]=256[/tex]
