Split up the integral over the bounded region [tex]D[/tex] as
[tex]\displaystyle\iiint_D\mathrm dV=\left\{\iiint_{\text{region between }z=4\text{ and }z=x^2+y^2}+\iiint_{\text{region between }z=4x^2+4y^2\text{ and }z=x^2+y^2}\right\}\mathrm dV[/tex]
Converting to cylindrical coordinates, you then have a volume of
[tex]\displaystyle\left\{\int_{\theta=0}^{\theta=2\pi}\int_{r=1}^{r=2}\int_{z=r^2}^{z=4}+\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=r^2}^{z=4r^2}\right\}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle2\pi\left(\int_{r=1}^{r=2}rz\bigg|_{z=r^2}^{z=4}\,\mathrm dr+\int_{r=0}^{r=1}rz\bigg|_{z=r^2}^{z=4r^2}\,\mathrm dr\right)[/tex]
[tex]=\displaystyle2\pi\left(\int_{r=1}^{r=2}(4r-r^3)\,\mathrm dr+\int_{r=0}^{r=1}(4r^3-r^3)\,\mathrm dr\right)[/tex]
[tex]=6\pi[/tex]
