$z=2(x^2+y^2) ,z=4 -\sqrt{x^2 +y2}$
I sketched the graph, substituted $x^2 +y^2=r^2$ and calculated radius of the widest point : $r_1 = \frac{-1-\sqrt{33}}{4}, r_2 = \frac{-1+\sqrt{33}}{4}$. I think $\theta \in[0,2\pi]$ and $r\in[0,\frac{-1+\sqrt{33}}{4}]$. From that point volume V is $\int_0^{2\pi}\int_0^{\frac{-1+\sqrt{33}}{4}} r(4 -\sqrt{r^2} -2r^2)drd\theta = \int_0^{2\pi}\int_0^{\frac{-1+\sqrt{33}}{4}} r(4 -r -2r^2)drd\theta$. I would like to stick to $r,\theta$. Is it correct?
Forget $r=\frac{-1-\sqrt{33}}4$; $r$ cannot be negative. So, $\theta$ can take any value in $[0,2\pi]$ (as you wrote), $r$ can take any value in $\left[0,\frac{-1+\sqrt{33}}4\right]$ and $z$ can take any value from $2r^2$ to $4-r$. So, compute$$\int_0^{2\pi}\int_0^\frac{-1+\sqrt{33}}4\int_{2r^2}^{4-r}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta.$$You should get $\dfrac\pi{96}\left(433+33 \sqrt{33}\right)$.