Let $$D=\{(x,y,z): 2(x^2+y^2)<z<3\}$$ and the vector field $$F(x,y,z)=(x,y,z^2)$$
I wanto to compute the flux of $F$ through $\partial D$
Applying the divergence theorem, I should compute
$$\int_D2+2z \ dxdydz$$ To evaluate the integral I pass to cylindrical coordinates but I have some problems at this point. We have $$2r^2<z<3,$$ $$0<r<\sqrt{3/2}$$ so then it becomes
$$\int_D2z \ dxdydz=2\pi\int_0^{\sqrt{3/2}}r \ dr\int_{2r^2}^3 \ z \ dz=2\pi(\sqrt{3/2}9/2-\int_0^{\sqrt{3/2}}2r^5 \ dr)=2\pi(\sqrt{3/2}9/2-\sqrt{3/2}^6/3)$$$$+$$$$\int_D2dz=2\pi(\int_0^{\sqrt{3/2}}6r-4r^2\ dr)$$
But this does not look good at all. So maybe I'm doing something wrong?
If I try computing the flux directly I first have to compute the boundary, which I am not sure about $$\partial D=\{(x,y,z):x^2+y^2=\frac{z}{2}, z<3\} \cup \{(x,y,z):x^2+y^2\leq \frac{3}{2}, \ z=0\}=B_1 \cup B_2$$ I can parametrize $B_1$ as $$\phi:(u,v)\mapsto (u,v,2(u^2+v^2))$$ so that $$\phi_u \land \phi_v= (-4u,-4v,1)$$ and gives, for $B_1,$ the integral $$\int_{2u^2+2v^2<3} \langle F(\phi(u,v)),\phi_u \land \phi_v \rangle dudv= \int_{2u^2+2v^2<3}-4(u^3+v^3)+4(u^2+v^2)^2 \ dudv$$ which is very ugly.
By using the divergence theorem, we have to evaluate $$\begin{align}2\pi\int_{r=0}^{\sqrt{3/2}}\int_{z=2r^2}^3(2+2z)r\,dz dr&=2\pi\int_{r=0}^{\sqrt{3/2}}(15r-4r^3-4r^5) dr\\&= \left[\frac{15r^2}{2}-r^4-\frac{2r^6}{3}\right]_{r=0}^{\sqrt{3/2}}=\frac{27\pi}{2}. \end{align}$$ As regards the direct computation, the outward flux through $B_1$ is $$\begin{align}\int_{2u^2+2v^2<3}4(u^2+v^2)-4(u^2+v^2)^2 \ dudv&=2\pi\int_{0}^{\sqrt{3/2}}(4r^2-4r^4)r\,dr\\ &=\left[r^4-\frac{2r^6}{3}\right]_{r=0}^{\sqrt{3/2}}=0 \end{align}$$ whereas the outward flux through $B_2$ is $$\int_{z=3,r\leq \sqrt{3/2}}z^2 r\,dr=9\cdot \frac{3\pi}{2}=\frac{27\pi}{2}.$$ Their sum is again $\frac{27\pi}{2}$.