I have to evaluate the flux of the vector field $$F(x,y,z)=(y^2,x,x^2)$$ through the surface $$\sigma_{0}\colon z=4-x^2-y^2$$ and above the plane $$\sigma_{1}\colon z=1$$ (but not to include the bottom).
What I thought was: if $\sigma$ is the surface bounded by the paraboloid and the plane above, so $$\iint_{\sigma}\vec{F}\cdot\vec{n}dS=\iint_{\sigma_{0}}\vec{F}\cdot\vec{n}dS+\iint_{\sigma_{1}}\vec{F}\cdot\vec{n}dS$$
A parametrization for $\sigma_{0}$ is $$\sigma_{0}(u,v)=(u,v,1),\quad\forall(u,v)\in\mathbb{R}^{2}$$ then $$\frac{\partial\sigma_{1}}{\partial u}\wedge\frac{\partial\sigma_{1}}{\partial v}=(0,0,1)$$ and since $$F(\sigma_{1})=(v^2,u,u^2)$$ I got $$\iint_{\sigma_{1}}\vec{F}\cdot\vec{n}dS=\iint_{R}u^{2}dudv$$ where $R$ is intersection between $\sigma_{0}$ and $\sigma_{1}$: $$x^{2}+y^{2}=3\Leftrightarrow u^{2}+v^{2}=3$$ which, in polar coordinates is equivalent to $$\iint_{\sigma_{1}}\vec{F}\cdot\vec{n}dS=\int_{0}^{2\pi}\int_{0}^{\sqrt{3}}r^{3}\cos^2(\theta)drd\theta=\frac{9\pi}{4}$$
And since $$div\vec{F}=0$$ from Gauss' theorem $$\iint_{\sigma_{0}}\vec{F}\cdot\vec{n}dS=\iiint_{B}div\vec{F}dV=\iiint_{B}0dV=0$$ Therefore, the flux is $-\dfrac{9\pi}{4}$. Is it correct?