Compute the Flux of the vector field The vector field is this $F(x,y,z)=\left\langle z^2-y^2e^z,z\ln(1-2x^2),3\right\rangle$.
S is a portion of the graph $z=5-x^2-y^2$ which sits above the plane $z=0$, and orientation upwards. Compute the flux of $F$ across $S$ (that is, $\iint_SF\cdot\,dS$)
I got $\nabla \circ F=0$, where do I go on from here?
 A: The divergence should indeed be $0$ since:
\begin{align}
&\frac{\partial}{\partial x}(z^2 - y^2) =0 \\
&\frac{\partial}{\partial y}(z\ln(1 - 2x^2) = 0 \\
&\frac{\partial}{\partial z}3 = 0 \\
\end{align}
This means that $\oint_{\partial V} \vec{F}\circ d\vec{A} = \int_V \nabla \circ \vec{F} dV = 0$.
We can break up the integral you want into two parts, the one you want ($\partial V_1$) and then the $z = 0$ plane ($\partial V_2$):
$$
\oint_{\partial V} \vec{F}\circ d\vec{A} = \int_{\partial V_1} \vec{F}\circ d\vec{A} + \int_{\partial V_2} \vec{F}\circ d\vec{A} = 0
$$
The second integral ($\partial V_2$, the $z = 0$ plane part), we can take easily:
$$
\int_{\partial V_2} \vec{F}\circ d\vec{A} = \int_{\partial V_2} \vec{F}\circ (-\hat{z})dA = -\int_{\partial V_2} \vec{F}\circ \hat{z}dA
$$
But $\vec{F}\circ \hat{z} = 3$, thus we have:
$$
-\int_{\partial V_2} \vec{F}\circ \hat{z}dA = -3\int_{\partial V} dA
$$
But we know this is the area of the circle of radius $\sqrt{5}$, so $\int_{\partial V} dA = 5\pi$.  We finally have:
\begin{align}
\oint_{\partial V} \vec{F}\circ d\vec{A} =& \int_{\partial V_1} \vec{F}\circ d\vec{A} + \int_{\partial V_2} \vec{F}\circ d\vec{A} = 0\\
=&\ \int_{\partial V_1} \vec{F}\circ d\vec{A} -3(5\pi)  = 0
\end{align}
$$
\int_{\partial V_1} \vec{F}\circ d\vec{A} = 15\pi
$$
