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?

  • $\begingroup$ I am taking you at your word about the divergence of $F$ being zero and the region sitting above the $z = 0$ plane. If this is the case then the flux through the entire region is $0$...but that's not what you are computing (I believe). This means that what you want to compute will be the negative of the flux through the $z = 0$ plane (I believe oriented "down")...so the circle $x^2 + y^2 = 5$ (circle of radius $\sqrt{5}$)...can you take that integral?? $\endgroup$ – Jared Apr 23 '16 at 2:41

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}

$$ \int_{\partial V_1} \vec{F}\circ d\vec{A} = 15\pi $$


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