Say we are given a vector field $$F=(-x^2/2+xy,xy+y^2,-3yz-3)$$ with the property $\nabla\cdot F=0$. If we would like to find the flux through the part of the surface $x^2+y^2+2z^2=3$ that lies above $z=0$, wouldn't the net total flux be zero?

My reasoning Applying the divergence theorem, the flux turns out to be zero.

$$\iint\limits_D {\vec F \cdot \hat NdS = \{ {\text{divergence thm}}{\text{.}}\} = \iiint\limits_{D'} {\operatorname{div} \vec FdV} = 0}$$

Key The key argues that the answer is $-9\pi$. The argument is that the flux through the surface (given above) is equal to the flux through a disk $x^2+y^2\le 3$ in $xy$-plane, because of $div \vec F =0$ and the divergence theorem?

Question My question is how it can turn out to be $-9 \pi$. Do they assume that we only consider the flux in a certain direction and then use the fact that we know that it should be zero in total?


The surface you integrate over in the divergence theorem has to be closed. The two surfaces you mention, the "hemiellipse" and the disk, are not individually closed, but together form a closed surface. Hence the integral over both is zero, by the divergence theorem, so the integral over the hemiellipse is the negative of the integral over the disk.

  • $\begingroup$ Thank you for your answer! $\endgroup$ – Artem May 30 '15 at 12:00

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