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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?

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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.

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  • $\begingroup$ Thank you for your answer! $\endgroup$
    – Artem
    May 30, 2015 at 12:00

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