= Polynomial flux through a half-ellipsoid
{title2=$\Phi=\pi abC+2\pi ab^3c/15$}
For the upper curved half of $x^2/a^2+y^2/b^2+z^2/c^2=1$, take the <vector field> $(xy^2,y^2,C)$ and outward <unit normal>. Its <divergence> is $y^2+2y$. The odd term integrates to zero over the half-ellipsoid. Scaling a unit half-ball gives $\int y^2dV=2\pi ab^3c/15$. The equatorial closing disk has downward flux $-C\pi ab$. The <divergence theorem> therefore gives the displayed curved-surface flux. The constant vertical component is not detected by the divergence, but is recovered by accounting for the closing disk.
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