Polynomial flux through a half-ellipsoid ($\Phi =\pi abC+2\pi a{b}^{3}c/15$)

For the upper curved half of $x^2/a^2+y^2/b^2+z^2/c^2=1$, take the vector field $(x{y}^2,y^2,C)$ 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 $\int y^{2}dV=2\pi a{b}^{3}c/15$. 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.