Flux calculation through half ellipsoid.

Question

Determine the total flux of the vector field $$ F(x, y, z)=\frac{r}{|r|^{3}}+z i-x k $$ through the half ellipsoid in the direction $N \cdot j>0$.

Answer 1

Assuming we can fill in the blank on what exactly is the equation of the half ellipsoid, we can proceed by splitting up the vector field into two pieces and using two separate divergence theorems of each piece.

First, consider closing the ellipsoid by the plane $y=0$ and take the not radial piece of the vector field. Then divergence theorem tells us that

$$\iiint_V \nabla\cdot(z,0,-x)dV = \iint_{\text{half-ellipsoid}}(z,0,-x)\cdot NdS + \iint_{y=0} (z,0,-x)\cdot(0,-1,0) dS$$

$$\implies 0 = \iint_{\text{half-ellipsoid}}(z,0,-x)\cdot NdS + 0$$

so that contribution completely vanishes.

Next, consider instead closing the ellipsoid from "underneath" by a half sphere of small enough radius $R$ such that it touches the ellipsoid (assuming the half ellipsoid intersects the $xz$ plane in a circle; if not this question gets a bit more complicated). Then divergence theorem tells us

$$\iiint_V \nabla\cdot\frac{r}{|r|^3}dV = \iint_{\text{half-ellipsoid}}\frac{r}{|r|^3}\cdot NdS + \iint_{\text{half-sphere}} \frac{r}{R^3}\cdot-\frac{r}{R} dS$$

$$\implies 0 = \iint_{\text{half-ellipsoid}}\frac{r}{|r|^3}\cdot NdS - \iint_{\text{half-sphere}} \frac{R^2}{R^4} dS = \iint_{\text{half-ellipsoid}}\frac{r}{|r|^3}\cdot NdS - 2\pi$$

So the contribution from this piece of the vector field is $2\pi$. Seeing as it is the only contribution, the answer is $2\pi$. But hopefully you can give us the exact equation of the half ellipsoid so we can verify that this is the correct approach.

Notice, though, we were able to get a full answer without even knowing the exact equation of the half-ellipsoid, only assuming that it was sliced in half by the $xz$ plane, and that it had a circular base (though even this assumption is unnecessary, but it would be more work to prove it other wise). The exact size of the ellipsoid was of no consequence.