= Rotational flux through an ellipsoidal graph patch
{title2=$\Phi=\frac{c(a^2-b^2)}3[1-(1-A)^{3/2}-(1-B)^{3/2}+(1-A-B)^{3/2}]$}
Take the upward-oriented graph $z=c\sqrt{1-x^2/a^2-y^2/b^2}$ over $0\leq x\leq L_x$, $0\leq y\leq L_y$, where $a,b,c,L_x,L_y>0$, $A=L_x^2/a^2$, $B=L_y^2/b^2$ and $A+B<1$. The <surface flux> of the <vector field> $F=(-y,x,0)$ is the displayed formula. Indeed the <vector surface element> is $(-z_x,-z_y,1)dxdy$, so
$$
\Phi=c\left(\frac1{b^2}-\frac1{a^2}\right)\int_0^{L_x}\int_0^{L_y}\frac{xy}{\sqrt{1-x^2/a^2-y^2/b^2}}\,dy\,dx.
$$
The <change of variables> $u=x^2/a^2$, $v=y^2/b^2$ changes $xy\,dxdy$ to $a^2b^2\,dudv/4$. Integrating $(1-u-v)^{-1/2}$ over $[0,A]\times[0,B]$ gives $\tfrac43[1-(1-A)^{3/2}-(1-B)^{3/2}+(1-A-B)^{3/2}]$, proving the result. When $a=b$, the value of the <vector field> is a <tangent vector> to the <ellipsoid> at every point, and the <surface flux> vanishes pointwise.
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