Rotational flux through an ellipsoidal graph patch ($\Phi =\frac{c(a^2-b^2)}{3}[1-(1-A{)}^{3/2}-(1-B{)}^{3/2}+(1-A-B{)}^{3/2}]$)

Take the upward-oriented graph $z=c\sqrt{1-x^2/a^2-y^2/b^2}$ over $0\le x\le L_x$, $0\le y\le L_y$, 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 $(-z_x,-z_y,1)dxdy$, soThe change of variables $u=x^2/a^2$, $v=y^2/b^2$ changes $xydxdy$ to $a^2{b}^{2}dudv/4$. Integrating $(1-u-v{)}^{-1/2}$ over $[0,A]\times [0,B]$ gives $\frac{4}{3}[1-(1-A{)}^{3/2}-(1-B{)}^{3/2}+(1-A-B{)}^{3/2}]$, 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.