I'm not sure how to compute the integral $$\int_{s} (a^2x^2+b^2y^2+c^2z^2)^{-1/2} d\vec{S}\cdot \vec{n}$$ over the surface of the ellipsoid $$ax^2+by^2+cz^2=1$$, $$z>0$$
Where $\vec{n}$ is the unitary normal vector to the surface.
Any help is appreciated. Thank you.
We propose to attack the problem with a blunt instrument: on the surface $$\vec{r}=\langle x,y,z\rangle=\left\langle x,y,\pm\sqrt{\frac{1-ax^2-by^2}{c}}\right\rangle$$ so $$\begin{align}d\vec{r}&=\left\langle1,0,\mp\frac{\frac{ax}c}{\sqrt{\frac{1-ax^2-by^2}{c}}}\right\rangle\,dx+\left\langle0,1,\mp\frac{\frac{by}c}{\sqrt{\frac{1-ax^2-by^2}{c}}}\right\rangle\,dy\\ &=\left\langle1,0,-\frac{ax}{cz}\right\rangle\,dx+\left\langle0,1,-\frac{by}{cz}\right\rangle\,dy\end{align}$$ So that the vector areal element is $$\begin{align}d^2\vec{A}&=\pm\left\langle1,0,-\frac{ax}{cz}\right\rangle\,dx\times\left\langle0,1,-\frac{by}{cz}\right\rangle\,dy\\ &=\pm\left\langle\frac{ax}{cz},\frac{by}{cz},1\right\rangle\,dx\,dy=\frac{\langle ax,by,cz\rangle}{c|z|}dx\,dy\end{align}$$ Then $$\hat{n}=\frac{d^2\vec{A}}{\left\lVert d^2\vec{A}\right\rVert}=\frac{\langle ax,by,cz\rangle}{\left(a^2x^2+b^2y^2+c^2z^2\right)^{1/2}}$$ $$\begin{align}I&={\large\bigcirc}\kern-1.55em\iint_S\frac1{\left(a^2x^2+b^2y^2+c^2z^2\right)^{1/2}}\frac{\langle ax,by,cz\rangle}{\left(a^2x^2+b^2y^2+c^2z^2\right)^{1/2}}\cdot d^2\vec{A}\\ &=2\int_{-\frac1{\sqrt a}}^{\frac1{\sqrt a}}\int_{-\sqrt{\frac{1-ax^2}b}}^{\sqrt{\frac{1-ax^2}b}}\frac1{\left(a^2x^2+b^2y^2+c^2z^2\right)^{1/2}}\frac{\langle ax,by,cz\rangle}{\left(a^2x^2+b^2y^2+c^2z^2\right)^{1/2}}\cdot\frac{\langle ax,by,cz\rangle}{c|z|}dy\,dx\\ &=2\int_{-\frac1{\sqrt a}}^{\frac1{\sqrt a}}\int_{-\sqrt{\frac{1-ax^2}b}}^{\sqrt{\frac{1-ax^2}b}}\frac1{\sqrt c\sqrt{1-ax^2-by^2}}\,dy\,dx=2\int_{-\frac1{\sqrt a}}^{\frac1{\sqrt a}}\int_{-\frac{\pi}2}^{\frac{\pi}2}\frac1{\sqrt{bc}}\,d\theta\,dx=\frac{4\pi}{\sqrt{abc}}\end{align}$$ Where we have used the substitution $$y=\sqrt{\frac{1-ax^2}b}\sin\theta$$ to solve the $y$-integral. Note that if $a=b=c$ then $a^2x^2+b^2y^2+c^2z^2=a$ on the surface and $d^2\vec{A}\cdot\hat{n}=d^2A$ and the integral is over the surface of a sphere of radius $1/\sqrt a$ so we should get $$I=\frac1{\sqrt a}\frac{4\pi}{a}$$ so this checks.