How find this $I=\iint_{\Sigma}(x^2+y^2+z^2)^{-\frac{3}{2}}(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4})^{-\frac{1}{2}}dS$ Find this Surface integral
$$I=\iint_{\Sigma}(x^2+y^2+z^2)^{-\frac{3}{2}}\left(\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}\right)^{-\frac{1}{2}}dS$$
where
$$\Sigma:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1,a>0,b>0,c>0$$
My try: let
$$x=a\sin{\alpha}\cos{\beta},y=b\sin{\alpha}\sin{\beta},z=c\cos{\alpha},\alpha\in[0,\pi],\beta\in[0,2\pi]$$
$$E=x''_{\alpha}+y''_{\alpha}+z''_{\alpha}=a^2\cos^2{\alpha}\cos^2{\beta}+b^2\cos^2{\alpha}\sin^2{\beta}+c^2\sin^2{\alpha}$$
$$G=x''_{\beta}+y''_{\beta}+z''_{\beta}=a^2\sin^2{\alpha}\sin^2{\beta}+b^2\sin^2{\alpha}\cos^2{\beta}$$
$$F=x'_{\alpha}x'_{\beta}+y'_{\alpha}y'_{\beta}+z'_{\alpha}z'_{\beta}=-a^2\sin{\alpha}\cos{\alpha}\sin{\beta}\cos{\beta}+b^2\sin{\alpha}\cos{\alpha}\sin{\beta}\cos{\beta}$$
so
$$EG-F^2=(a^2b^2\cos^2{\alpha}+a^2c^2\sin^2{\alpha}\sin^2{\theta}+b^2c^2\sin{\alpha}\cos^2{\beta})\sin^2{\alpha}$$
then
$$\int_{\Sigma}f(x,y,z)dS=\int_{\Delta}f(a\sin{\alpha}\cos{\beta},b\sin{\alpha}\sin{\beta},c\cos{\alpha})\sqrt{EG-F^2}d\alpha d\beta$$
where $\Delta:0\le\alpha\le \pi,0\le \beta\le 2\pi$
$$f(x,y,z)=(x^2+y^2+z^2)^{-\frac{3}{2}}\left(\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}\right)^{-\frac{1}{2}}$$
and follow I fell very ugly,Thank you very much
 A: Let $\vec{r} = (x,y,z)$ and $\varphi(\vec{r}) = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}$, we have
$$\vec{\nabla}\varphi(\vec{r}) = (\frac{2x}{a^2}, \frac{2y}{b^2},\frac{2z}{c^2})
\quad\implies\quad
\begin{cases}
\vec{r}\cdot \vec{\nabla}\varphi(\vec{r}) &= 2 \left(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}\right) = 2\varphi(\vec{r})\\
\left|\vec{\nabla}\varphi(\vec{r})\right|^2 &= 4\left(\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4}\right)
\end{cases}$$
On the surfaces $\Sigma$, $\varphi(\vec{r}) = 1$, we can rewrite the factor
$$\frac{1}{\sqrt{\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4}}} d\text{S}
\quad\text{ as }\quad
\vec{r} \cdot \frac{\vec{\nabla}\varphi(\vec{r})}{\left|\vec{\nabla}\varphi(\vec{r})\right|} d\text{S}
$$
But $\;\displaystyle \frac{\vec{\nabla}\varphi(\vec{r})}{\left|\vec{\nabla}\varphi(\vec{r})\right|}$ is nothing but the normal vector $\hat{n}{(\vec{r})}\;$ for the surfaces $\Sigma$. This means
$$I = \int_{\Sigma} \frac{\vec{r}}{|\vec{r}|^3} \cdot \hat{n} d\text{S}
    = - \int_{\Sigma} \vec{\nabla}\frac{1}{|\vec{r}|} \cdot \hat{n} d\text{S}
$$
Since $\;\displaystyle \nabla^2 \frac{1}{|\vec{r}|} = 0\;$ for $\vec{r} \ne \vec{0}$,
we can use Divergence Theorem to deform the surface $\Sigma$ to the unit sphere $S^2$ centered at $\vec{0}$ without changing the value of $I$. As a result,
$$I = \int_{S^2} \frac{\vec{r}}{|\vec{r}|^3} \cdot \hat{n} d\text{S} = 4\pi$$
A: Lets write the integral
$$I=\iint_{\Sigma}(x^2+y^2+z^2)^{-\frac{3}{2}}\left(\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}\right)^{-\frac{1}{2}}dS$$
in the form
$$I=\iint_{\Sigma} \frac{(x, y, z)}{(x^2+y^2+z^2)^{\frac{3}{2}}} \cdot \left(\left(\frac{x}{a^2}, \frac{y}{b^2}, \frac{z}{c^2}\right)\left(\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}\right)^{-\frac{1}{2}}\right)dS$$
and notice the integral now reads
$$I=\iint_{\Sigma}(\mathbf{F\cdot n})dS$$
with 
$$\mathbf{F}= \frac{(x, y, z)}{(x^2+y^2+z^2)^{\frac{3}{2}}}, \mathbf{n}=\left(\left(\frac{x}{a^2}, \frac{y}{b^2}, \frac{z}{c^2}\right)\left(\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}\right)\right)^{-\frac{1}{2}}$$
Notice how $\nabla\cdot \mathbf{F}=0$ on $\mathbb{R}\backslash\{(0, 0, 0)\}$, which means you can integrate $\mathbb{F}$ over any surface $S$ with the origin in its interior. For simplicity, take $S: \{(x,y,z)\in \mathbb{R}^3|\ x^2+y^2+z^2 = 1\}$. The integral is now simply
$$I=\iint_{S} \frac{(x, y, z)}{(x^2+y^2+z^2)^{\frac{3}{2}}} \cdot \frac{(x, y, z)}{(x^2+y^2+z^2)^{\frac{1}{2}}}dS = \iint_{S} \frac{dS}{r^{2}} $$
Therefore, $$I= \int_{0}^{1} \frac{4\pi r^2 dr}{r^2} \Rightarrow I=4\pi$$
