Surface integral over ellipsoid I've problem with this surface integral:
$$
\iint\limits_S {\sqrt{ \left(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\right)}}{dS}
$$, where
$$
S = \{(x,y,z)\in\mathbb{R}^3: \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}= 1\}
$$

Let the ellipsoid $S$ be given by
$${\bf x}(\theta,\phi)=(a\cos\theta\cos\phi,b\cos\theta\sin\phi,c\sin\theta)\ .$$
Then for all points $(x,y,z)\in S$ one has
$$Q^2:={x^2\over a^4}+{y^2\over b^4}+{z^2\over c^4}={1\over a^2b^2c^2}\left(\cos^2\theta(b^2c^2\cos^2\phi+a^2c^2\sin^2\phi)+a^2b^2\sin^2\theta\right)\ .$$
On the other hand
$${\rm d}S=|{\bf x}_\theta\times{\rm x}_\phi|\>{\rm d}(\theta,\phi)\ ,$$
and one computes
$$\eqalign{|{\bf x}_\theta\times{\rm x}_\phi|^2&=\cos^4\theta(b^2c^2\cos^2\phi+a^2c^2\sin^2\phi)+a^2b^2\cos^2\theta\sin^2\theta\cr
&=\cos^2\theta\ (a^2b^2c^2\ \ Q^2)\ \cr}$$
It follows that your integral ($=:J$) is given by
$$\eqalign{J&=\int\nolimits_{\hat S} Q\ |{\bf x}_\theta\times{\rm x}_\phi|\>{\rm d}(\theta,\phi)=\int\nolimits_{\hat S}abc\  Q^2\ \cos\theta\ {\rm d}(\theta,\phi) \cr &={1\over abc}\int\nolimits_{\hat S}\cos\theta\left(\cos^2\theta(b^2c^2\cos^2\phi+a^2c^2\sin^2\phi)+a^2b^2\sin^2\theta\right)\ {\rm d}(\theta,\phi)\ ,\cr}$$
where $\hat S=[-{\pi\over2},{\pi\over2}]\times[0,2\pi]$. Using
$$\int_{-\pi/2}^{\pi/2}\cos^3\theta\ d\theta={4\over3},\quad \int_{-\pi/2}^{\pi/2}\cos\theta\sin^2\theta\ d\theta={2\over3},\quad \int_0^{2\pi}\cos^2\phi\ d\phi=\int_0^{2\pi}\sin^2\phi\ d\phi=\pi$$
we finally obtain
$$J={4\pi\over3}\left({ab\over c}+{bc\over a}+{ca\over b}\right)\ .$$
