<s>[INST] 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
$$
S = \{(x,y,z)\in\mathbb{R}^3: \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}= 1\}
$$
 [/INST] What an interesting integral.  I had to resort to referring to the first differential form of the spherical parametrization, but doing that, I am amazed at how this turns out.
We parametrize in the usual way:
$$x=a\sin{u} \cos{v}$$
$$y=b\sin{u} \sin{v}$$
$$z=c \cos{u}$$
where $u \in [0,\pi)$ and $v \in [0,2 \pi)$.  The coefficients of the first differential form are
$$E=(a^2 \sin^2{v}+b^2 \cos^2{v}) \sin^2{u}$$
$$F=(b^2-a^2) \sin{u} \cos{u} \sin{v} \cos{v}$$
$$G=(a^2 \cos^2{v}+b^2 \sin^2{v}) \cos^2{u}+c^2 \sin^2{u}$$
The stated integral is equal to
$$\int_0^{\pi} du \: \int_0^{2 \pi} dv \: \sqrt{E G-F^2} \sqrt{\frac{\sin^2{u} \cos^2{v}}{a^2} + \frac{\sin^2{u} \sin^2{v}}{b^2} + \frac{\cos^2{u}}{c^2}}$$
There is an enormous amount of algebra involved in simplifying the integrand. Miraculously, it simplifies a lot, and the integral is equal to
$$\frac{1}{a b c}  \int_0^{\pi} du \: \sin{u} \int_0^{2 \pi} dv\: (a^2 b^2 \cos^2{u} + b^2 c^2 \sin^2{u} \cos^2{v} + a^2 c^2 \sin^2{u} \sin^2{v})$$
I really couldn't believe this myself at first, but it does check out for the case $a=b=c$.  In any case, these integrals are much easier than one would expect from first seeing this problem, and the reader should have no trouble evaluating them by hand.  The result is
$$\frac{4 \pi}{3} \left ( \frac{a\, b}{c} + \frac{a\, c}{b} + \frac{b\, c}{a} \right)$$
</s>