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
$$
S = \{(x,y,z)\in\mathbb{R}^3: \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}= 1\}
$$

Note that the outward unit normal ${\bf n}$ to the ellipsoid is given by ${\displaystyle {({x \over a^2}, {y \over b^2}, {z \over c^2}) \over \sqrt{ \left(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\right)}}}$. 
So if ${\bf F} = ({x \over a^2}, {y \over b^2}, {z \over c^2})$, your integral is $\int_S {\bf F} \cdot {\bf n}\,dS$. By the divergence theorem, this is equal to $\int_E div\,\mathbf{F}$, where
$E$ is the ellipsoid's interior. But $ div \,\mathbf{F}$ is the constant ${1 \over a^2} + {1 \over b^2} + {1 \over c^2}$ and the ellipsoid has volume ${4\pi \over 3}abc$, so the integral will evaluate to 
$${4\pi \over 3}abc \times \bigg({1 \over a^2} + {1 \over b^2} + {1 \over c^2}\bigg)$$
$$= {4\pi \over 3} \bigg({bc \over a} + {ac \over b} + {ab \over c}\bigg)$$
