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Evaluating this surface integral

$\int_{S}(a^2x^2+b^2y^2+c^2z^2)^{1/2}dS$ over the ellipsoid $S:ax^2+by^2+cz^2=1$

I am well aware of evaluating surface integral over any given sphere $x^2+y^2+z^2=a^2$ using the transformation $R(u,v)=\langle a\cos(u)\sin(v), a\sin(u)\sin(v), a\cos(v) \rangle$ and obtaining $dS =|R_u \times Rv| = a^2\sin(v)$.

So for the given surface integral, I want to be able to convert the ellipsoid to a sphere and then use the above transformation.

The transformation that seems reasonable to me for this task would be $R(u,v)=\langle \frac1a\cos(u)\sin(v), \frac1b\sin(u)\sin(v), \frac1c\cos(v)\rangle$. But evaluating $dS=|R_u\times R_v|$ is not yielding any simple manageable form.

So I thought, I would first use the transformation to convert into sphere first using

$x=X/a, \;y=Y/b,\;z=Z/c$ But I don't know how to evaluate $dS$ in this case since it has three unknowns and something like $|R_u \times R_v|$ doesn't works.

Can you please help me how can solve this problem?

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Let $f(x, y, z) = a x^2 + b y^2 + c z^2 - 1$. Notice that the integrand is $|\nabla f|/2$. Therefore, $$\int_{\partial \mathcal E} \frac {|\nabla f|} 2 \,dS = \frac 1 2 \int_{\partial \mathcal E} |\nabla f| \,\hat {\mathbf n} \cdot d\mathbf S = \frac 1 2 \int_{\partial \mathcal E} |\nabla f| \,\frac {\nabla f} {|\nabla f|} \cdot d\mathbf S = \\ \frac 1 2 \int_{\partial \mathcal E} \nabla f \cdot d\mathbf S = \frac 1 2 \int_{\mathcal E} \nabla^2 f \,dV = \\ (a + b + c) \operatorname{Vol}(\mathcal E) = \frac {4 \pi (a + b + c)} {3 \sqrt {a b c \,}}.$$

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