Computing a double integral over a surface S, where S is the unit sphere, $$ \int \int_S (x^2+y^2)d\sigma$$
Where S is the unit sphere centered at (0,0,0), and $\sigma$ is surface area.
I arrived at the correct answer of $\large \frac{8\pi}{3}$, but I took an (educated?) guess on the set-up of the integral.
Instead of the usual volume element of the sphere, in spherical coordinates,
$$dV=\rho^2sin(\phi)d\rho d\theta d\phi$$
I just simply omitted the $\rho^2 d\rho$ factor of the Jacobian and guessed that the remaining factor $$sin(\phi) d\theta d\phi$$
represents the "surface area element" of the sphere. 
Integration gave me the correct answer, and I didn't have to think about the tricky ndS factor to compute.
Is my work ok - or did I arrive at the correct answer with a wrong method?
Any suggestions or alternative solutions are welcome -- especially if you would like to comment on the computation of the n.dS factor in this problem.
EDIT:  I think I may have completely missed the point of the question, too.  My trying to get an n.dS factor is only relevant for flux integrals of vector fields.  This question is an "ordinary" double integral over the surface area of a sphere -- it is not considered a "surface integral".  Do I have it correct now?  So, what I really should be doing is to compute a Jacobian determinant factor.  But would the Jacobian be a determinant of a 2x2 matrix or 3x3?  I feel that if I compute it for a 3x3 matrix, then I am computing the Jacobian determinant for a volume integral, and not an surface area integral -- which is a bit confusing.
Thanks,
 A: The correct surface element for spherical coordinates is given by $dS = \rho^2 \sin \phi \cdot d\theta d\phi$ (whose derivation is widely available, e.g. in textbooks), i.e. the surface element you used is missing a factor of $\rho^2$.
You were asked to integrate over the unit sphere, i.e. $\rho = 1$ everywhere, so in this case your omission of $\rho$ did not have an impact on your answer. (So in fact, the surface element you used is correct--you happened to correctly simplify the expression for $dS$ in this special case!)
If you were asked to integrate over some other shape, say a cylinder (for some reason...), with the coordinates you used, you would have gotten a wrong answer.
A: Using spherical coordinates as follows $$y=r\sin\theta\cos\phi$$ $$x=r\sin\theta\sin\phi$$ 
$$d\sigma=\text{Elementary surface area of sphere}$$$$=(r\sin\theta d\phi)(rd\theta)=r^2\sin\theta d\theta d\phi$$
Now, $$\int\int_S(x^2+y^2)d\sigma$$
$$=\int_{0}^{2\pi}\int_{0}^{\pi}(r^2\sin^2\theta\cos^2\phi+r^2\sin^2\theta\sin^2\phi)r^2\sin\theta d\theta d\phi$$
$$=\int_{0}^{2\pi}\int_{0}^{\pi}(r^2\sin^2\theta(\cos^2\phi+\sin^2\phi))r^2\sin\theta d\theta d\phi$$
$$=\int_{0}^{2\pi}\int_{0}^{\pi}r^4\sin^3\theta d\theta d\phi$$
$$=r^4\int_{0}^{2\pi}\int_{0}^{\pi}\sin^3\theta d\theta d\phi$$
Setting $r=\text{radius of unit sphere}=1$
$$=\int_{0}^{2\pi}\left[\int_{0}^{\pi}\frac{3\sin\theta-\sin 3\theta}{4}d\theta \right]d\phi$$
$$=\int_{0}^{2\pi}\left[-\frac{3}{4}\cos\theta+\frac{1}{12}\cos 3\theta \right]_{0}^{\pi}d\phi$$
$$=\int_{0}^{2\pi}\left[\frac{3}{2}-\frac{1}{6} \right]d\phi$$
$$=\int_{0}^{2\pi}\left[\frac{4}{3}\right]d\phi$$
$$=\frac{4}{3}\int_{0}^{2\pi}d\phi$$
$$=\frac{4}{3}[\phi]_{0}^{2\pi}$$
$$=\frac{4}{3}[2\pi]=\frac{8\pi}{3}$$
