$$ \int \int_S (x^2+y^2)d\sigma$$
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 2×2 matrix or 3×3? I feel that if I compute it for a 3×3 matrix, then I am computing the Jacobian determinant for a volume integral, and not an surface area integral — which is a bit confusing.
Thanks,
Best Answer
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 $r=\text{radius of unit sphere}=1$ $$=\int_{0}^{2\pi}\left[\int_{0}^{\pi}\frac{3\sin\theta-\sin 3\theta}{4}d\theta \right]d\phi$$ $$=\int_{0}^{2\pi}\left[-\frac{3}{4}\cos\theta+\frac{1}{12}\cos 3\theta \right]_{0}^{\pi}d\phi$$ $$=\int_{0}^{2\pi}\left[\frac{3}{2}-\frac{1}{6} \right]d\phi$$ $$=\int_{0}^{2\pi}\left[\frac{4}{3}\right]d\phi$$ $$=\frac{4}{3}\int_{0}^{2\pi}d\phi$$ $$=\frac{4}{3}[\phi]_{0}^{2\pi}$$ $$=\frac{4}{3}[2\pi]=\frac{8\pi}{3}$$