How do you integrate surface area in spherical coordinates? A single-valued function of spherical coordinates $r(\theta,\phi)$ (where $(\theta,\phi)\in[0,\pi]\otimes[0,2\pi]$) naturally defines a surface in 3D space.
How does one calculate the surface area of this object? I was able to deduce from the Pythagorean theorem that 
$$S=\int_0^\pi \sin(\theta) d\theta\int_0^{2\pi}d\phi\,R(\theta,\phi)^2\sqrt{1+m(\theta,\phi)^2}$$
where $m(\theta,\phi)$ is the norm of the gradient of the surface at $(\theta,\phi)$.
Assuming that this is the correct expression for the integrated surface area, how does one compute the norm of the gradient, $m$? It's obviously not $|\nabla r|=|(\partial_\theta r,\partial_\phi r)^\mathsf{T}|$, and I suspect the metric tensor $g$ is involved, but I'm having trouble working it out. 
I have a conjecture that the correct expression is 
$$m=\frac{\sqrt{(\nabla r)\mathbf{g}^{-1}(\nabla r)}}{r},$$
and this seems to agree with numerical tests. If this is correct, is there a way to prove this?
 A: The general strategy to find length or area or volume or hypervolume in an alternative coordinate system is as follows:
Find the Jacobian of the coordinate transformation. In this case:
$$J = \left[\begin{matrix}
\frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi} & \frac{\partial x}{\partial r} \\
\frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial \phi} & \frac{\partial y}{\partial r} \\
\frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial \phi} & \frac{\partial z}{\partial r}
\end{matrix}\right] = \left[ \begin{matrix}
\sin \theta \cos \phi & r \cos \theta \cos \phi & -r \sin \theta \sin \phi \\
\sin \theta \sin \phi & r \cos \theta \sin \phi & \;\;\;r \sin \theta \cos \phi \\
\cos \theta & - r \sin \theta & 0
\end{matrix}\right]$$
Now take the determinant of the Jacobian to obtain the volume element in the new coordinate system:
$$\vert J \vert = r^2 \sin \theta$$
And insert this into the integral which covers the surface, line, or volume of your choosing:
$$\int_0^{2\pi} \int_0^\pi \Big(r(\theta,\phi)^2 \sin \theta \Big)\; \;\mathrm d \theta  \;\mathrm d \phi$$
I think I understand where you got the $\sqrt{1-m^2}$ term from, but it shouldn't be there. Probably you were thinking of something analagous to integrating the path length of a graph $y = f(x)$. Everything is accounted for by setting the $r^2$ in the determinant as a function of $\theta, \phi$.
