I am trying to calculate the mean curvature $J=\kappa_1+\kappa_2$ of a surface in the $(r, z)$ plane, the surface has cylindrical symmetry around the $z$ axis. Using Monge parameterization, getting the second fundamental form and taking the derivative with respect to the arc-length $s$ I have arrived at the values for the two principal curvature: $$\kappa_1=\frac{\sin\phi}{r} ; \kappa_2=\frac{d\phi}{ds}$$ Noting that $\phi=\arctan(\frac{dz}{ds}/\frac{dr}{ds})$
Resulting with a mean curvature: $J=\frac{\sin\phi}{r}+\frac{d\phi}{ds}$
I also found on a website (that proves this but I can't find it again) showing that $J$ can be contracted to the form of: $J=\frac{1}{r}\frac{d(r\sin\phi)}{dr}$ by a change of variables from $s$ to $r$.
Differentiating gives me: $J=\frac{\sin\phi}{r}+\frac{d(\sin\phi)}{dr}$.
So I concluded that $\frac{d(\sin\phi)}{dr}=\frac{d\phi}{ds}$ but I don't manage to prove it myself. Generally, I need to find a way to change variables so that $\phi(s)\rightarrow\phi(r)$.
The shape revolves around the $z$ axis, and decays to infinity.
I followed the derivation of the principal curvatures given in a paper by Markus Deserno on pages 15-16 he gives the results. It may be that I misunderstood his derivation.
I believe this solves my problem, and I do feel a little ashamed I didn't use the chain rule to solve it: $$\frac{d\phi}{ds}\cdot\frac{dr}{dr}=\frac{d\phi}{dr}\cdot\frac{dr}{ds}$$ Now we know that $\cos\phi=\frac{dr}{ds}$ from the way we defined $ds$ and $\phi$, and using the general solution for the derivative of a sine of a function $\frac{d\sin(f(x))}{dx}=f'(x)\cos(f(x))$: $$\frac{d\phi}{dr}\cos\phi=\frac{d\sin\phi}{dr}$$