I'm having some trouble with the following exercise:
Let $S$ be a surface that admits a global parametrization $\phi(u,v)$ such that the coefficients of the first fundamental form are $$E:=\left<\phi_u,\phi_u\right>=h(v)^2$$ $$F:=\left<\phi_u,\phi_v\right>=0$$ $$G:=\left<\phi_v,\phi_v\right>=1$$ where $h$ is an $C^\infty$ function such that $h(v)>0$ and $|h'(v)|<1$. Prove that $S$ is locally isometric to a surface of revolution.
My first attempt was to define the following surface of revolution: Let $\alpha(t)=(h(t), 0, z(t))$ be an curve in $\mathbb R^3$ such that $(z'(t))^2 = 1 - (h'(t))^2$, so that $||\alpha'(t)|| = 1$ for all $t$. Consider the surface of revolution $K$ with global parametrization $$\psi(u,v)=(h(v)\cos(u), h(v)\sin(u), z(v))$$
Then the coefficients of the first fundamental form for $K$ are $E=h(v)^2$, $F=0$, and $G=1$.
So $K$ is a surface of revolution with the same coefficients as $S$ and it seems a good candidate to be locally isometric to $S$. I tried defining the following map: $f:S\to K$ such that: $$f(\phi(u,v))=\psi(u,v)$$
But now I'm having some trouble proving that $f$ is a local isometry. Is this the correct approach, or is there an easier way?