Computing the Weingarten map/shape operator Let $M$ be a surface of revolution of the form
$F(t,s)=(r(t)\cos(s),r(t)\sin(s),z(t))$
where $\gamma(t)=(r(t),z(t))$ is a curve with unit speed and $r(t)>0$.
I know the unit normal to $M$ is $\xi(s,t)=(-z'(t)\cos(s),-z'(t)\sin(s),r'(t))$ (or written as $-z'(t)\cos(s) \partial_x -z'(t)\sin(s) \partial_y+  r'(t) \partial_z$)
and now I want to compute the Weingarten map $S^{\xi}$.
I know that for $X \in \mathfrak{X}(M)$, $S^{\xi}(X)=-\nabla_X \xi=-X(\xi)$.
$X$ is then of the form $X(s,t)=X_s \partial_s +X_t \partial_t$. Now I have problems how to plug in $\xi$ in $X$. Can I somehow compute $\partial_x$ in terms of $\partial_s, \partial_t$?
I always have big problems doing such computations on concrete examples, so I'm sorry if something I wrote here is complete nonsense..
 A: I think the best way to calculate the matrix $S$ for the shape operator in the $(\partial t,\partial s)$ basis is to use
$$h=g\cdot S
$$
where $g$ and $h$ are the matrices corresponding to the first and second fundamental form:
$$g=\begin{pmatrix}\langle\frac{\partial F}{\partial t},\frac{\partial F}{\partial t}\rangle&\langle\frac{\partial F}{\partial t},\frac{\partial F}{\partial s}\rangle\\\langle\frac{\partial F}{\partial t},\frac{\partial F}{\partial s}\rangle&\langle\frac{\partial F}{\partial s},\frac{\partial F}{\partial s}\rangle\end{pmatrix}\space\space\space\space
h=\begin{pmatrix}\langle\frac{\partial^2 F}{\partial t\partial t},\xi\rangle&\langle\frac{\partial^2 F}{\partial t \partial s},\xi\rangle \\\langle\frac{\partial^2 F}{\partial t \partial s},\xi\rangle &\langle\frac{\partial^2 F}{\partial s\partial s},\xi\rangle \end{pmatrix}$$
You should arrive at
$$g=\begin{pmatrix}1&0\\0&r^2\end{pmatrix}\space\space\space\space h=\begin{pmatrix}r'z''-r''z'&0\\0&rz'\end{pmatrix}\space\space\space\space
S=\begin{pmatrix}r'z''-r''z'&0\\0&z'/r\end{pmatrix}$$
