Showing that a mapping is an isometry Can anyone please help with this question? I have tried substituting but I'm not sure if that's correct so think I am missing something.
Let S denote the surface of revolution
$(x, y, z) = (\cos{\theta} \cosh {v}, \sin {\theta} \cosh {v}, v),$  $\ 0 < \theta < 2\pi$, $−1 < v < 1$,
and $S'$ the surface
$(x', y', z') = (u \cos{\phi}, u \sin{\phi}, \phi),$  $\ 0 < \phi < 2\pi$, $\ −1 < u < 1$.$\\$
Let $f$ be the mapping which takes the point $(x, y, z)$ on $S$ to the point
$(x', y', z')$ on $S'$ where $\theta = \phi$ and $u = \sinh {v}$. Show that f is an isometry from $S$ onto $S'$.
 A: I just change your notations a bit in order to be more convenient. Let us call:
$$\sigma_1\,(u_1,u_2) := \left(\begin{array}{c} \cosh\left(u_1\right)\,\cos\left(u_2\right)\\ \cosh\left(u_1\right)\,\sin\left(u_2\right)\\ u_1
\end{array}\right)$$ 
where $u_1\in\left(-1,1\right)$ and $u_2\in\left(0,2\,\pi\right)$. Furthermore, let us call the surface that you want to prove isometry to by
$$\sigma_2\,(u_1,u_2) := f\circ\sigma_1\,(u_1,u_2) = \left(\begin{array}{c} \sinh\left(u_1\right)\,\cos\left(u_2\right)\\ \sinh\left(u_1\right)\,\sin\left(u_2\right)\\ u_2
\end{array}\right)$$
In order to prove the isometry you have to check the equality of the following matrices:
$$\left(\begin{array}{cc} \langle\frac{\partial\sigma_1}{\partial\,u_1},\frac{\partial\sigma_1}{\partial\,u_1}\rangle &&
 \langle\frac{\partial\sigma_1}{\partial\,u_1},\frac{\partial\sigma_1}{\partial\,u_2}\rangle\\
\langle\frac{\partial\sigma_1}{\partial\,u_2},\frac{\partial\sigma_1}{\partial\,u_1}\rangle &&
 \langle\frac{\partial\sigma_1}{\partial\,u_2},\frac{\partial\sigma_1}{\partial\,u_2}\rangle \end{array}\right) = 
\left(\begin{array}{cc} \langle\frac{\partial\sigma_2}{\partial\,u_1},\frac{\partial\sigma_2}{\partial\,u_1}\rangle && \langle\frac{\partial\sigma_2}{\partial\,u_1},\frac{\partial\sigma_2}{\partial\,u_2}\rangle\\
\langle\frac{\partial\sigma_2}{\partial\,u_2},\frac{\partial\sigma_2}{\partial\,u_1}\rangle && \langle\frac{\partial\sigma_2}{\partial\,u_2},\frac{\partial\sigma_2}{\partial\,u_2}\rangle \end{array}\right)
$$
which reduces to checking the following three equtions
$$
\langle \frac{\partial\sigma_1}{\partial u_1},\frac{\partial\sigma_1}{\partial u_1} \rangle = \langle \frac{\partial\sigma_2}{\partial u_1},\frac{\partial\sigma_2}{\partial u_1} \rangle ,\\
\langle \frac{\partial\sigma_1}{\partial u_1},\frac{\partial\sigma_1}{\partial u_2} \rangle = \langle \frac{\partial\sigma_2}{\partial u_1},\frac{\partial\sigma_2}{\partial u_2} \rangle ,\\
\langle \frac{\partial\sigma_1}{\partial u_2},\frac{\partial\sigma_1}{\partial u_2} \rangle = \langle \frac{\partial\sigma_2}{\partial u_2},\frac{\partial\sigma_2}{\partial u_2} \rangle , 
$$
which themselves by a straight forward calculation are verifiable.
