Showing that a mapping is an isometry

Question

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 $S'$.

Answer 1

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.