This is the question I have:
Let $S$ denote the surface of revolution $$(x,y,z)=(\cos\theta \cosh v, \sin \theta \cosh v, v)$$ $0 < \theta < 2 \pi$ and $-\infty < v <\infty$
and $S'$ the surface $$(x',y',z')=(u \cos \phi, u \sin \phi, \phi)$$
$0 < \phi < 2\pi$ and $ -\infty < u < \infty$
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 $S'$
This is my proof.
I reparametrise the helicoid as $$S'=(\sinh v \cos \theta, \sinh v \sin \theta ,\theta )$$
I then find the first fundamental forms of the reparametrized helicoid and the first fundamental forms of the catenoid and show that they are the same.
Is the outline of the proof correct?
Have I re-parametrized the helicoid correctly?
Do I need to reparametrise the catenoid?