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?
Best Answer
You have not shown that they share the same first fundamental form coefficients in your brief outline. You have correctly given the result.
EDIT1:
This is quite standard and can be found in several DG text books, and other sites. including code.
Finding surface of revolution isometric to helicoid
http://xahlee.info/surface/helicoid-catenoid/helicoid-catenoid.html