Is my proof of showing a helicoid and catenoid are isometric, correct? 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?

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

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