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.


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


  • $\begingroup$ This was a brief outline of the proof, But is the general outline of the proof correct? Do we not need to parametrise the catenoid? $\endgroup$ – user232183 May 8 '16 at 13:50
  • $\begingroup$ You already parametrised catenoid in first line! Hope you understand what you are doing. $\endgroup$ – Narasimham May 8 '16 at 14:00
  • $\begingroup$ Lets say I am now working out the first fundamental form and I calculate $E=\langle S'_\theta , S'_\theta \rangle$. I then get $\sinh^2 v \cos^2 \theta + \sinh^2 v \sin^2 \theta+ \theta^2$ which becomes just $\sinh^2 v + \theta^2$. Is this correct? $\endgroup$ – user232183 May 8 '16 at 14:33
  • $\begingroup$ Ok,is it done? what about $F$ and $ G?$ $\endgroup$ – Narasimham May 8 '16 at 14:38
  • $\begingroup$ Before I work out the rest I want to check something, does $S'_\theta =(-\sinh v\ sin \theta, \sinh v \cos \theta, 1)$ and does $S'_v = (\cosh v \cos \theta, \cosh v \sin \theta, \theta)$? So if these are correct then $F=\theta$? $\endgroup$ – user232183 May 8 '16 at 18:19

This is good. You should find that the first fundamental form for both $S$ and $S^{\prime}$ are the same which is equivalent to being locally isometric and are equal to: $$ds^2 = \cosh^{2}(v)\left(dv^2 + d\theta^2\right)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.