I am trying to find all the surfaces of revolution with Gaussian curvature $K \equiv 0$.

This is what I got so far. If we assume the surface of revolution is parametrized by $(\varphi(v) \cos u, \varphi(v) \sin u, \delta(v))$. Then since $\varphi'' + 0\cdot\varphi = 0$, $\varphi(v) = C\cdot v$ which implies that $\delta(v) = \int_0^v \sqrt{1 - C^2}dv = \sqrt{1 - C^2}\cdot v$. I do not know how to continue from this point. Any ideas? Thanks!


Assuming what you've done is correct, a generating curve (i.e., a longitude of your surface of revolution) is given by $u = 0$, so we have

$$ s(v) = (Cv, 0, \sqrt{1-C^2} v) = v(C, 0, \sqrt{1-C^2}) $$ which is a straight line.

That means that the corresponding surface of revolution is a cone, and you're done.

  • $\begingroup$ But if $C = 0$, the surface of revolution is a cylinder right?, and if $C = 1$ the surface is a plane right? $\endgroup$ – Claudia Prune Nov 21 '17 at 7:50
  • $\begingroup$ I carefully wrote "assuming what you've done is correct". But it's not correct, because $\phi'' + 0 \cdot \phi = 0$ implies that $\phi(v) = Cv + D$; the $C = 0$ case, with $D$ nonzero, gets you the cylinder. But you'll want to check out what that revised answer tells you about $\delta$ as well. $\endgroup$ – John Hughes Nov 21 '17 at 12:50
  • $\begingroup$ BTW, you seem to have used the standard formulas for curvature of a surface of revolution, but also seem to have assumed that $\phi'(v)^2 + \delta'(v)^2 = 1$, i.e., that the parameterization of the longitude is unit-speed. There's nothing wrong with that, but it's not necessary. You cou.d instead have just said "let's let $\delta(v) = v$ (i..e, treated the longitude curve as the graph of a function of $z$ in the $zy$-plane) and simplified a little. $\endgroup$ – John Hughes Nov 21 '17 at 13:08
  • $\begingroup$ Thank you so much @JohnHughes! $\endgroup$ – Claudia Prune Nov 21 '17 at 17:34

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.