# Surfaces of revolution with curvature 0

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.

• But if $C = 0$, the surface of revolution is a cylinder right?, and if $C = 1$ the surface is a plane right? – Claudia Prune Nov 21 '17 at 7:50
• 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. – John Hughes Nov 21 '17 at 12:50
• 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. – John Hughes Nov 21 '17 at 13:08
• Thank you so much @JohnHughes! – Claudia Prune Nov 21 '17 at 17:34