What is the Surface of Revolution with Constant Curvature +1?

[mooshasta]

Homework Statement

I'm trying to find a surface of revolution with Gauss curvature K of +1 at all points, which doesn't lie in a sphere.

Homework Equations

The surface is parametrized as $\psi (t, \theta ) = ( x(t), y(t) cos \theta , y(t) sin \theta )$

I have the equation
$$K = \frac{x' (x'' y' - x' y'')}{y(x'^2 + y'^2)^2}$$

The Attempt at a Solution

I am thinking it has to do with the curve $\alpha (t) = (x(t),y(t))$ not having unit speed, but I am kind of stuck as to where to go from there.


Thanks!

 

[Dick]

Is there one? The only way out I can think of is to make it disconnected. A surface of constant guassian curvature is locally isometric to a sphere.

 

[mooshasta]

I know that for a unit-speed $\alpha (t)$, the equation reduces to $K = \frac{-y''}{y}$, which does clearly represent a sphere.

The way the question is worded on my homework seems to point to the fact that that reduction only applies to unit-speed curves, which is why I think that perhaps an $\alpha (t)$ that doesn't have unit-speed perhaps can give a surface of revolution with constant curvature +1 that isn't a sphere... but maybe there's another "gimmick" that I'm overlooking...

Anyways, thanks for your help!