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.