# Surfaces of revolution perpendicular to the plane xOy

Let $S$ be a surface of revolution parametrized by $(\varphi(v) \cos u, \varphi(v) \sin u, \delta(v))$. Assume $S$ has constant Gaussian curvature equal to 1. Since $K = -\frac{\varphi''}{\varphi}$, $\varphi'' + \varphi = 0$. If we solve this differential equation we obtain that \begin{equation} \varphi(v) = c_1 \cos(v) + c_2 \sin(v). \end{equation} My question is the following: if we know that $S$ intersects perpendicularly the plane $xOy$, can we say that $\varphi(v) = c_1 \cos(v)$?

Apparently, this is true, but I don't know why. I found the Gaussian map and at points $p$, where $p$ is in the intersection of $S$ and the plane $xOy$, $\varphi'(v) = 0$. I try to play with this but nothing came up. Please if you can help me, I will really appreciate it. Thanks!

Anyway, the condition means that $(\varphi'(v)\cos u, \varphi'(v)\sin u, \delta'(v))$ is a vertical vector for the points that $\delta(v) = 0$, which you can assume happens for $v=0$. This means that $\varphi'(0) = 0$.
Now, by your previous work we have $\varphi'(v) = -c_1 \sin(v) + c_2\cos(v)$, and so $c_2 = \varphi'(0) = 0$. So $\varphi(v) = c_1 \cos (v)$, as wanted.