Finding the integral curve of a vector field on cylinder 
Find the integral curve of the vector field $X = \{-y, x, 1\}$ that is given on circular cylinder.

I have looked through the definition of integral curve. But I cannot find any beginner-friendly explanation of finding the curve itself. After a some research on MSE I found this post but still I don't know what to do with the cylinder. I quite new to all of these and self-learner. Some hints and explanations are greatly appreciated.
So, am I to solve just
$$\begin{cases}\dot x = -y \\ \dot y = x \\ \dot z = 1\end{cases}$$
?
 A: Short answer: yes, this is exactly the system you have to solve. Finding an integral curve of a vector field $X$ means finding $\gamma\colon I \to M$ (where $M$ is the manifold the vector field lives on --- in your case, the cylinder) such that $\dot{\gamma}(t)= X(\gamma(t))$. Since here $\gamma(t) = (\dot{x}(t),\dot{y}(t),\dot{z}(t))$ and $X(x,y,z) = (-y,x,1)$, the relation $\dot{\gamma}(t) = X(\gamma(t))$ reads $$(\dot{x}(t),\dot{y}(t), \dot{z}(t)) = (-y(t),x(t), 1),$$which gives your system. People usually skip this step and just write the system directly. Clearly $z(t) = t+z_0$ will do, and for the others, note that you get $\ddot{x}(t) = -\dot{y}(t) = -x(t)$, so $\ddot{x}(t)+x(t) = 0$, hence $$x(t) = x_0\cos(t)-y_0\sin(t), \qquad y(t) = x_0\sin(t) - y_0\cos(t).$$The flow $\Phi_X\colon \Bbb R \times C \to C$ of $X$ is complete and given by $$\Phi_X(t,(x,y,z)) = (x\cos(t)-y\sin(t), x\sin(t)-y\cos(t), t+z),$$where $C$ denotes the cylinder --- and salvo arithmetic mistakes on my part.
