Constructing piecewise smooth integral curve of $\ker(x \,dy - y \,dx + \,dz)$ connecting the origin to $(a, b, c) $

Question

I am trying to construct a piecewise curve $\gamma: [0, 1] \to \mathbb R^3$ such that $\gamma(0) = (0,0,0)$ and $\gamma(1) = (a, b, c)$ and each smooth piece $\eta$ is an integral curve to the distribution $$\ker (x \,dy - y \,dx + \,dz) ,$$ i.e. the curve follows the flow of some vector field in the distribution. More explicitly, \begin{equation*} \dot \eta = X(\eta) \end{equation*} for some vector field $X \in \ker (x \,dy - y \,dx + \,dz)$. Writing $X = f \partial_x + g \partial_y + h \partial_z$, I know that $f, g, h$ have to satisfy the equation $xg - yf + h = 0$, which we obtain by evaluating the $1$-form on $X$. In particular, \begin{equation*} \ker (x \,dy - y \,dx + \,dz) = \operatorname{span} \{ \partial_x + y \partial_z, \partial_y - x \partial_z \}. \end{equation*} The issue I am having is that I don't have a good heuristic for which vector fields and curves to choose. So far I have been trying to brute force a solution by trying out a number of different curves, e.g. the ones generated by the vector fields above are lines, though controlling the movement in each direction has been a difficult problem.

Answer 1

Here's a major HINT: If an integral curve $\gamma$ projects to a (simple) closed curve $\bar\gamma$ in the $xy$-plane, then the change of the $z$ coordinate along $\gamma$ is given by twice the area enclosed by $\bar\gamma$. (Why?)

Answer 2

Hint The distribution is invariant under rotations about and translations parallel to the $z$-axis, which suggests rewriting it in cylindrical coordinates $(r, \theta, z)$, for which these two types of transformations are given by the coordinate translations of $\theta$ and $z$, respectively.

Additional hint In the usual cylindrical coordinates $(r, \theta, z)$, the distribution is $$D := \ker(r^2 \,d\theta + dz) .$$ (1) Comparing this expression for $D$ with the usual formula for the area of a region in the plane bounded by a polar curve effectively recovers Ted's excellent hint. (2) What is the geometric interpretation of the zero coefficient of $dr$ in the $1$-form defining $D$ in these coordinates?