 Solution

In the negligible bead-mass limit the bead contribution to $\mathcal{Q}$ vanishes. Since the particle starts at rest, $\mathcal{Q}=0$, and hence $r^2\dot{\theta }+b\dot{z}=0$. Assume $b\ne 0$ and a regular massless limit. Integrating along the trajectory giveswhere $C$ is the closed projection onto the plane perpendicular to the coil axis. The given final $r$ and $\theta$ ensure that this projection closes. Green's theorem identifiesThus $A$ is the signed area enclosed by the transverse projection, positive for counterclockwise traversal viewed from positive $z$. Self-intersecting projections contribute their winding-weighted areas; it is neither an unsigned area nor an area of a surface in three dimensions. For $b>0$, a positive signed area gives negative axial displacement. This is signed-area displacement under a helical constraint: returning to the same projected coordinates can leave a nonzero axial displacement.

The plotted loop illustrates the geometric integral only; it is not asserted to be a particular spring trajectory.