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Assume a simple, piecewise smooth closed curve oriented counterclockwise around its enclosed region $D$. Apply Green's theorem with $P=-y$ and $Q=x$:The orientation assumption is essential. A clockwise unit circle, $x=\cos \theta$, $y=-\sin \theta$ for $0\le \theta \le 2\pi$, gives $-2\pi$ although its enclosed area is $\pi$. For a self-intersecting curve the same line integral measures twice the signed area, with multiplicity, rather than necessarily twice an ordinary enclosed area.

For positive semiaxes $a,b$, the ellipse has the counterclockwise parametrization $x=a\cos \theta$, $y=b\sin \theta$. ThenThereforeIf nonzero signed constants are allowed in place of positive semiaxes, the geometric area is $\pi |ab|$.

The value $f(\mathbf{a},\mathbf{b},\mathbf{c})$ is twice the signed area of the triangle traversed in that order; it is positive for counterclockwise orientation. Choose points on the parabolaFor $i<j<k$, subtracting the first row in the determinant and factoring givesEach of the three factors is positive, proving the required property for every integer $n\ge 3$.

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.

If zero conserved helical momentum reduces to $r^2\dot{\theta }+b\dot{z}=0$, a closed transverse projection $C$ has axial displacement $\Delta z=-b^{-1}{\oint }_C(xdy-ydx)=-2A/b$. Here $b\ne 0$ and $A$ is the signed area, with winding multiplicity for self-intersections. This is a path-dependent displacement even when the transverse coordinates return to their initial values. It follows from the momentum constraint and Green's theorem; it is not a prediction that every chosen closed curve is dynamically realizable.