Prove that the static moment of an arbitrary arc

Prove that the static moment of an arbitrary arc of a parabola with respect to its axis is proportional to the difference of the radii of curvature at the endpoints of that arc, with proportionality coefficient $\dfrac p3$, where $p$ is the parabola's parameter.

Step 1: Set up the static moment

For $x^2=2py$: $y=\dfrac{x^2}{2p}$, $y'=\dfrac xp$, so $ds=\sqrt{1+\dfrac{x^2}{p^2}}\,dx$, and

\[M_y=\int_{x_1}^{x_2}x\sqrt{1+\frac{x^2}{p^2}}\,dx\]

Step 2: Evaluate via substitution

With $u=1+\dfrac{x^2}{p^2}$, $x\,dx=\dfrac{p^2}2\,du$:

\[M_y=\frac{p^2}3\left[\left(1+\frac{x_2^2}{p^2}\right)^{3/2}-\left(1+\frac{x_1^2}{p^2}\right)^{3/2}\right]\]

Step 3: Radius of curvature

With $y''=\dfrac1p$: \[\rho(x)=\frac{[1+(y')^2]^{3/2}}{|y''|}=p\left(1+\frac{x^2}{p^2}\right)^{3/2}\]

Step 4: Combine

So $\left(1+\dfrac{x_i^2}{p^2}\right)^{3/2}=\dfrac{\rho_i}p$ for $i=1,2$, giving

\[M_y=\frac{p^2}3\left(\frac{\rho_2}p-\frac{\rho_1}p\right)=\frac p3(\rho_2-\rho_1)\]

\[\boxed{M_y=\frac p3(\rho_2-\rho_1)}\]

Step 5: Remark

This holds for any arc of the parabola, independent of its location — a clean illustration of how the arc-length weighting in $M_y=\int x\,ds$ happens to produce exactly the same power of $\left(1+x^2/p^2\right)$ that appears in the curvature formula.