### Mamikon’s Theorem and the area under a cycloid arch

The Cycloid

The cycloid is the locus of a point fixed to the rim of a circular disk that is rolling along a straight line (see figure). The parametric equations for the cycloid are $\displaystyle x = r (\theta - \sin\theta)\,, \qquad y = r (1 - \cos\theta ) \ \ \ \ \ (1)$

where ${\theta}$ is the angle through which the disk has rotated. The centre of the disk is at ${(x_0,y_0) = (r\theta, r)}$. * * * * *

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* * * * * The cycloid, generated by a rolling disk. A single arch of the cycloid is shown, bounded within a rectangle of area ${4\pi r^2}$.

The differentials of the coordinates (1) are $\displaystyle \mathrm{d}x = r (1 - \cos\theta)\mathrm{d}\theta\,, \qquad \mathrm{d}y = r \sin\theta\,\mathrm{d}\theta \,. \ \ \ \ \ (2)$

From these we find the increment of arc-length ${\mathrm{d}\ell}$, $\displaystyle \mathrm{d}\ell = \sqrt{ \mathrm{d}x^2 + \mathrm{d}y^2 } = 2r\sin\textstyle{\frac{1}{2}}\theta\,\mathrm{d}\theta \,, \ \ \ \ \ (3)$

and the increment of area ${\mathrm{d}A}$, $\displaystyle \mathrm{d}A = y\,\mathrm{d}x = r^2(1-\cos\theta)^2\mathrm{d}\theta \,. \ \ \ \ \ (4)$

The length ${L}$ of an arch is easily computed by integrating (3): $\displaystyle L = \int_{0}^{2\pi} 2r\sin\textstyle{\frac{1}{2}}\theta\,\mathrm{d}\theta = 8 r \,.$

We see that the arc-length does not depend on ${\pi}$. The area under an arch is the integral of (4) over the interval ${\theta\in[0,2\pi]}$, which is slightly more tricky: $\displaystyle A = \int_{0}^{2\pi} r^2(1-\cos\theta)^2\mathrm{d}\theta = 3\pi r^2 \,. \ \ \ \ \ (5)$

Thus, the area under an arch is three times the area of the generating circle or three-quarters of the area of the surrounding rectangle.

Using Mamikon’s Theorem

To get the area ${A}$ under the cycloid arch, we required the parametric equations (1) for the cycloid and the evaluation of a definite integral (5). We will now show, by using Mamikon’s Theorem, that the area can be found by simple geometric reasoning, without any equations or integrations (Apostol, 2000).

For rolling motion, the instantaneous centre of rotation of the disk is the point of contact P (figure above) and the vertical line PQ is the diameter of the disk. All points T, T’, T” on the boundary of the disk move in directions orthogonal to the chords joining them to P. In particular, the angle PTQ is a right angle so TQ is a segment of the tangent to the cycloid. As the disk rolls, the point Q moves and the chords tangent to the cycloid sweep out the region above this curve.

We draw a set of tangents for the left-hand side of the arch in the figure below (left panel). Now moving all these segments horizontally and parallel to themselves so that they have a common upper point O (figure below, right panel), we see that they fill a semi-circle of radius ${r}$. The tangents for the right-hand side of the arch complete the circle. Thus, the area of the tangent cluster is ${\pi r^2}$, the same area as the generating disk.

By Mamikon’s theorem, the area of the rectangle above the arch, the tangent sweep, is equal to the area of the tangent cluster or ${\pi r^2}$. But the total area of the rectangle is ${4\pi r^2}$, so the area of the arch must be ${3\pi r^2}$, or three times the area of the generating disk.

Sources ${\bullet}$ Apostol, Tom M., 2000: A visual approach to calculus problems, Engineering and Science, vol. LXIII, no. 3, 22–31. A copy of this article can be found here: PDF. ${\bullet}$ Mamikon Mnatsakanian, 1997: Annular rings of equal area. Math Horizons, 5 (2), 5–8.