Area under Arc of Cycloid

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Theorem

Let $C$ be a cycloid generated by the equations:

$x = a \paren {\theta - \sin \theta}$
$y = a \paren {1 - \cos \theta}$


Then the area under one arc of the cycloid is $3 \pi a^2$.


That is, the area under one arc of the cycloid is three times the area of the generating circle.


Proof

Let $A$ be the area under of one arc of the cycloid.

From Area under Curve, $A$ is defined by:

\(\ds A\) \(=\) \(\ds \int_0^{2 \pi a} y \rd x\)
\(\ds \) \(=\) \(\ds \int_0^{2 \pi} a \paren {1 - \cos \theta} \frac {\d x} {\d \theta} \rd \theta\)


But:

$\dfrac {\d x} {\d \theta} = a \paren {1 - \cos \theta}$

and so:

\(\ds A\) \(=\) \(\ds a^2 \int_0^{2 \pi} \paren {1 - \cos \theta}^2 \rd \theta\)
\(\ds \) \(=\) \(\ds a^2 \int_0^{2 \pi} \paren {1 - 2 \cos \theta + \cos^2 \theta} \rd \theta\)
\(\ds \) \(=\) \(\ds a^2 \int_0^{2 \pi} \paren {1 - 2 \cos \theta + \frac 1 2 + \frac {\cos 2 \theta} 2} \rd \theta\)
\(\ds \) \(=\) \(\ds a^2 \intlimits {\theta + 2 \sin \theta + \frac {\theta} 2 + \frac {\sin 2 \theta} 4} 0 {2 \pi}\)
\(\ds \) \(=\) \(\ds 3 \pi a^2\)

$\blacksquare$


Also see


Historical Note

The proof of the Area under Arc of Cycloid was published in $1644$ by Evangelista Torricelli.

Galileo, by cutting out templates and measuring their masses, had previously estimated the area as being approximately $3$ times the area of the generating circle.

It is believed that Gilles Personne de Roberval found this result earlier than Torricelli, but lost priority when he failed to publish his proof.


Sources

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