Finding area under the cycloid without parametrizing

Question

I tried to calculate the area under the cycloid without parametrizing the $x$ & $y$ coordinate in angle $\theta$.

Let's say the radius of circle is $R$ and we are rolling $2\pi$ rad.

If I assume that $y(\theta) = R(1-\cos(\theta))$ can be written as $y(x) = R(1-\cos(\theta(x)))$, the area under the cycloid of $2\pi$ cycle should be the following integral:

$$\int_0^{2\pi R} R(1-\cos(\theta(x))dx = 3\pi R^2 $$

Then,

$$\int_0^{2\pi R} R(1-\cos(\theta(x))dx = \int_0^{2\pi R} \ R dx - \int_0^{2\pi R} R\cos(\theta(x))dx $$ $$ = 2\pi R^2 - \int_0^{2\pi R} R\cos(\theta(x))dx$$ $$ = 3\pi R^2$$

Which gives us: $\int_0^{2\pi R} \cos(\theta(x))dx = -\pi R$

But, since $\cos(\theta(x)) \in [-1,1]$, this restricts the radius R to be: $-\frac{1}{\pi} \le R \le \frac{1}{\pi}$.

(I find this interesting because setting $\theta(x)$ now lead us to some constraint on radius $R$)

We know $x(\theta) = R(\theta - \sin(\theta))$.

Taylor expanding $\sin(\theta)$ then subtracting $\theta$ gives: $$\frac{x}{R} = \sum_{i=0}^\infty \frac{(-1)^n \theta^{2n+3}}{(2n+3)!} $$ If there is a function $F(\theta) = \sum_{i=0}^\infty \frac{(-1)^n \theta^{2n+3}}{(2n+3)!}$, then $\theta(x) = F^{-1}(\frac{x}{R})$.

Is there such a function $F$ that allows us to integrate this cycloid situation in a closed form?

If not, why?

Answer 1

Recall the Cartesian equation for one half-period of the cycloid with radius $R$,

$$x(y) = R \cos^{-1}\left(1-\frac yR\right) - \sqrt{2Ry-y^2}$$

I've plotted a few other half-periods below $(n\in\Bbb Z)$:

Then the area under a full period, by symmetry, is given by

$$2 \int_0^{2R} \int_{x(y)}^{R\pi} dx \, dy$$

I think you should be able to take it from here.

Answer 2

Notice that reindexing the sum we get\begin{align*} \sum_{n\ge 0}\frac{(-1)^n \theta^{2n+3}}{(2n+3)!} &\overset{\color{blue}{n \to n-1}}{=} \sum_{n \ge 1} \frac{(-1)^{n-1}\theta^{2n+1}}{(2n+1)!} \\ &= \underbrace{\color{purple}{\frac{(-1)^0\theta^{2(0)+1}}{(2(0)+1)!}} \color{green}{- \frac{(-1)^0\theta^{2(0)+1}}{(2(0)+1)!}}}_{0} - \sum_{n \ge 1} \frac{(-1)^{n}\theta^{2n+1}}{(2n+1)!} \\ &= \color{purple}{\theta} - \sum_{n \ge \color{green}{0}} \frac{(-1)^{n}\theta^{2n+1}}{(2n+1)!} \\ & = \theta - \sin(\theta) \end{align*} where on the last step you recognize the Taylor series of $\sin(\theta)$. As for the closed form of the inverse, @user170231 linked a useful resource in the comments where they discuss this. Some other possibly useful posts about this are this and this.