Find the area bounded by one arc of the cycloid x = a(t

Find the area bounded by one arc of the cycloid $x = a(t - \sin t)$, $y = a(1 - \cos t)$, and the segment of the $x$-axis.

We will apply the parametric area formula directly over one period of the parameter.

Step 1: Understand the cycloid

The cycloid is traced by a point on the circumference of a circle of radius $a$ rolling along the $x$-axis. At $t=0$: $x=0,y=0$. At $t=2\pi$: $x=2\pi a,y=0$ (end of one arch). Maximum height $y=2a$ occurs at $t=\pi$.

Step 2: Set up the area formula

For a parametric curve, $A=\int_{t_1}^{t_2} y\,\frac{dx}{dt}\,dt$. Here $\frac{dx}{dt}=a(1-\cos t)$, so:

\[A = \int_0^{2\pi} a(1-\cos t)\cdot a(1-\cos t)\,dt = a^2\int_0^{2\pi}(1-\cos t)^2\,dt\]

Step 3: Expand the integrand

\[(1-\cos t)^2 = 1 - 2\cos t + \cos^2 t = 1-2\cos t+\frac{1+\cos2t}{2} = \frac32-2\cos t+\frac{\cos2t}{2}\]

Step 4: Integrate

\[A = a^2\left[\frac{3t}{2} - 2\sin t + \frac{\sin 2t}{4}\right]_0^{2\pi} = a^2(3\pi-0) = 3\pi a^2\]

\[\boxed{A = 3\pi a^2 \text{ square units}}\]

Step 5: Geometric interpretation

The area under one arch of the cycloid is exactly three times the area $\pi a^2$ of the rolling circle — a result discovered by Roberval in the 17th century.

Step 6: Numerical check

For $a=1$: $A=3\pi\approx9.42$, and the rolling circle's area is $\pi\approx3.14$; indeed $9.42\approx3\times3.14$. ✓

Summary

\[\boxed{A = 3\pi a^2 \text{ square units}}\]

Key insight: The cycloid encloses an area exactly three times that of its generating circle, a landmark result in the early history of calculus.