Answer
The area under one arch of the cycloid is $A = 3\pi {R^2}$, which is equal to three times the area of the generating circle of radius $R$.
Work Step by Step
The parametrization of the cycloid:
$c\left( t \right) = \left( {Rt - R\sin t,R - R\cos t} \right)$.
We have
$x\left( t \right) = Rt - R\sin t$, $x'\left( t \right) = R - R\cos t$.
$y\left( t \right) = R - R\cos t$.
The $x$-interval for one arch of the cycloid is $0 \le x \le 2\pi R$ as is shown in Figure 25. The corresponding $t$-interval is $0 \le t \le 2\pi $.
Using Eq. (11) the area under one arch of the cycloid is
$A = \mathop \smallint \limits_{t = 0}^{2\pi } \left( {R - R\cos t} \right)\left( {R - R\cos t} \right){\rm{d}}t = \mathop \smallint \limits_{t = 0}^{2\pi } {R^2}{\left( {1 - \cos t} \right)^2}{\rm{d}}t$
$A = {R^2}\mathop \smallint \limits_{t = 0}^{2\pi } \left( {1 - 2\cos t + {{\cos }^2}t} \right){\rm{d}}t$
$A = {R^2}\left( {\mathop \smallint \limits_{t = 0}^{2\pi } {\rm{d}}t - 2\mathop \smallint \limits_{t = 0}^{2\pi } \cos t{\rm{d}}t + \mathop \smallint \limits_{t = 0}^{2\pi } {{\cos }^2}t{\rm{d}}t} \right)$
Using Eq. (6) of Section 8.1 we get
$A = {R^2}\left( {t|_0^{2\pi } - 2\sin t|_0^{2\pi } + \frac{1}{2}\cos t\sin t|_0^{2\pi } + \frac{1}{2}t|_0^{2\pi }} \right)$
$A = 3\pi {R^2}$
The area is equal to three times the area of the generating circle $\pi {R^2}$.
