Example 2.52. Calculate the area of the figure bounded by one arch of the cycloid $x=2(t-\sin t), y=$ $=2(1-\cos t)$.
Solution. One arch of the cycloid is obtained for \(0 \leqslant t \leqslant 2 \pi\) (Fig. 2.6). The required area is conveniently calculated using the first formula from (2.22).

\[
\begin{aligned}
S= & \left|-\int_{0}^{2 \pi} y(t) x^{\prime}(t) d t\right|=\left|-\int_{0}^{2 \pi} 2(1-\cos t) 2(1-\cos t) d t\right|= \\
= & \left|4 \int_{0}^{2 \pi}(1-\cos t)^{2} d t\right|=\left|4 \int_{0}^{2 \pi}\left(1-2 \cos t+\cos ^{2} t\right) d t\right|= \\
& =\left|4 \int_{0}^{2 \pi}(1-2 \cos t+0.5(1+\cos 2 t)) d t\right|= \\
& =\left.4|(t-2 \sin t+0.5 t+0.25 \sin 2 t)|\right|_{0} ^{2 \pi}=12 \pi
\end{aligned}
\]

![](https://cdn.mathpix.com/cropped/2024_05_22_0f84e17422d7c0df8d68g-082.jpg?height=290&width=488&top_left_y=133&top_left_x=129)

Fig. 2.7.

![](https://cdn.mathpix.com/cropped/2024_05_22_0f84e17422d7c0df8d68g-082.jpg?height=292&width=429&top_left_y=130&top_left_x=641)

Fig. 2.8.