## Problem Statement

Calculate the area of the figure bounded by the lines given by the equations.

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=t-\sin t \\
y=1-\cos t
\end{array}\right. \\
& y=1(0<x<2 \pi, y \geq 1)
\end{aligned}
$$
## Solution

Let's find the points of intersection:

![](https://cdn.mathpix.com/cropped/2024_05_22_fa23e2deef6f8d90c6d5g-44.jpg?height=683&width=1330&top_left_y=1086&top_left_x=637)

$$
\begin{aligned}
& y=1-\cos t=1 \\
& \cos t=1-1=0 \\
& t=\frac{\pi}{2}+\pi n, n \in \mathbb{Z}
\end{aligned}
$$

From the problem statement, the interval is \(0 < x < 2\pi\).

$$
\begin{aligned}
& x=0 \Rightarrow t=0 \\
& x=2\pi \Rightarrow t=2\pi
\end{aligned}
$$

Then the x-coordinates of the points of intersection will be:

$$
\begin{aligned}
& t=\frac{\pi}{2} \Rightarrow x=\frac{\pi}{2}-\sin \frac{\pi}{2}=\frac{\pi}{2}-1 \\
& t=\frac{\pi}{2}+\pi \Rightarrow x=\frac{3\pi}{2}-\sin \left(\frac{\pi}{2}+\pi\right)=\frac{3\pi}{2}+1
\end{aligned}
$$

Calculate the area:

$$
\begin{aligned}
& S=S_{0}-S_{1} \\
& \Gamma \text{ where } \\
& S_{1}=\left(y_{2}-y_{1}\right) \cdot\left(x_{2}-x_{1}\right)=(1-0) \cdot\left(\frac{3\pi}{2}+1-\frac{\pi}{2}+1\right)=1 \cdot\left(\frac{2\pi}{2}+2\right)=\pi+2 \\
& S_{0}=\int_{\pi / 2}^{3 \pi / 2}(1-\cos t) \cdot(t-\sin t)^{\prime} d t= \\
& =\int_{\pi / 2}^{3 \pi / 2}(1-\cos t) \cdot(1-\cos t) d t= \\
& =\int_{\pi / 2}^{3 \pi / 2}\left(1-2 \cos t+\cos ^{2} t\right) d t= \\
& =\left.(t-2 \sin t)\right|_{\pi / 2} ^{3 \pi / 2} \quad+\int_{\pi / 2}^{3 \pi / 2} \frac{1}{2}(1+\cos 2 t) d t= \\
& =\left(\frac{3\pi}{2}-2 \sin \frac{3\pi}{2}\right)-\left(\frac{\pi}{2}-2 \sin \frac{\pi}{2}\right)+\left.\frac{1}{2} \cdot\left(t+\frac{1}{2} \sin 2 t\right)\right|_{\pi / 2} ^{3 \pi / 2}= \\
& =\frac{3\pi}{2}-2\left(-\sin \frac{\pi}{2}\right)-\frac{\pi}{2}+2 \cdot 1+\frac{1}{2} \cdot\left(\frac{3\pi}{2}+\frac{1}{2} \sin \frac{2 \cdot 3\pi}{2}\right)-\frac{1}{2} \cdot\left(\frac{\pi}{2}+\frac{1}{2} \sin \frac{2\pi}{2}\right)= \\
& =\frac{3-1}{2} \cdot \pi+2 \cdot 1+2+\frac{3\pi}{4}+\frac{1}{2} \cdot \sin 3\pi-\frac{\pi}{4}-\frac{1}{2} \cdot \sin \pi= \\
& =\pi+4+\frac{2\pi}{4}+\frac{1}{2} \cdot 0-\frac{1}{2} \cdot 0=\frac{3\pi}{2}+4 \\
& S=S_{0}-S_{1}=\frac{3\pi}{2}+4-(\pi+2)=\frac{\pi}{2}+2
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+15-21 »$ Categories: Kuznetsov's Problem Book Integrals Problem 15 | Integrals

- Last edited: 09:23, June 21, 2010.
- Content is available under CC-BY-SA 3.0.