Chapter 11.1, Problem 39E

To determine

To calculate: The area of the shaded region.

Answer to Problem 39E

The required area is $3\pi a^2$ square units.

Explanation of Solution

Calculation:

The parametric equations of a Cycloid is $x=a\left(t-\sin t\right)$ , $y=a\left(1-\cos t\right)$ .

The graph of the Cycloid is shown below:

  

Differentiate the above equations with respect to t .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left[a\left(1-\sin t\right)\right]\\ =a\left(1-\cos t\right)\end{array}$

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left[a\left(1-\cos t\right)\right]\\ =a\sin t\end{array}$

The area of the region can be written as:

  $\begin{array}{c}A={\displaystyle \underset{0}{\overset{2a\pi }{\int }}y}dx\\ ={\displaystyle \underset{0}{\overset{2\pi }{\int }}y}\left(\frac{dx}{dt}\right)dt\\ ={\displaystyle \underset{0}{\overset{2\pi }{\int }}a\left(1-\cos t\right)a\left(1-\cos t\right)dt}\\ =a^2{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(1-2\cos t+{\cos }^{2}t\right)}dt\end{array}$

Simplify the above expression further.

  $\begin{array}{c}{a}^2{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(1-2\cos t+{\cos }^{2}t\right)}dt={a^2\left(t-2\sin t+\frac{t}{2}+\frac{1}{4}\sin 2t\right)|}_0^{2\pi }\\ =3\pi a^2\end{array}$

Hence, the required area is $3\pi a^2$ square units.