Chapter 11.1, Problem 40E

To determine

To calculate: The volume swept out by revolving the region about the x -axis.

Answer to Problem 40E

The required volume is $5{\pi }^2{a}^3$ cube 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 volume swept out by the bounded region about the x -axis can be written as:

  $\begin{array}{c}V={\displaystyle \underset{0}{\overset{2a\pi }{\int }}\pi y^2}dx\\ ={\displaystyle \underset{0}{\overset{2a\pi }{\int }}\pi y^2}\left(\frac{dx}{dt}\right)dt\\ ={\displaystyle \underset{0}{\overset{2a\pi }{\int }}\pi {\left[a\left(1-\cos t\right)\right]}^{2}a\left(1-\cos t\right)dt}\\ =\pi a^3{\displaystyle \underset{0}{\overset{2a\pi }{\int }}\pi {\left[a\left(1-\cos t\right)\right]}^{2}a\left(1-\cos t\right)dt}\end{array}$

Simplify the above expression further.

  $\begin{array}{c}\pi a^3{\displaystyle \underset{0}{\overset{2a\pi }{\int }}\pi {\left[a\left(1-\cos t\right)\right]}^{2}a\left(1-\cos t\right)dt}=\pi a^3{\left(t-3\sin t+\frac{3t}{2}+\frac{3}{4}\sin 2t-\left(\sin t-\frac{1}{3}{\sin }^{3}t\right)\right)|}_0^{2\pi }\\ =5{\pi }^2{a}^3\end{array}$

Hence, the required volume is $5{\pi }^2{a}^3$ cube units.