Chapter 8, Problem 16RE

To determine

To find the area of the region enclosed by the lines and curves.

Answer to Problem 16RE

The area of larger bead shaped region enclosed by curves $y=1+\cos x$ and $y=2-\cos x$ is

  $7.65$ square units.

Explanation of Solution

Given:

One of the larger bead-shaped regions enclosed by the graphs of $y=1+\cos x$ and $y=2-\cos x$

Calculation:

Using graphing calculator to graph the given equations

  $y=1+\cos x$ and $y=2-\cos x$

Graph:

  

Now, solving equations $y=1+\cos x$ and $y=2-\cos x$ simultaneously to find the point of intersections of both the curve

  $\begin{array}{l}1+\cos x=2-\cos x\\ 2\cos x=1\\ \cos x=\frac{1}{2}\\ \cos x=\cos \frac{\pi }{3}\\ x=2k\pi \pm \frac{\pi }{3},\text{where }k\text{ is an integer}\end{array}$

Now, to find area of one of smaller bead put $k=0$ and $k=1$

  $\begin{array}{l}x=2k\pi \pm \frac{\pi }{3}\\ x=\pm \frac{\pi }{3}\end{array}$

  $\begin{array}{l}x=2k\pi \pm \frac{\pi }{3}\\ x=\frac{7\pi }{3},\frac{5\pi }{3}\end{array}$

The area of region is

  $A={\displaystyle \underset{\pi /3}{\overset{5\pi /3}{\int }}\left(2-\cos x-\left(1+\cos x\right)\right)}dx={\displaystyle \underset{\pi /3}{\overset{5\pi /3}{\int }}\left(1-2\cos x\right)dx}={\left|x-2\sin x\right|}_{\pi /3}^{5\pi /3}=7.65$

Therefore,

the area of larger bead shaped region enclosed by curves $y=1+\cos x$ and $y=2-\cos x$ is

  $7.65$ square units.