Chapter 7, Problem 15RE

To determine

To find: the area of the similar bead-shaped regions enclosed by the given functions.

Answer to Problem 15RE

The required area is $1.370$ sq. unit.

Explanation of Solution

Given information:

The functions are, $y=1+\cos x;y=2-\cos x$ .

Formula used:

Area of a curve $f(x)$ from $x_1$ to $x_2$ can be determined as, ${\displaystyle \underset{x_1}{\overset{x_2}{\int }}f(x)dx}$ .

Calculation:

The graph of the given functions can be drawn as,

  

Solve the given curves to find the point of intersections of the curves.

  $\begin{array}{c}1+\cos x=2-\cos x\\ 2\cos x=2-1\\ \cos x=\frac{1}{2}\\ x=2\pi ,2\pi +\frac{\pi }{3}\\ x=2\pi ,\frac{7\pi }{3}\end{array}$

As the area is symmetric, the required area is,

  $\begin{array}{c}2{\displaystyle \underset{2\pi }{\overset{\frac{7\pi }{3}}{\int }}[(1+\cos x)-(2-\cos x)]dx}=2{\displaystyle \underset{2\pi }{\overset{\frac{7\pi }{3}}{\int }}[2\cos x-1]dx}\\ =2{\left[2\sin x-x\right]}_{2\pi }^{\frac{7\pi }{3}}\\ =2\left(2\sin \left(\frac{7\pi }{3}\right)-\left(\frac{7\pi }{3}\right)-2\sin \left(2\pi \right)+\left(2\pi \right)\right)\\ =2\sqrt{3}-\frac{2\pi }{3}\\ =1.370\text{sq}\text{.unit}\end{array}$