Chapter 11, Problem 37RE

To determine

To calculate: The area of the region described.

Answer to Problem 37RE

The required area is $\frac{8+\pi }{4}$ .

Explanation of Solution

Given information:

The region enclosed inside the “figure eight”: $r=1+\cos 2\theta$

The region enclosed outside the circle: $r=1$

Calculation:

The region enclosed inside the “figure eight” is $r=1+\cos 2\theta$ and the region enclosed outside the circle is $r=1$

The graph of the figure eight and the circle is covered for $0\le \theta \le \frac{\pi }{4}$ .

The graph is shown below:

  

To find the area of the region, integrate the given region for $0\le \theta \le \frac{\pi }{4}$ .

  $\begin{array}{c}A={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{1}{2}\left(r_2{}^2\right)d\theta }-{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{1}{2}\left(r_1{}^2\right)d\theta }\\ =4\left[{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{1}{2}{\left(1+\cos 2\theta \right)}^{2}d\theta }-{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{1}{2}{\left(1\right)}^{2}d\theta }\right]\\ =2\left[{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(1+2\cos 2\theta +{\cos }^{2}2\theta \right)d\theta }-{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}d\theta }\right]\\ =2\left[{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(1+2\cos 2\theta +\frac{1+\cos 4\theta }{2}\right)d\theta }-{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}d\theta }\right]\end{array}$

Simplify the above expression further.

  $\begin{array}{c}2\left[{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(1+2\cos 2\theta +\frac{1+\cos 4\theta }{2}\right)d\theta }-{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}d\theta }\right]=2{\left[\theta +\frac{2\sin 2\theta }{2}+\frac{\theta }{2}+\frac{\sin 4\theta }{8}-\theta \right]}_0^{\frac{\pi }{4}}\\ =2{\left[\sin 2\theta +\frac{\theta }{2}+\frac{\sin 4\theta }{8}\right]}_0^{\frac{\pi }{4}}\\ =2\left[\sin 2\left(\frac{\pi }{4}\right)+\frac{\pi }{4}+\frac{\sin \pi }{8}-\sin 2\left(0\right)+\frac{0}{2}+\frac{\sin 4\left(0\right)}{8}\right]\\ =2\left[1+\frac{\pi }{8}+\frac{0}{8}-0+0+\frac{0}{8}\right]\end{array}$

Simplify further.

  $\begin{array}{c}2\left[1+\frac{\pi }{8}+\frac{0}{8}-0+0+\frac{0}{8}\right]=\cancel{2}\left[\frac{8+\pi }{{\cancel{8}}_4}\right]\\ =\frac{8+\pi }{4}\end{array}$

Hence, the required area is $\frac{8+\pi }{4}$ .