Chapter 10, Problem 37RE

To determine

To Find: The area inside the figure eight and outside the circle.

Answer to Problem 37RE

The area is $\frac{\pi }{4}+2$

Explanation of Solution

Given information: $r=1+\cos (2\theta )$

Calculation:

The graph can be made as:

  

  $\begin{array}{l}r=1+\cos 2\theta \text{ ; r = 1}\\ \text{1+}\cos 2\theta =1\\ \cos 2\theta =0\\ 2\theta \in \left\{\pm \frac{\pi }{2}\right\}\\ \theta \in \left\{\pm \frac{\pi }{4}\right\}\end{array}$

Now, from the graph it is clear that both $r=1+\cos 2\theta$ and $\text{ r = 1}$ have symmetrical region with respect to y-axis. Hence, the area inside the $r=1+\cos 2\theta$ and outside the circle $\text{ r = 1}$ will be two times of the area of the region inside the polar curve.

So, are can be written as;

  $\begin{array}{l}A=2.2{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{1}{2}.\left({\left(1+\cos 2\theta \right)}^2-1^2\right)}d\theta \\ A=\frac{4}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(1+2\cos 2\theta +co{s}^{2}2\theta -1\right)d\theta }\\ A=2{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(2\cos 2\theta +{\cos }^{2}2\theta \right)}d\theta \\ A={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}4\cos }2\theta d\theta +{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}2{\cos }^2}2\theta d\theta \\ \text{Let }u=2\theta \\ du=2d\theta \\ d\theta =\frac{du}{2}\end{array}$

As, when $u=0;\theta =\frac{\pi }{4}\text{ and u=}\frac{\pi }{2}$

Solve the above equation one by one:

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}4\cos 2\theta }d\theta =\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}4\cos u.du}\\ {\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}4\cos 2\theta }d\theta ={\left.2\sin u\right|}_0^{\frac{\pi }{2}}\\ {\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}4\cos 2\theta }d\theta =2\left(\sin \left(\frac{\pi }{2}\right)-\sin 0\right)\\ {\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}4\cos 2\theta }d\theta =2\left(1-0\right)\\ {\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}4\cos 2\theta }d\theta =2\end{array}$

Now, solve the second part of the equation:

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}2{\cos }^{2}2\theta d\theta ={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}2.\frac{1+\cos \left(2.2\theta \right)}{2}}}d\theta \\ \text{ = }{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}1.d\theta +{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\cos 4\theta d\theta }}\\ \text{ = 1}\text{.}{\left.\theta \right|}_o^{\frac{\pi }{4}}+{\left[\frac{\sin 4\theta }{4}\right]}_0^{\frac{\pi }{4}}\\ \text{ = }\frac{\pi }{4}+0\\ A=2+\frac{\pi }{4}+0\\ A=\frac{\pi }{4}+2\end{array}$