Chapter 11, Problem 36RE

To determine

To calculate: The area of the region described.

Answer to Problem 36RE

The required area is $\frac{9\pi }{2}$ .

Explanation of Solution

Given information:

The region enclosed by the limacon: $r=2-\cos \theta$

Calculation:

The given region is $r=2-\cos \theta$ .

The graph of the limacon is covered for $0\le \theta \le 2\pi$ .

To find the area of the region, integrate the given region for $0\le \theta \le 2\pi$ .

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

Simplify the above expression further.

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

Simplify further.

  $\begin{array}{c}\frac{1}{2}\left[8\pi -2\left(0\right)+\pi -0-+0\right]=\frac{1}{2}\left[8\pi +\pi \right]\\ =\frac{9\pi }{2}\end{array}$

Hence, the required area is $\frac{9\pi }{2}$ .