Chapter 10, Problem 35RE

To determine

To Explain: The area of the region.

Answer to Problem 35RE

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

Explanation of Solution

Given information: The equation is $r=2-\cos \theta$

Calculation:

The graph of the equation is

  

Now, area can be find out as:

  $\begin{array}{l}\theta \in \left[0,2\pi \right]\\ A={\displaystyle \underset{0}{\overset{2\pi }{\int }}\frac{1}{2}.{\left(2-\cos \theta \right)}^{2}d\theta }\\ A={\displaystyle \underset{0}{\overset{2\pi }{\int }}\frac{1}{2}\left(4-4\cos \theta +{\cos }^2\theta \right)}d\theta \\ A={\displaystyle \underset{0}{\overset{2\pi }{\int }}2d\theta -{\displaystyle \underset{0}{\overset{2\pi }{\int }}2\cos \theta d\theta +\frac{1}{2}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\cos }^2\theta d\theta }}}\\ A=2{\left[\theta \right]}_0^{2\pi }-2{\left[\sin \theta \right]}_0^{2\pi }+\frac{1}{2}{\displaystyle \underset{0}{\overset{2\pi }{\int }}\frac{1+\cos 2\theta }{2}}d\theta \\ A=4\pi -2\left[\sin 2\pi -\sin 0\right]+\frac{1}{2}{\displaystyle \underset{0}{\overset{2\pi }{\int }}\frac{1}{2}d\theta }+\frac{1}{2}{\displaystyle \underset{0}{\overset{2\pi }{\int }}\frac{\cos 2\theta }{2}d\theta }\\ A=4\pi -2\times 0+\frac{1}{4}{\left[\theta \right]}_0^{2\pi }+\frac{1}{4}{\left[\frac{\sin 4\theta }{4}\right]}_0^{2\pi }\\ A=4\pi -0+\frac{1}{4}\left[2\pi \right]+\frac{1}{16}\left[\sin 8\pi -\sin 0\right]\\ A=4\pi +\frac{\pi }{2}+0\\ A=\frac{9\pi }{2}\end{array}$