Chapter 11.3, Problem 54E

To determine

To calculate: The area of the region described.

Answer to Problem 54E

The required area of the region is $6\pi -16$ .

Explanation of Solution

Given information:

The curves of the circles: $r=2\left(1+\cos \theta \right)$ , $r=2\left(1-\cos \theta \right)$

Calculation:

The given curves are $r=2\left(1+\cos \theta \right)$ and $r=2\left(1-\cos \theta \right)$ .

It is known that in a polar coordinate system, if $r=f\left(\theta \right)$ is a curve; then the area of the region within the interval $\alpha \le \theta \le \beta$ is $A={\displaystyle {\int }_{\alpha }^{\beta }\frac{1}{2}{r}^{2}d\theta }={\displaystyle {\int }_{\alpha }^{\beta }\frac{1}{2}{\left(f\left(\theta \right)\right)}^{2}d\theta }$ .

The graph of the given curves is shown below:

  

It is seen that the shaded region between the curves lie within the interval $0\le \theta \le \frac{\pi }{2}$ .

Write the area of the given region as:

  $\begin{array}{c}A={\displaystyle {\int }_{\alpha }^{\beta }\frac{1}{2}{\left(f\left(\theta \right)\right)}^{2}d\theta }\\ =4{\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{1}{2}{\left(2\left(1-\cos \theta \right)\right)}^{2}d\theta }\\ ={\displaystyle {\int }_0^{\frac{\pi }{2}}8{\left(1-\cos \theta \right)}^{2}d\theta }\\ ={\displaystyle {\int }_0^{\frac{\pi }{2}}8\left(1+{\cos }^2\theta -2\cos \theta \right)d\theta }\end{array}$

Simplify the above expression further.

  $\begin{array}{c}{\displaystyle {\int }_0^{\frac{\pi }{2}}8\left(1+{\cos }^2\theta -2\cos \theta \right)d\theta }={\displaystyle {\int }_0^{\frac{\pi }{2}}8\left(1+\left(\frac{1+\cos 2\theta }{2}\right)-2\cos \theta \right)d\theta }\\ ={\displaystyle {\int }_0^{\frac{\pi }{2}}\left(12+4\cos 2\theta -16\cos \theta \right)d\theta }\\ =\left(6\pi +0-16\right)-\left(0\right)\\ =6\pi -16\end{array}$

Hence, the required area of the region is $6\pi -16$ .