Chapter 11.3, Problem 56E

To determine

To calculate: The area of the region described.

Answer to Problem 56E

The required area of the region is $\frac{8\pi }{3}+4\sqrt{3}$ .

Explanation of Solution

Given information:

The curves of the circles: $r=4\cos 2\theta$ , $r=2$

Calculation:

The given curves are $r=4\cos 2\theta$ and $r=2$ .

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 $-\frac{\pi }{6}\le \theta \le \frac{\pi }{6}$ .

Write the area of the given region as:

  $\begin{array}{c}A={\displaystyle {\int }_{\alpha }^{\beta }\frac{1}{2}{\left(f\left(\theta \right)\right)}^{2}d\theta }\\ =4{\displaystyle {\int }_0^{\frac{\pi }{6}}\frac{1}{2}\left({\left(4\cos 2\theta \right)}^2-2^2\right)d\theta }\\ =8{\displaystyle {\int }_0^{\frac{\pi }{6}}\frac{1}{2}\left(16{\cos }^{2}2\theta -4\right)d\theta }\\ ={\displaystyle {\int }_0^{\frac{\pi }{6}}\left(64\left(\frac{1+\cos 4\theta }{2}\right)-16\right)d\theta }\end{array}$

Simplify the above expression further.

  $\begin{array}{c}{\displaystyle {\int }_0^{\frac{\pi }{6}}\left(64\left(\frac{1+\cos 4\theta }{2}\right)-16\right)d\theta }={\displaystyle {\int }_0^{\frac{\pi }{6}}\left(32\left(1+\cos 4\theta \right)-16\right)d\theta }\\ ={\displaystyle {\int }_0^{\frac{\pi }{6}}\left(16+32\cos 4\theta \right)d\theta }\\ =\left[\left(\frac{8\pi }{3}+8\sin \frac{2\pi }{3}\right)-\left(0+8\sin 0\right)\right]\\ =\frac{8\pi }{3}+4\sqrt{3}\end{array}$

Hence, the required area of the region is $\frac{8\pi }{3}+4\sqrt{3}$ .