Chapter 11.3, Problem 47E

To determine

To calculate: The area of the region described.

Answer to Problem 47E

The required area of the region is 2.

Explanation of Solution

Given information:

The one loop of the lemniscate: $r^2=4\cos 2\theta$

Calculation:

The given one loop of the lemniscate is $r^2=4\cos 2\theta$ .

The one loop of the lemniscate is covered in the region $-\frac{\pi }{4}\le \theta \le \frac{\pi }{4}$ .

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 $A={\displaystyle {\int }_{\alpha }^{\beta }\frac{1}{2}{r}^{2}d\theta }={\displaystyle {\int }_{\alpha }^{\beta }\frac{1}{2}{\left(f\left(\theta \right)\right)}^{2}d\theta }$ .

Write the area of the given region as:

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

Simplify the above expression further.

  $\begin{array}{c}{\displaystyle {\int }_{\frac{\pi }{4}}^{\frac{\pi }{4}}2\cos 2\theta d\theta }=2{\displaystyle {\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\cos 2\theta d\theta }\\ ={\sin 2\theta |}_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\\ =1+1\\ =2\end{array}$

Hence, the required area of the region is 2.