Chapter 11, Problem 38RE

To determine

To calculate: The area of the region described.

Answer to Problem 38RE

The required area is $5\pi$ .

Explanation of Solution

Given information:

The region enclosed inside the cardioid: $r=2\left(1+\sin \theta \right)$

The region enclosed outside the circle: $r=2\sin \theta$

Calculation:

The region enclosed inside the cardioid is $r=2\left(1+\sin \theta \right)$ and the region enclosed outside the circle is $r=2\sin \theta$

The graph of the figure eight and the circle is covered for $0\le \theta \le 2\pi$ .

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

  $\begin{array}{c}A={\displaystyle \underset{0}{\overset{2\pi }{\int }}\frac{1}{2}\left(r_2{}^2\right)d\theta }-{\displaystyle \underset{0}{\overset{2\pi }{\int }}\frac{1}{2}\left(r_1{}^2\right)d\theta }\\ =\left[{\displaystyle \underset{0}{\overset{2\pi }{\int }}\frac{1}{2}{\left(2\left(1+\sin \theta \right)\right)}^{2}d\theta }-{\displaystyle \underset{0}{\overset{2\pi }{\int }}\frac{1}{2}{\left(2\sin \theta \right)}^{2}d\theta }\right]\\ ={\displaystyle \underset{0}{\overset{2\pi }{\int }}2\left(1+2\sin \theta +{\sin }^2\theta \right)d\theta }-{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\left(1-\cos \theta \right)}^{2}d\theta }\end{array}$

Simplify the above expression further.

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{2\pi }{\int }}2\left(1+2\sin \theta +{\sin }^2\theta \right)d\theta }-{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\left(1-\cos \theta \right)}^{2}d\theta }={\left[3\theta -4\cos \theta -\frac{1}{2}\sin 2\theta \right]}_0^{2\pi }-{\left[\theta -\frac{1}{2}\sin 2\theta \right]}_0^{2\pi }\\ =\left[\begin{array}{l}3\left(2\pi \right)-4\cos \left(2\pi \right)-\frac{1}{2}\sin 2\left(2\pi \right)\\ -3\left(0\right)-4\cos \left(0\right)-\frac{1}{2}\sin 2\left(0\right)\end{array}\right]-\left[\pi -\frac{1}{2}\sin 2\pi -0+\frac{1}{2}\sin 2\left(0\right)\right]\\ =6\pi -4-0+4-\pi +0\\ =5\pi \end{array}$

Hence, the required area is $5\pi$ .