Chapter 11.3, Problem 44E

To determine

To calculate: The area of the region described.

Answer to Problem 44E

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

Explanation of Solution

Given information:

The cardioid: $r=2+2\sin \theta$

Calculation:

The given cardioid is $r=2+2\sin \theta$ .

The cardioid is covered in the region $0\le \theta \le 2\pi$ .

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:

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

Simplify the above expression further.

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

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