Chapter 11.3, Problem 49E

To determine

To calculate: The area of the region described.

Answer to Problem 49E

The required area of the region is $11\pi$ .

Explanation of Solution

Given information:

The dimpled limacon: $r=3-2\cos \theta$

Calculation:

The given dimpled limacon is $r=3-2\cos \theta$ .

The one loop of the six-petaled rose 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:

  $\begin{array}{c}A={\displaystyle {\int }_{\alpha }^{\beta }\frac{1}{2}{\left(f\left(\theta \right)\right)}^{2}d\theta }\\ ={\displaystyle {\int }_0^{2\pi }\frac{1}{2}{\left(3-2\cos \theta \right)}^{2}d\theta }\\ ={\displaystyle {\int }_0^{2\pi }\frac{1}{2}\left(9-12\cos \theta +4{\cos }^2\theta \right)d\theta }\\ ={\displaystyle {\int }_0^{2\pi }\left(\frac{9}{2}-6\cos \theta +1+\cos 2\theta \right)d\theta }\end{array}$

Simplify the above expression further.

  $\begin{array}{c}{\displaystyle {\int }_0^{2\pi }\left(\frac{9}{2}-6\cos \theta +1+\cos 2\theta \right)d\theta }={\left[\frac{11\theta }{2}-6\sin \theta +\frac{1}{2}\sin 2\theta \right]}_0^{2\pi }\\ =11\pi -0\\ =11\pi \end{array}$

Hence, the required area of the region is $11\pi$ .