Chapter 11.3, Problem 58E

To determine

To graph: The polar curves. Also, find the area that lies inside the circle and outside the cardioid.

Answer to Problem 58E

The required area that lies inside the circle and outside the cardioid is $8-\pi$ .

Explanation of Solution

Given information:

The polar curves: $r=2$ , $r=2$

Calculation:

The given curves are $r=4\cos 2\theta$ and $r=2\left(1-\sin \theta \right)$ .

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 $0\le \theta \le \pi$ .

Write the area of the given region as:

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

Simplify the above expression further.

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

Again, simplify the above expression further.

  $\begin{array}{c}{\left[-4\cos \theta -\theta +\frac{1}{2}\sin 2\theta \right]}_0^{\pi }=\left[\left(-4\cos \pi -\pi +\frac{1}{2}\sin 2\pi \right)-\left(-4\cos \left(0\right)-0+\frac{1}{2}\sin \left(0\right)\right)\right]\\ =\left[\left(\left(-4\right)\left(-1\right)-\pi +0\right)-\left(\left(-4\right)\left(1\right)-0\right)\right]\\ =\left[\left(4-\pi \right)-\left(-4\right)\right]\\ =8-\pi \end{array}$

Hence, the required area that lies inside the circle and outside the cardioid is $8-\pi$ .