Chapter 11.3, Problem 55E

To determine

To calculate: The area of the region described.

Answer to Problem 55E

The required area of the region is $8-\pi$ .

Explanation of Solution

Given information:

The curves of the circles: $r=2$ , $r=2\left(1-\sin \theta \right)$

Calculation:

The given curves are $r=2$ 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 $0\le \theta \le \frac{\pi }{2}$ .

Write the area of the given region as:

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

Simplify the above expression further.

  $\begin{array}{c}4{\left[\theta \right]}_0^{\frac{\pi }{2}}-{\displaystyle {\int }_0^{\frac{\pi }{2}}\left(4-8\sin \theta +4\left(\frac{1-\cos 2\theta }{2}\right)\right)d\theta }=4\left(\frac{\pi }{2}\right)-{\displaystyle {\int }_0^{\frac{\pi }{2}}\left(6-8\sin \theta -2\cos 2\theta \right)d\theta }\\ =2\pi -{\left[6\theta +8\cos \theta -2\sin 2\theta \right]}_0^{\frac{\pi }{2}}\\ =2\pi -3\pi +8\\ =8-\pi \end{array}$

Hence, the required area of the region is $8-\pi$ .