Chapter H.2, Problem 20E

To determine

To find: the area of the region

Answer to Problem 20E

The area of the region is $\frac{\pi }{4}+2$

Explanation of Solution

Given:

  $r=1-\sin \theta$ and $r=1$

  

Calculation:

The two curves intersects when

  $\begin{array}{l}1-\sin \theta =1\\ \Rightarrow \sin \theta =0\\ \Rightarrow \text{the}\text{intersecting}\text{points}\text{are}\pi ,2\pi \end{array}$

  $\begin{array}{l}\therefore \text{The}\text{desired}\text{area}A={\displaystyle \underset{a}{\overset{b}{\int }}\frac{1}{2}}{r}^{2}d\theta \\ A={\displaystyle \underset{\pi }{\overset{2\pi }{\int }}\frac{1}{2}{\left[1-\sin \theta \right]}^{2}d\theta }-{\displaystyle \underset{\pi }{\overset{2\pi }{\int }}\frac{1}{2}{\left[1\right]}^{2}d\theta }\\ A={\displaystyle \underset{\pi }{\overset{2\pi }{\int }}\frac{1}{2}{\left[1+{\sin }^2\theta -2\sin \theta \right]}^{2}d\theta }-{\displaystyle \underset{\pi }{\overset{2\pi }{\int }}\frac{1}{2}{\left[1\right]}^{2}d\theta }\\ A={\displaystyle \underset{\pi }{\overset{2\pi }{\int }}\frac{1}{2}\left[1+\frac{1-\cos 2\theta }{2}-2\sin \theta -1\right]d\theta }\\ A={\displaystyle \underset{\pi }{\overset{2\pi }{\int }}\left[\frac{1}{4}-\frac{\cos 2\theta }{4}-\sin \theta \right]d\theta }\\ A={\left[\frac{1}{4}\theta -\frac{1}{4}\frac{\sin 2\theta }{2}+\cos \theta \right]}_{\pi }^{2\pi }\\ A=\frac{1}{4}\left(2\pi -\pi \right)-\frac{1}{8}\left(\sin 4\pi -\sin 2\pi \right)+\left(\cos 2\pi -\cos \pi \right)\\ A=\frac{\pi }{4}-\frac{1}{8}\left(0\right)+\left(1+1\right)\left[\because \sin 4\pi =\sin 2\pi =0,\cos 2\pi =1,\cos \pi =-1\right]\\ A=\frac{\pi }{4}+2\end{array}$

Conclusion:

Therefore, the area of the region is $\frac{\pi }{4}+2$