Chapter 11.3, Problem 52E

To determine

To calculate: The area of the region described.

Answer to Problem 52E

The required area of the region is $\frac{2\pi }{3}-\frac{\sqrt{3}}{2}$ .

Explanation of Solution

Given information:

The curves of the circles: $r=1$ , $r=2\sin \theta$

Calculation:

The given curves are $r=1$ and $r=2\sin \theta$ .

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 $0\le \theta \le \frac{\pi }{6}$ .

Write the area of the given region as:

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

Simplify the above expression further.

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

Hence, the required area of the region is $\frac{2\pi }{3}-\frac{\sqrt{3}}{2}$ .