Chapter 10.3, Problem 52E

To determine

To find: The area of the common region enclosed by the circles $r=1$ and $r=2\sin \theta$ .

Answer to Problem 52E

The area of the region enclosed by the given circles is $\frac{2\pi }{3}-\frac{\sqrt{3}}{2}$ .

Explanation of Solution

Given information: Equations of two circles are $r=1$ and $r=2\sin \theta$ .

Formula used: The double-angle formula for cosine function is given by:

  ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$

Calculation:

Equate the given equations of circles and determine the value of $\theta$ .

  $\begin{array}{c}2\sin \theta =1\\ \sin \theta =\frac{1}{2}\\ \theta \in \left\{\frac{\pi }{6},\frac{5\pi }{6}\right\}\end{array}$

According to the fact $\sin \left(\pi -\alpha \right)=\sin \alpha$ for $\theta \le \alpha \le \frac{\pi }{2}$ .

  $\begin{array}{c}\sin \left(\frac{5\pi }{6}\right)=\sin \left(\pi -\frac{\pi }{6}\right)\\ =\sin \left(\frac{\pi }{6}\right)\\ =\frac{1}{2}\end{array}$

  $\begin{array}{l}\theta =\frac{\pi }{6}\\ \theta =\frac{5\pi }{6}\end{array}$

Draw the graph of region enclosed by the circles as shown below.

  

Figure (1)

Here, $\theta$ is the angle between the ray containing $\left(r,\theta \right)$ and the $x$ -axis.

From the above graph it can be observed that the common region enclosed by the given circles is same as the region enclosed by the polar curve $r=2\sin \theta$ for $0\le \theta \le \frac{\pi }{6}$ , the polar curve $r=1$ for $\frac{\pi }{6}\le \theta \le \frac{5\pi }{6}$ , and the polar curve $r=2\sin \theta$ for $\frac{5\pi }{6}\le \theta \le \pi$ .

Also, the common region enclosed by the given circles is symmetrical with respect to the positive $y$ -axis or $\theta =\frac{\pi }{2}$ . Consequently, it is clear that the area enclosed by the circles $r=1$ and $r=2\sin \theta$ equals twice the areas of the region inside the polar curves $r=2\sin \theta$ for $0\le \theta \le \frac{\pi }{6}$ and $r=1$ for $\frac{\pi }{6}\le \theta \le \frac{\pi }{2}$ .

Calculate the area of the region enclosed by the circles.

  $A=2\cdot \left({\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{1}{2}{\left(2\sin \theta \right)}^{2}d\theta }+{\displaystyle \underset{\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{2}\cdot 1^{2}d\theta }\right)$

Evaluate the obtained integral.

  $\begin{array}{c}A={\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}4{\sin }^2\theta d\theta }+{\displaystyle \underset{\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}1d\theta }\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}4{\sin }^2\theta d\theta }+1\cdot {\left[\theta \right]}_{\frac{\pi }{6}}^{\frac{\pi }{2}}\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}4{\sin }^2\theta d\theta }+\frac{\pi }{2}-\frac{\pi }{6}\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}4{\sin }^2\theta d\theta }+\frac{3\pi -\pi }{6}\end{array}$

Further simplify.

  $A={\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}4{\sin }^2\theta d\theta }+\frac{\pi }{3}$

Apply the identity ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$ and evaluate the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}4{\sin }^2\theta d\theta }$ .

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}4{\sin }^2\theta d\theta }={\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}4\cdot \frac{1-\cos 2\theta }{2}d\theta }\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}2d\theta }-{\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}2\cos 2\theta d\theta }\\ =2\cdot {\left[\theta \right]}_0^{\frac{\pi }{6}}-{\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}2\cos 2\theta d\theta }\\ =\frac{\pi }{3}-{\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}2\cos 2\theta d\theta }\end{array}$

Evaluate the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}2\cos 2\theta d\theta }$ . Consider $u=2\theta$ , so $du=2d\theta$ or $d\theta =\frac{du}{2}$ .

For $\theta =0$ , $u=0$ and for $\theta =\frac{\pi }{6}$ , $u=\frac{\pi }{3}$ .

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}2\cos 2\theta d\theta }={\displaystyle \underset{0}{\overset{\frac{\pi }{3}}{\int }}2\cos u\frac{du}{2}}\\ ={\left[\sin u\right]}_0^{\frac{\pi }{3}}\\ =\sin \left(\frac{\pi }{3}\right)-\sin 0\\ =\frac{\sqrt{3}}{2}\end{array}$

Using the above results determine the integral $A={\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}4{\sin }^2\theta d\theta }+\frac{\pi }{3}$ .

  $\begin{array}{c}A={\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}4{\sin }^2\theta d\theta }+\frac{\pi }{3}\\ =\frac{\pi }{3}-{\displaystyle \underset{0}{\overset{\frac{\pi }{6}}{\int }}2\cos 2\theta d\theta }+\frac{\pi }{3}\\ =\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\end{array}$

Therefore, the area of the region enclosed by the given circles is $\frac{2\pi }{3}-\frac{\sqrt{3}}{2}$ .