Chapter 10.3, Problem 51E

To determine

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

Answer to Problem 51E

The area of the region enclosed by the given circles is $\frac{\pi }{2}-1$ .

Explanation of Solution

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

Formula used: The double-angle formula of cosine function is given by ${\sin }^2\theta =\frac{1}{2}\left(1-\cos 2\theta \right)$ .

The quotient identity of tangent function is $\tan \theta =\frac{\sin \theta }{\cos \theta }$ .

Calculation:

Find the intersection point of $r=2\cos \theta$ and $r=2\sin \theta$ by equating them.

  $\begin{array}{c}2\sin \theta =2\cos \theta \\ \frac{\sin \theta }{\cos \theta }=\frac{2}{2}\\ \tan \theta =1\\ \theta =\frac{\pi }{4}\end{array}$

Draw the region enclosed by the given circles as shown below.

  

Figure (1)

Area of the shaded region as shown in the figure is given by the integral.

  $A=\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}{\left(2\sin \theta \right)}^{2}d\theta }$

So, the area of the common region enclosed by the circles $r=2\cos \theta$ and $r=2\sin \theta$ will be twice the area $A=\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}{\left(2\sin \theta \right)}^{2}d\theta }$ , that is, $A=2\cdot \frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}{\left(2\sin \theta \right)}^{2}d\theta }$ .

To find the required area, solve the integral.

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

Further simplify.

  $\begin{array}{c}A=2\left(\frac{\pi }{4}-\frac{1}{2}\right)\\ =\frac{\pi }{2}-1\end{array}$

Therefore, the required area enclosed by the circles is $\frac{\pi }{2}-1$ .