Question

Find the area of the region that lies inside both of the circles \(r = 2\sin \theta \) and \(r = \sin \theta + \cos \theta .\)

Short answer

The area of the region that lies inside both of the circles \(r = 2\sin \theta \) and \(r = \sin \theta + \cos \theta \) is\(\frac{1}{2}(\pi - 1)\) square units.

Step-by-step solution

Definition of graph

A diagram depicting the relationship between two or more variables, each measured along one of a pair of axes at right angles.

Explain graph

In the graph below:

Mark \(r = 2\sin \theta \) by blue circle. And mark \(r = \sin \theta + \cos \theta \) by red circle.

The dotted lines are \(y = x\)and \(y = - x\).

We are asked to find the area of the white region, which is common to both the circles.

Calculating the limits

The required area is sum of area \(A\)and \(B\).

To find limits of area \(A\) we need to find point of intersection of those two curves. To find that equate equations:

\(\begin{aligned}{c}2\sin \theta = \sin \theta + \cos \theta \sin \theta \\ = \cos \theta \tan \theta \\ = 0 \Rightarrow \;\;\;\theta = \frac{\pi }{4}\end{aligned}\)

The second limit is when red curve meets the origin:

\(\begin{aligned}{c}\sin \theta + \cos \theta = 0\sin \theta \\ = - \cos \theta \tan \theta \\ = - 1\\ \Rightarrow \;\;\theta = \frac{{3\pi }}{4}\end{aligned}\)

Calculate area of region \({\bf{A}}\)

Find the Area of region A:

\(\begin{aligned}{c}Are{a_A} = \frac{1}{2}\int_{\pi /4}^{3\pi /4} {{{(\sin \theta + \cos \theta )}^2}} d\theta \\ = \frac{1}{2}\int_{\pi /4}^{3\pi /4} {\left( {1 + 2\sin \theta \cos \theta } \right)} d\theta \\ = \frac{1}{2}\int_{\pi /4}^{3\pi /4} {\left( {1 + \sin 2\theta } \right)} d\theta \\ = \frac{1}{2}\left( {\theta - \frac{1}{2}\cos 2\theta } \right)_{\pi /4}^{3\pi /4}\\ = \frac{1}{2}\left( {\frac{{3\pi - \pi }}{4}} \right)\\ = \frac{\pi }{4}\end{aligned}\)

Therefore, area calculated for region A is \(\frac{1}{2}\left( {\frac{{3\pi - \pi }}{4}} \right) = \frac{\pi }{4}\).

Calculate area of region B

For area \(B\) we already have one limit: \(\theta = \frac{\pi }{4}\). The second limit is when blue curve is in the origin:

\(\begin{aligned}{c}2\sin \theta = 0\\\sin \theta = 0\\ \Rightarrow \;\;\;\theta = 0\end{aligned}\)

Area of region B is,

\(\begin{aligned}{c}{\rm{Are}}{{\rm{a}}_B}{\rm{ }} = \frac{1}{2}\int_0^{\pi /4} {{{(2\sin \theta )}^2}} d\theta \\ = \int_0^{\pi /4} 2 {\sin ^2}\theta d\theta \end{aligned}\)

Use \({\sin ^2}\theta = \frac{{1 - \cos 2\theta }}{2}\).

\(\begin{aligned}{c}Are{a_B} = \int_0^{\pi /4} {\left( {1 - \cos 2\theta } \right)} d\theta \\ = \left( {\theta - \frac{1}{2}\sin 2\theta } \right)_0^{\pi /4}\\ = \frac{\pi }{4} - \frac{1}{2}\end{aligned}\)

Therefore, area calculated for region B is \(\frac{\pi }{4} - \frac{1}{2}\).

Total area will be, \(A = Are{a_A} + Are{a_B}\).

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

Hence, the area of the region that lies inside both of the circles \(r = 2\sin \theta \) and \(r = \sin \theta + \cos \theta \) is \(\frac{1}{2}(\pi - 1)\) square units.