Q. 49. - Calculus

Question

Find the area interior to two circles with the same radius

if each circle passes through the center of the other. (Hint:

Consider the circles $r=a and r=r=2a\cos \theta$

Step-by-Step Solution

Answer

Therefore the required area is $a^2\left(\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\right)$

Step 1: Given information

Take a look at the polar function 

$r=a,r=2a\cos \theta$

Step 2: The objective is to find the area interior of two circles with the same radius.

To find the limits, equal the functions

$$\begin{gathered} a=2a\cos \theta \\ \Rightarrow \frac{a}{2a}=\cos \theta \\ \Rightarrow \cos \theta =\frac{1}{2} \\ \Rightarrow \theta =\frac{\pi }{3},\frac{2\pi }{3} \end{gathered}$$

Step 3: Find the area interior of the circles

The region's corresponding limits are $0$ to $\frac{2\pi }{3}$

The interval is $\left[0,\frac{2\pi }{3}\right]$

Formula to find the area is $A={\int }_{\alpha }^{\beta }\frac{1}{2}\left(f\right(\theta ){)}^{2}d\theta$or $A={\int }_{\alpha }^{\beta }\frac{1}{2}{r}^{2}d\theta$

The area between the circles,

$$\begin{gathered} A=2·\frac{1}{2}{\int }_0^{\frac{2\pi }{3}}\left((2a\cos \theta {)}^2-a^2\right)d\theta \\ \Rightarrow A={\int }_0^{\frac{2\pi }{3}}\left(4a^2{\cos }^2\theta -a^2\right)d\theta \\ \Rightarrow A=a^2{\int }_0^{\frac{2\pi }{3}}\left(4{\cos }^2\theta -1\right)d\theta \\ \Rightarrow A=a^2{\int }_0^{\frac{2\pi }{3}}\left(4\frac{1+\cos 2\theta }{2}-1\right)d\theta \left[Since \left.\cos 2\theta =2{\cos }^2\theta -1\Rightarrow {\cos }^2\theta =\frac{1+\cos 2\theta }{2}\right] \right. \\ \Rightarrow A=a^2{\int }_0^{\frac{2\pi }{3}}\left(2\right(1+\cos 2\theta )-1)d\theta \\ \Rightarrow A=a^2{\int }_0^{\frac{2\pi }{3}}(2+2\cos 2\theta -1)d\theta \end{gathered}$$

On integration,

$A=a^2{\left(\theta +2\frac{\sin 2\theta }{2}\right)}_0^{\frac{2\pi }{3}}$

Step 4: Find the area by applying the limits

By applying the limits,
$$\begin{gathered} A=a^2\left(\frac{2\pi }{3}+\sin 2·\frac{2\pi }{3}-0\right) \\ \Rightarrow A=a^2\left(\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\right)\left[since\sin \frac{4\pi }{3}=-\frac{\sqrt{3}}{2}\right] \end{gathered}$$

Hence, the required area is $a^2\left(\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\right)$