Chapter 10.P, Problem 1P

The outer circle in the figure has radius 1 and the centers of the interior circular arcs lie on the outer circle. Find the area of the shaded region.

To determine

To find:

The area of the shaded region

Answer to Problem 1P

Solution:

$A=\frac{2\pi }{3}+2-2\sqrt{3}$

Explanation of Solution

1) Concept:

Draw the figure on $xy-$ plane and use the polar integration to find the area of the shaded portion

2) Calculation:

Consider the given figure on $xy-$ plane is

Consider the circle with center $\left(-1, 0\right)$ and radius as $\sqrt{2}$

So the equation of the circle is

${\left(x+1\right)}^2+y^2=2$

The figure above describes the circular arc from $\left(0, -1\right)\mathrm{ }\mathrm{t}\mathrm{o} \left(0, 1\right)$

To write the equation of circle in the form of polar coordinates

Substitute $x=r\cos \theta , y=r\sin \theta$

${\left(r\cos \theta +1\right)}^2+{\left(r\sin \theta \right)}^2=2$

$r^2{\cos }^2\theta +2r\cos \theta +1+r^2{\sin }^2\theta =2$

$r^2\left({\cos }^2\theta +{\sin }^2\theta \right)+2r\cos \theta +1-2=0$

$r^2+2r\cos \theta -1=0$

Use the quadratic formula,

$r=\frac{-2\cos \theta \pm \sqrt{4{\cos }^2\theta -4\left(-1\right)}}{2}$

$r=\frac{-2\cos \theta \pm 2\sqrt{{\cos }^2\theta +1}}{2}$

$r=-\cos \theta \pm \sqrt{{\cos }^2\theta +1}$

For $r>0$,

$r=-\cos \theta +\sqrt{{\cos }^2\theta +1}$

To find the total area of shaded region A

The area of green region is $\frac{1}{8}$ of the total area of shaded region A

That is,

$\frac{1}{8}A=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{1}{2}{r}^{2}d\theta$

$\frac{1}{8}A=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(1-2r\cos \theta \right)d\theta$

$\frac{1}{4}A=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(1-2\left(-\cos \theta +\sqrt{{\cos }^2\theta +1}\right)\cos \theta \right)d\theta$

$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(1+2{\cos }^2\theta -2\cos \theta \sqrt{{\cos }^2\theta +1}\right)d\theta$

$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(1+2\left(\frac{1+\cos 2\theta }{2}\right)-2\cos \theta \sqrt{1-{\sin }^2\theta +1}\right)d\theta$

$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(2+\cos 2\theta -2\cos \theta \sqrt{2-{\sin }^2\theta }\right)d\theta$

$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(2+\cos 2\theta \right)d\theta -2\underset{0}{\overset{\frac{\pi }{4}}{\int }}\cos \theta \sqrt{2-{\sin }^2\theta }d\theta$

$={\left[2\theta +\frac{\sin 2\theta }{2}\right]}_0^{\frac{\pi }{4}}-2\underset{0}{\overset{\frac{\pi }{4}}{\int }}\cos \theta \sqrt{2-{\sin }^2\theta }d\theta$

$=\left(2\left(\frac{\pi }{4}\right)+\frac{\sin \frac{\pi }{2}}{2}-0-0\right)-2\underset{0}{\overset{\frac{\pi }{4}}{\int }}\cos \theta \sqrt{2-{\sin }^2\theta }d\theta$

$=\left(\frac{\pi }{2}+\frac{1}{2}\right)-2\underset{0}{\overset{\frac{\pi }{4}}{\int }}\cos \theta \sqrt{2-{\sin }^2\theta }d\theta$

To evaluate the second integral by using substitution method,

Substitute $t=\sin \theta \Rightarrow dt=\cos \theta d\theta$

$\theta =\frac{\pi }{4}\Rightarrow t=\sin \frac{\pi }{4}\Rightarrow t=\frac{1}{\sqrt{2}} and \theta =0\Rightarrow t=\sin 0\Rightarrow t=0$

$=\left(\frac{\pi }{2}+\frac{1}{2}\right)-2\underset{0}{\overset{\frac{1}{\sqrt{2}}}{\int }}\sqrt{2-t^2}dt$

Integrate the above equation, by using the following formula:

$\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)$

$=\left(\frac{\pi }{2}+\frac{1}{2}\right)-2{\left[\frac{t}{2}\sqrt{2-t^2}+{\sin }^{-1}\frac{t}{\sqrt{2}}\right]}_0^{\frac{1}{\sqrt{2}}}$

$=\frac{\pi }{2}+\frac{1}{2}-2\left(\frac{1}{2\sqrt{2}}\sqrt{2-\frac{1}{2}}+{\sin }^{-1}\frac{1}{2}\right)$

$=\frac{\pi }{2}+\frac{1}{2}-2\left(\frac{1}{2\sqrt{2}}\sqrt{\frac{3}{2}}+\frac{\pi }{6}\right)$

$=\frac{\pi }{2}+\frac{1}{2}-\frac{1}{2}\sqrt{3}-\frac{\pi }{3}$

$\frac{1}{4}A=\frac{\pi }{2}+\frac{1}{2}-\frac{1}{2}\sqrt{3}-\frac{\pi }{3}$

$A=2\pi +2-2\sqrt{3}-\frac{4\pi }{3}$

$A=\frac{2\pi }{3}+2-2\sqrt{3}$

Conclusion:

$A=\frac{2\pi }{3}+2-2\sqrt{3}$