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  • Using method of integration, find the area of the region enclosed between two circles 

Using method of integration, find the area of the region enclosed between two circles x^2+y^2=4\:\:and\:\:(x-2)^2+y^2=4.

 

 

 

 
 
 
 
 

Answers (1)

Equations of the given circles are: 

x^2+y^2=4\:\:(i)\\ (x-2)^2+y^2=4\:\:(ii)

Equation (i) is a circle with centre O at the origin and radius 2. Equation (ii) is a circle with centre C(2,0) and radius 2.

Solving equation (i) and (ii) we have 

(x-2)^2+y^2=x^2+y^2 \:\:or \\ x^2-4x+4+y^2=x^2+y^2\:\:or\:\:x=1 \:\:which\:gives\: y= \pm \sqrt{3}

Thus, the points of intersection of the given circles are 

A(1,\sqrt3)\:\:A^1(1,-\sqrt3)

The required area of the enclosed region OACA1O between circles 

=2\left [ area\:\:of\:region\:\:ODCAO \right ]\\ =2[area \:\:of\:region\:\:ODAO+\:area\:of\:region\:DCAD]\\ =2 \left [ \int_{0}^{1}ydx+\int_{1}^{2}ydx \right ]\\ 2\left [ \int_{0}^{1}\sqrt{4-(x-2)^2dx} +\int_{1}^{2}\sqrt{4-x^2}dx\right ]\:\:from(i) \\ 2\left [ \frac{1}{2}(x-2)\left ( \sqrt{4-(x-2)^2}+\frac{1}{2}\times 4 \sin^{-1}\left ( \frac{x-2}{2} \right ) \right ) \right ] _{0}^{1}+2\left [ \frac{1}{2}\times \sqrt{4-x^2}+\frac{1}{2}\times 4 \sin^{-1} + \frac{x}{2}\right ]_{1}^{2}

=\left [ (x-2)\sqrt{4-(x-2)^2}+4\sin^{-1}\left ( \frac{x-2}{2} \right ) \right ]_{1}^{2}+\left [ x\sqrt{4-x^2}+4\sin^{-1}\frac{x}{2} \right ]_{1}^{2} \\ = \left [ -\sqrt3 +4 \sin^{-1}\left ( \frac{-1}{2} \right )-4\sin^{-1}(-1)\right ]+\left [ 4\sin^{-1}1-\sqrt3 - 4\sin^{-1}\frac{1}{2} \right ] \\ = \left [ \left ( -\sqrt3 -4\times \frac{\pi}{6}\right )+4\times \frac{\pi}{2} \right ]+\left [ 4\times \frac{\pi}{2}- \sqrt3 -4\times \frac{\pi}{6}\right ] \\ =\frac {8\pi}{3}-2\sqrt3

Posted by

Ravindra Pindel

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