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  • Using integration, find the area of the region bounded by the curves and <img alt="x^2 + y^2 = 1" src="https://

Using integration, find the area of the region bounded by the curves (x - 1)^{2} + y^2 = 1 and x^2 + y^2 = 1.

 

Answers (1)

Given:

(x - 1)^{2} + y^2 = 1 ....(i)

x^2 + y^2 = 1....(ii)

For intersection point

Put value of y2 from equation (ii) to in equation (i)

(x-1)^2 +1- x^2 = 1

(x-1)^2 =x^2

x-1 =\pm x

x = \frac{1}{2}

y = \pm \frac{\sqrt{3}}{2}

$Required Area $ = 2\times $Area of OAM$

$Required Area $ = 2 \left[ \int _0^\frac{1}{2}\sqrt{1-(x-1)^2}dx+\int _\frac{1}{2}^1\sqrt{1-x^2}dx\right]

= 2 \left[ \frac{1}{2} \sin^{-1}\frac{x-1}{1} + \frac{x-1}{2}\sqrt{1-(x-1)^2}\right ]_0^\frac{1}{2} + 2 \left[ \frac{1}{2} \sin^{-1}\frac{x}{1} + \frac{x}{2}\sqrt{1-x^2}\right ]_\frac{1}{2}^1

= 2 \left[ \frac{1}{2} \sin^{-1}\frac{-1}{2} - \frac{1}{8}\sqrt{3} - \frac{1}{2}\sin^{-1}(-1) -0\right ] + 2 \left[ \frac{1}{2} \sin ^{-1}(1) +0 - \frac{1}{2}\sin^{-1}\frac{1}{2} - \frac{1}{8}\sqrt{3}\right ]

\\= \left[ -\frac{\pi}{6} - \frac{\sqrt{3}}{4} + \frac{\pi}{2} + \frac{\pi}{2} - \frac{\pi}{6} - \frac{\sqrt{3}}{4}\right ] \\\\ = \left[ \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \right ] \\\\

 

Posted by

Safeer PP

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