# Using integration, find the area of the region bounded by the curves and .

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)

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

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

$x-1 =\pm x$

$x = \frac{1}{2}$

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

$Required Area = 2\times Area of OAM$

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

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

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

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

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