Find the area bounded by the curves (x-1)^{2} | vedclass

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(A) The area bounded by the curves $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$ is represented by the shaded region.On solving the equations $(x-1)^{2}+y^{

Vedclass · Accepted Answer

$\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right)$

Vedclass · Answer

(A) The area bounded by the curves $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$ is represented by the shaded region.On solving the equations $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$,we get:$x^2 - 2x + 1 + y^2 = 1$Since $x^2 + y^2 = 1$,we have $1 - 2x + 1 = 1$,which gives $2x = 1$,so $x = \frac{1}{2}$.Substituting $x = \frac{1}{2}$ in $x^2 + y^2 = 1$,we get $y^2 = 1 - \frac{1}{4} = \frac{3}{4}$,so $y = \pm \frac{\sqrt{3}}{2}$.The intersection points are $A\left(\frac{1}{2}, \frac{\sqrt{3}}{2}ight)$ and $B\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}ight)$.The required area is symmetric about the $x$-axis.Area $= 2 	imes \int_{0}^{1/2} \sqrt{1 - (x-1)^2} dx + 2 	imes \int_{1/2}^{1} \sqrt{1 - x^2} dx$.Using the formula $\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}ight)$:Area $= 2 \left[ \frac{x-1}{2}\sqrt{1-(x-1)^2} + \frac{1}{2}\sin^{-1}(x-1) ight]_0^{1/2} + 2 \left[ \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}(x) ight]_{1/2}^1$$= 2 \left[ (-\frac{1}{4}\frac{\sqrt{3}}{2} + \frac{1}{2}\sin^{-1}(-1/2)) - (0 + \frac{1}{2}\sin^{-1}(-1)) ight] + 2 \left[ (0 + \frac{1}{2}\sin^{-1}(1)) - (\frac{1}{4}\frac{\sqrt{3}}{2} + \frac{1}{2}\sin^{-1}(1/2)) ight]$$= 2 \left[ -\frac{\sqrt{3}}{8} - \frac{\pi}{12} + \frac{\pi}{4} ight] + 2 \left[ \frac{\pi}{4} - \frac{\sqrt{3}}{8} - \frac{\pi}{12} ight]$$= 2 \left[ \frac{\pi}{6} - \frac{\sqrt{3}}{8} ight] + 2 \left[ \frac{\pi}{6} - \frac{\sqrt{3}}{8} ight] = 4 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{8} ight) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$ sq. units.

Find the area bounded by the curves $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$.

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