The area (in square units) of the region encl | vedclass

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(D) The intersection points of the two circles $x^2+y^2=1$ $(i)$ and $(x-1)^2+y^2=1$ (ii) are found by substituting $y^2=1-x^2$ from $(i)$ into (ii):

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$\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}$

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(D) The intersection points of the two circles $x^2+y^2=1$ $(i)$ and $(x-1)^2+y^2=1$ (ii) are found by substituting $y^2=1-x^2$ from $(i)$ into (ii): $(x-1)^2+(1-x^2)=1$ $x^2-2x+1+1-x^2=1$ $2-2x=1 \Rightarrow x=\frac{1}{2}$ Substituting $x=\frac{1}{2}$ into $(i)$,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 $C\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}ight)$. The total area is symmetric about the $x$-axis,so 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 $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$,Area $= 2 \left[ \frac{x-1}{2}\sqrt{1-(x-1)^2} + \frac{1}{2}\sin^{-1}(x-1) ight]_0^{1/2} + 2 \left[ \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}(x) ight]_{1/2}^1$ $= 2 \left[ (-\frac{1}{4}\cdot\frac{\sqrt{3}}{2} + \frac{1}{2}\sin^{-1}(-\frac{1}{2})) - (0 + \frac{1}{2}\sin^{-1}(-1)) ight] + 2 \left[ (0 + \frac{1}{2}\sin^{-1}(1)) - (\frac{1}{4}\cdot\frac{\sqrt{3}}{2} + \frac{1}{2}\sin^{-1}(\frac{1}{2})) ight]$ $= 2 \left[ -\frac{\sqrt{3}}{8} - \frac{\pi}{12} + \frac{\pi}{4} ight] + 2 \left[ \frac{\pi}{4} - \frac{\sqrt{3}}{8} - \frac{\pi}{12} ight]$ $= 2 \left[ \frac{\pi}{6} - \frac{\sqrt{3}}{8} ight] + 2 \left[ \frac{\pi}{6} - \frac{\sqrt{3}}{8} ight] = 4 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{8} ight) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$.

The area (in square units) of the region enclosed by the two circles $x^2+y^2=1$ and $(x-1)^2+y^2=1$ is

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