The area of the region above the X-axis inclu | vedclass

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(B) The given equations are $y^{2}=x$ (parabola) and $x^{2}+y^{2}=2x$ (circle).Equation of circle can be written as $(x-1)^{2}+y^{2}=1$,which has cent

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$\frac{\pi}{4}-\frac{2}{3}$

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(B) The given equations are $y^{2}=x$ (parabola) and $x^{2}+y^{2}=2x$ (circle).Equation of circle can be written as $(x-1)^{2}+y^{2}=1$,which has center $(1,0)$ and radius $1$.Solving the two equations: $x^{2}+x=2x \implies x^{2}-x=0 \implies x(x-1)=0$. Thus,$x=0$ and $x=1$.The points of intersection are $(0,0)$ and $(1,1)$.We need the area above the $X$-axis,so we consider $y = \sqrt{x}$ for the parabola and $y = \sqrt{1-(x-1)^{2}}$ for the upper semi-circle.The required area is $\int_{0}^{1} (\sqrt{1-(x-1)^{2}} - \sqrt{x}) dx$.$= \left[ \frac{x-1}{2}\sqrt{1-(x-1)^{2}} + \frac{1}{2}\sin^{-1}(x-1) \right]_{0}^{1} - \left[ \frac{2}{3}x^{3/2} \right]_{0}^{1}$$= \left( 0 + \frac{1}{2}\sin^{-1}(0) \right) - \left( 0 + \frac{1}{2}\sin^{-1}(-1) \right) - \frac{2}{3}$$= 0 - (-\frac{\pi}{4}) - \frac{2}{3} = \frac{\pi}{4} - \frac{2}{3}$ square units.

The area of the region above the $X$-axis included between the parabola $y^{2}=x$ and the circle $x^{2}+y^{2}=2x$ in square units is

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