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If the area lying in the first quadrant and bounded by the circle $x^2+y^2-4 x=0$, the parabola $y^2=x$ and the $X$-axis is $A$, then $6 A-9 \sqrt{3}=$
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$4 \pi$
Given curve
$x^2+y^2-4 x=0 \Rightarrow(x-2)^2+y^2=4 \text { and } y^2=x$
intersecting point of circle and parabola is $(0,0)$ and $(3, \sqrt{3})$
Required area
$A=\int_0^3 \sqrt{x} d x+\int_3^4 \sqrt{4-(x-2)^2} d x$
$\begin{aligned} & A=\left[\frac{2}{3}(x)^{3 / 2}\right]_0^3+\left[\frac{x-2}{2} \sqrt{4-(x-2)^2}+2 \sin ^{-1} \frac{x-2}{2}\right]_3^4 \\ & A=2 \sqrt{3}+\left[\left(0+2 \sin ^{-1}(1)\right)-\left(\frac{\sqrt{3}}{2}+2 \sin ^{-1}\left(\frac{1}{2}\right)\right)\right] \\ & A=2 \sqrt{3}+\pi-\frac{\sqrt{3}}{2}-\frac{\pi}{3}=\frac{3 \sqrt{3}}{2}+\frac{2 \pi}{3} \\ & 6 A-9 \sqrt{3}=4 \pi\end{aligned}$
$x^2+y^2-4 x=0 \Rightarrow(x-2)^2+y^2=4 \text { and } y^2=x$
intersecting point of circle and parabola is $(0,0)$ and $(3, \sqrt{3})$
Required area
$A=\int_0^3 \sqrt{x} d x+\int_3^4 \sqrt{4-(x-2)^2} d x$
$\begin{aligned} & A=\left[\frac{2}{3}(x)^{3 / 2}\right]_0^3+\left[\frac{x-2}{2} \sqrt{4-(x-2)^2}+2 \sin ^{-1} \frac{x-2}{2}\right]_3^4 \\ & A=2 \sqrt{3}+\left[\left(0+2 \sin ^{-1}(1)\right)-\left(\frac{\sqrt{3}}{2}+2 \sin ^{-1}\left(\frac{1}{2}\right)\right)\right] \\ & A=2 \sqrt{3}+\pi-\frac{\sqrt{3}}{2}-\frac{\pi}{3}=\frac{3 \sqrt{3}}{2}+\frac{2 \pi}{3} \\ & 6 A-9 \sqrt{3}=4 \pi\end{aligned}$
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