If the area lying in the first quadrant and b | vedclass

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(D) Given curves are $x^2+y^2-4x=0 \Rightarrow (x-2)^2+y^2=4$ and $y^2=x$.To find the intersection points, substitute $y^2=x$ into the circle equation

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$4\pi$

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(D) Given curves are $x^2+y^2-4x=0 \Rightarrow (x-2)^2+y^2=4$ and $y^2=x$.To find the intersection points, substitute $y^2=x$ into the circle equation: $x^2+x-4x=0 \Rightarrow x^2-3x=0 \Rightarrow x(x-3)=0$. Thus, $x=0$ or $x=3$.For $x=0$, $y=0$. For $x=3$, $y^2=3 \Rightarrow y=\sqrt{3}$ (in the first quadrant).The area $A$ in the first quadrant is bounded by the parabola from $x=0$ to $x=3$ and the circle from $x=3$ to $x=4$.$A = \int_0^3 \sqrt{x} \, dx + \int_3^4 \sqrt{4-(x-2)^2} \, dx$$A = \left[ \frac{2}{3}x^{3/2} \right]_0^3 + \left[ \frac{x-2}{2} \sqrt{4-(x-2)^2} + 2 \sin^{-1} \left( \frac{x-2}{2} \right) \right]_3^4$$A = \frac{2}{3}(3\sqrt{3}) + \left[ (0 + 2 \sin^{-1}(1)) - (\frac{1}{2} \sqrt{3} + 2 \sin^{-1}(1/2)) \right]$$A = 2\sqrt{3} + \pi - \frac{\sqrt{3}}{2} - \frac{\pi}{3} = \frac{3\sqrt{3}}{2} + \frac{2\pi}{3}$.Therefore, $6A - 9\sqrt{3} = 6(\frac{3\sqrt{3}}{2} + \frac{2\pi}{3}) - 9\sqrt{3} = 9\sqrt{3} + 4\pi - 9\sqrt{3} = 4\pi$.

If the area lying in the first quadrant and bounded by the circle $x^2+y^2-4x=0$, the parabola $y^2=x$ and the $X$-axis is $A$, then $6A-9\sqrt{3}=$

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