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

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$π$

$2 π$

$3 π$

$8 π$

answer is D.

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Detailed Solution

$x^{2} + y^{2} - 4 x = 0, y^{2} = x \Rightarrow x^{2} + x - 4 x = 0 \Rightarrow x^{2} - 3 x = 0 \Rightarrow x = 0, x = 3$

$\begin{array}{l} A = \int_{0}^{3} [\sqrt{4 x - x^{2}} - \sqrt{x}] d x = \int_{0}^{3} \sqrt{2 - (x - 2)^{2}} d x - \int_{0}^{3} \sqrt{x} d x \\ = [\frac{x - 2}{2} \sqrt{{2^{2} - (x - 2)^{2} + \frac{2^{2}}{2} \sin^{- 1} \frac{x - 2}{2}]}_{0}^{3} - {[\frac{x^{3 / 2}}{3 / 2}]}_{0}^{3}} \\ = \frac{1}{2} \sqrt{3} + 2 (\frac{π}{6}) + 2 (\frac{π}{2}) - \frac{2}{3} \times 3^{3 / 2} = \frac{\sqrt{3}}{2} + \frac{4 π}{3} - 2 \sqrt{3} = \frac{4 π}{3} - \frac{3 \sqrt{3}}{2} \\ 6 A + 9 \sqrt{3} = 6 (\frac{4 π}{3} - \frac{3 \sqrt{3}}{2}) + 9 \sqrt{3} = 8 π - 9 \sqrt{3} + 9 \sqrt{3} = 8 π \end{array}$