Question

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

• A. $\pi$
• B. $2\pi$
• C. $3\pi$
• D. $4\pi$

Answer 1

Answer: (D)

Given curve,
$x^2+y^2-4x=0$
$\Rightarrow (x-2{)}^2+y^2=4$
and $y^2=x$ intersecting point of circle and parabola is $(0,0)$ and $(3,\sqrt{3})$
Required area

$A={\left[\frac{2}{3}(x{)}^{3/2}\right]}_0^3+{\left[\frac{x-2}{2}\sqrt{4-(x-2{)}^2}+2{\sin }^{-1}\frac{x-2}{2}\right]}_3^4$
$A=2\sqrt{3}+\left[\left(0+2{\sin }^{-1}(1)\right)-\left(\frac{\sqrt{3}}{2}+2{\sin }^{-1}\left(\frac{1}{2}\right)\right)\right]$
$A=2\sqrt{3}+\pi -\frac{\sqrt{3}}{2}-\frac{\pi }{3}=\frac{3\sqrt{3}}{2}+\frac{2\pi }{3}$
$6A-9\sqrt{3}=4\pi$