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 $6 A-9 \sqrt{3}=$

The area (in sq. units) enclosed by the curves $y=2 x-x^2$ and $y=x^2-2 x-6$ is

The area (in sq. units) of the portion lying above the $X$-axis and enclosed between the curves $y^2=2 a x-x^2$ and $y^2=a x$ is

The area (in square units) of the region enclosed between the parabola $y^2=2 x$ and the line $y=4 x-1$