If the area of the region $\{(x, y): 0 \leq y \leq \min \{2x, 6x-x^2\}\}$ is $A$,then $12A$ is equal to

Let $A_1, A_2$ and $A_3$ be the regions on $\mathbb{R}^2$ defined by:
$A_1 = \{(x, y) : x \geq 0, y \geq 0, 2x + 2y - x^2 - y^2 > 1 > x + y\}$
$A_2 = \{(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^2 + y^2\}$
$A_3 = \{(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^3 + y^3\}$
Denote by $|A_1|, |A_2|$ and $|A_3|$ the areas of the regions $A_1, A_2$ and $A_3$ respectively. Then,

The area of the region enclosed by the curves $y^2=4(x+1)$ and $y^2=5(x-4)$ is

The area of the region bounded by the curves $x + 3y^2 = 0$ and $x + 4y^2 = 1$ is equal to: (in $/3$)

The area (in sq. units) of the smaller portion enclosed between the curves,$x^2 + y^2 = 4$ and $y^2 = 3x$,is

The volume of the solid generated by revolving about the $y-$axis the figure bounded by the parabolas $y = x^2$ and $x = y^2$ is