The area (in $sq. units$) of the region $A = \{(x,y) : \frac{y^2}{2} \le x \le y + 4\}$ is

Let $A_{1}$ be the area of the region bounded by the curves $y = \sin x$,$y = \cos x$ and the $y$-axis in the first quadrant. Also,let $A_{2}$ be the area of the region bounded by the curves $y = \sin x$,$y = \cos x$,the $x$-axis and $x = \frac{\pi}{2}$ in the first quadrant. Then ..... .

The area of the region $\{(x, y): xy \leq 8, 1 \leq y \leq x^2\}$ is

The area of the smaller portion between the curves $x^2 + y^2 = 8$ and $y^2 = 2x$ is

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,

If the area of the region $\{(x, y) : |x-5| \leq y \leq 4 \sqrt{x}\}$ is $A$,then $3A$ is equal to . . . . . . .