The parabolas $y^2 = 4x$ and $x^2 = 4y$ divide the square region bounded by the lines $x = 4$,$y = 4$ and the coordinate axes. If $S_1, S_2, S_3$ are respectively the areas of these parts numbered from top to bottom,then $S_1:S_2:S_3$ is

The area (in sq. units) bounded by the curve $y=2x-x^2$ and the line $y=-x$ is

The area (in sq. units) of the region bounded by the curve $x^2=4y$ and the straight line $x=4y-2$ 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,

The area (in sq. units) of the region bounded by the curves $y=x^2$ and $y=8-x^2$ 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 ..... .