Let $A = \{(x, y) \in R^2 : y \geq 0, 2x \leq y \leq \sqrt{4-(x-1)^2}\}$ and $B = \{(x, y) \in R \times R : 0 \leq y \leq \min \{2x, \sqrt{4-(x-1)^2}\}\}$. Then the ratio of the area of $A$ to the area of $B$ is

The area (in square units) in the first quadrant bounded by the curve $y=x^2+2$ and the lines $y=x+1, x=0$ and $x=3$,is

If the area of the region bounded by $y=\cos x$, $y=\sin x$, $x=\frac{\pi}{4}$ and $x=\pi$ is bisected by the line $x=a$, then $\sin \left(a+\frac{\pi}{4}\right)=$

The area (in square units) bounded by the curves $x = -2y^2$ and $x = 1 - 3y^2$ is

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

Let $AB$ be the latus rectum of the parabola $y^2 = 4ax$ in the $XY$-plane. Let $T$ be the region bounded by the finite arc $AB$ of the parabola and the line segment $AB$. $A$ rectangle $PQRS$ of maximum possible area is inscribed in $T$ with $P, Q$ on line $AB$,and $R, S$ on arc $AB$. Then,$\frac{\text{area}(PQRS)}{\text{area}(T)}$ equals