The area (in sq. units) of the part of the circle $x^{2} + y^{2} = 36$ which is outside the parabola $y^{2} = 9x$ is:

The area bounded by the curves $y = |x| - 1$ and $y = -|x| + 1$ is

The area enclosed by $y^{2}=8x$ and $y=\sqrt{2}x$ that lies outside the triangle formed by $y=\sqrt{2}x$,$x=1$,and $y=2\sqrt{2}$ is equal to

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

The area enclosed by the curves $y = \sqrt{x}$ and $x = -\sqrt{y}$,and the circle $x^2 + y^2 = 2$ above the $x$-axis,is

The area (in sq. units) of the region $\{(x, y) : x \geq 0, x+y \leq 3, x^2 \leq 4y \text{ and } y \leq 1+\sqrt{x}\}$ is