Find the area enclosed by the parabola $4y = 3x^{2}$ and the line $2y = 3x + 12$.

The area (in square units) enclosed between the curves $y=\sin x$ and $y=\cos x$ for $\frac{\pi}{4} \leq x \leq \frac{5 \pi}{4}$ is

The area of the region included between the parabolas $y^2=8x$ and $x^2=8y$ is

Let $A = \{(x, y) : y^2 \le 4x, y - 2x \ge -4\}$. The area of the region $A$ is

The area (in sq. units) bounded by the curves $y=\sqrt{x}$,$2y-x+3=0$,$X$-axis and lying in the first quadrant,is

Let $S = \{(x, y) \in R \times R : x \geq 0, y \geq 0, y^2 \leq 4x, y^2 \leq 12 - 2x \text{ and } 3y + \sqrt{8}x \leq 5\sqrt{8}\}$. If the area of the region $S$ is $\alpha \sqrt{2}$,then $\alpha$ is equal to