The area of the region bounded by $y^{2}=8x$ and $y^{2}=16(3-x)$ is equal to

Let $C_{1}$ be the curve obtained by the solution of the differential equation $2xy \frac{dy}{dx} = y^{2} - x^{2}, x > 0$. Let the curve $C_{2}$ be the solution of $\frac{2xy}{x^{2} - y^{2}} = \frac{dy}{dx}$. If both the curves pass through $(1, 1)$,then the area enclosed by the curves $C_{1}$ and $C_{2}$ is equal to:

The area bounded by the curves $y=x^2$ and $y-6=-|x|$ is

The area of the region $\{(x, y): x^2 \leq y \leq 8-x^2, y \leq 7\}$ is

The area enclosed by the curves $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$ and the lines $x=0, x=\frac{\pi}{2}$ is:

The area (in sq. units) of the region $\{(x, y) \in R^{2} | 4 x^{2} \leq y \leq 8 x+12\}$ is: