The area of the region bounded by the curves $x^{2}+y^{2}=8$ and $y^{2}=2x$ is

The area of the shorter region bounded by $|y| = 4 - x^2$ and $|y| = 3x$ is given by $\left( 3K + \frac{1}{3} \right)$ sq-unit,where $K$ is equal to:

The area (in sq. units) of the region enclosed by the curves $y=x^{2}-1$ and $y=1-x^{2}$ is equal to

Let $T$ and $C$ respectively be the transverse and conjugate axes of the hyperbola $16x^2 - y^2 + 64x + 4y + 44 = 0$. Then the area of the region above the parabola $x^2 = y + 4$,below the transverse axis $T$ and on the right of the conjugate axis $C$ is:

The area bounded by the curves $y-1=\cos x$,$y=\sin x$ and the $X$-axis between $x=0$ and $x=\pi$ is

The area (in sq. units) of the region described by $\{(x, y): y^2 \leq 2x\}$ and $\{y \geq 4x - 1\}$ is