The area of the region $\{(x, y): y^2 \leq 4x, x < 4, \frac{xy(x-1)(x-2)}{(x-3)(x-4)} > 0, x \neq 3\}$ is

The area (in sq. units) of the portion lying above the $X$-axis and enclosed between the curves $y^2=2ax-x^2$ and $y^2=ax$ is

If the area enclosed between the curves $y = kx^2$ and $x = ky^2$ $(k > 0)$ is $1$ square unit,then $k$ is

Given: $f(x) = \begin{cases} x, & 0 \leq x < \frac{1}{2} \\ \frac{1}{2}, & x = \frac{1}{2} \\ 1-x, & \frac{1}{2} < x \leq 1 \end{cases}$ and $g(x) = (x-\frac{1}{2})^2, x \in R$. Then the area (in sq. units) of the region bounded by the curves $y=f(x)$ and $y=g(x)$ between the lines $2x=1$ and $2x=\sqrt{3}$ is:

Area of the region $\{(x, y): x^2+(y-2)^2 \leq 4, x^2 \geq 2y\}$ is

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