Find the area of the region enclosed between the two circles $:$ $x^{2}+y^{2}=4$ and $(x-2)^{2}+y^{2}=4$.

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

The area (in sq. units) of the smaller region lying above the $X$-axis and bounded between the circle $x^2+y^2=2ax$ and the parabola $y^2=ax$ is

The area of the region bounded by the curves $y=x^3$, $y=x^2$ and the lines $x=0$ and $x=2$ is

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 of the region bounded by $y^2=x$ and $y=|x|$ is