The area enclosed between the curves $y^2=x$ and $y=|x|$ is:

The curves are $y^2=x$ (a parabola) and $y=|x|$ (V-shape).The parabola $y^2=x$ only exists for $x\ge 0$.In the region $x\ge 0$, $y=|x|$ becomes $y=x$.Intersection points: $x^2=x\Rightarrow x(x-1)=0\Rightarrow x=0,1$.The points are $(0,0)$ and $(1,1)$.Between $x=0$ and $x=1$, the parabola $y=\sqrt{x}$ is above the line $y=x$.
$$\text{Area}={\int }_0^1(\sqrt{x}-x)dx$$
$$\text{Area}={\left[\frac{2}{3}{x}^{3/2}-\frac{x^2}{2}\right]}_0^1=\left(\frac{2}{3}-\frac{1}{2}\right)-0$$
$$\text{Area}=\frac{4-3}{6}=\frac{1}{6}\text{sq. units}$$