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

The curves are $y^2=x$ (a rightward opening parabola) and $y=|x|$.Since $y^2=x⟹x\ge 0$, $y=|x|$ simplifies to $y=x$.Intersection points: $x^2=x⟹x(x-1)=0⟹x=0,1$.The points are $(0,0)$ and $(1,1)$.The area is ${\int }_0^1(\text{upper curve}-\text{lower curve})dx$.In $[0,1]$, $y=\sqrt{x}$ is the upper curve and $y=x$ is the lower curve.
$$A={\int }_0^1(\sqrt{x}-x)dx$$
$$A={\left[\frac{2}{3}{x}^{3/2}-\frac{x^2}{2}\right]}_0^1=\frac{2}{3}-\frac{1}{2}=\frac{4-3}{6}=\frac{1}{6}$$