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

$\frac{1}{6}$ $\frac{1}{3}$ $\frac{2}{3}$ 1

$\frac{1}{3}$

Let A be the area enclosed between the given curves, then $A=\int\limits_{0}^{1}(\sqrt{x}-x)dx=\left[\frac{2}{3}x^{3/2}-\frac{x^2}{2}\right]_{0}^{1}=\frac{1}{3}$