The area enclosed between the curves ${x}^{2}=y$ and ${y}^{2}=x$ is equal to

$\because$ Curves ${x}^{2}=y$ and ${y}^{2}=x$ intersect at $\left(0,0\right)$ and $\left(1,1\right)$.

$\therefore$Required Area$=\int_{0}^{1}{\left(\sqrt{x}-{x}^{2}\right)dx}$

; & ;                       $=\left[\dfrac{{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\dfrac{{x}^{3}}{3}\right]_{0}^{1}$

                         $=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}$.sq.unit.

Also, both curves ${x}^{2}=y$ and ${y}^{2}=x$ are symmetrical about $y=x$

$\therefore$ Required Area$=2\int_{0}^{1}{\left(x-{x}^{2}\right)dx}$

Option$\left(c\right):{x}^{2}\le y\le \left|x\right|$

$y={x}^{2},y=\left|x\right|$ point of intersection is $\left(0,0\right)$ and $\left(1,1\right)$.

$\therefore$ Required Area$=2\int_{0}^{1}{\left(x-{x}^{2}\right)dx}$

                               $=2\left[\dfrac{{x}^{2}}{2}-\dfrac{{x}^{3}}{3}\right]_{0}^{1}$

                             $=2\left(\dfrac{1}{2}-\dfrac{1}{3}\right)$

                             $=1-\dfrac{2}{3}=\dfrac{1}{3}$.sq.unit