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

$\frac{1}{3}$

So $y=x^2$, and $x=y^2$

finding area along x-axis

uppor cueve $y^2=x$

$\Rightarrow y=\sqrt{x}$

lower corve

$y=x^2$

intersecting points

$x^2=\sqrt{x}$

So x = 0, 1

y = 0, 1

So area

$=\int\limits_0^1 \sqrt{x}-x^2 d x$

area $=\int\limits_0^1 \sqrt{x}-x^2 d x= \left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{x^{2+1}}{2+1}\right]_0^1$

$=\left[\frac{2}{3} x^{3 / 2}-\frac{x^3}{3}\right]_0^1$

$=\frac{2}{3}-\frac{1}{3}$

$=\frac{1}{3}$