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

$\frac{1}{6}$ sq. units

The correct answer is Option (3) → $\frac{1}{6}$ sq. units

$y^2=x$, $y = |x|$

$y = |x|⇒y^2=x^2$

$y^2=x⇒x^2=x$

$x=0,1$ point of intersection

so area = $\int\limits_0^1\sqrt{x}-xdx$

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

$=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}$ sq. units