The area enclosed between $y^2=4x$ and $x^2=4y $ is :

$\frac{16}{3}$ sq.units

The correct answer is Option (4) → $\frac{16}{3}$ sq.units

they intersect when

$x^2=4y$

$⇒y^2=4x$

$⇒\frac{x^4}{16}=4x$

$x^3=64$

$x=4$

$⇒y=4$

area = $\int\limits_0^4\sqrt{4x}-\frac{x^2}{4}dx$

$\left[\frac{4}{3}x^{\frac{3}{2}}-\frac{x^3}{12}\right]_0^4$

$=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}$ sq. units