The area bounded between the curves $y^2=4x$ and $x^2=4y$ in the plane is 

The given curves are $y^2=4x$ and $x^2=4y$

Points of intersection of curves are as below

${\left(\frac{x^2}{4}\right)}^2=4x\Rightarrow x^3=64$

it implies that $x=0,x=4$

The rough sketch of required area is as below

Hence, the area of the shaded region is the area of the region bounded between the curves $y=2\sqrt{x},y=\frac{x^2}{4}$ and the lines $x=0,x=4$

Hence,

      $\begin{array}{rcl}A& =& {\int }_0^4\left(2\sqrt{x}-\frac{x^2}{4}\right)dx\\ & =& {\left(2\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}-\frac{x^3}{12}\right)}_0^4\\ & =& \frac{4}{3}\left(8\right)-\frac{1}{12}\left(64\right)\\ & =& \frac{32}{3}-\frac{16}{3}\\ & =& \frac{16}{3}\end{array}$

Therefore, the area of the region bounded between the curves  $y^2=4x$and $x^2=4y$ is $\boxed{\frac{16}{3}}$