The area bounded by the curves \(y^2=4x\) and \(x^2=4y\) is

16/3

$y^2=4x$ & $x^2=4y$

$⇒(\frac{x^2}{4})^2=4x⇒x^4=64x$

$⇒x(x^3-64)=0$

$⇒x=0,4$

Area bounded is

$A=\int\limits_0^4(y_1-y_2)dx=\int\limits_0^4(\sqrt{4x}-\frac{x^2}{4})dx$

$=2.\frac{x^{3/2}}{(\frac{3}{2})}-\frac{1}{4}.\frac{x^3}{3}|_0^4$

$=\frac{4}{3}.4^{3/2}-\frac{1}{12}.4^3=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}$