Curves: \(y^2=4x\) (right-opening) and \(x^2=4y\) (upward). Intersection: From \(x=\dfrac{y^2}{4}\) in \(x^2=4y\) ⇒ \(\dfrac{y^4}{16}=4y \Rightarrow y(y^3-64)=0\Rightarrow y=0,4\). Thus points are \((0,0)\) and \((4,4)\) in the first quadrant.

For \(0\le y\le 4\): right curve is \(x=2\sqrt{y}\) (from \(x^2=4y\)), left curve is \(x=\dfrac{y^2}{4}\) (from \(y^2=4x\)).

Area \(=\displaystyle \int_{0}^{4}\!\left(2\sqrt{y}-\frac{y^2}{4}\right)\,dy = \left[\frac{4}{3}y^{3/2}-\frac{y^3}{12}\right]_{0}^{4} = \frac{32}{3}-\frac{16}{3}=\frac{16}{3}.\)

Final Answer: \(\displaystyle \frac{16}{3}\) square units.