$\begin{aligned}y^2=x \text { and } 2 y=x \Rightarrow y^2 & =2 y \Rightarrow y=0,2 \\\therefore \text { Required area } & =\int_0^2\left(y^2-2 y\right) d y=\left(\frac{y^3}{3}-y^2\right)_0^2=\frac{4}{3}\end{aligned}$