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Solution

Curves: $y^2 = x$ (right-opening parabola) and $x^2 = y$ (upward parabola). Points of intersection: substitute $y = x^2$ into $y^2 = x$: $(x^2)^2 = x \Rightarrow x^4 - x = 0 \Rightarrow x(x^3 - 1) = 0.$ $\Rightarrow x = 0, 1.$ Between $x = 0$ and $1$: upper curve is $y = \sqrt{x}$, lower is $y = x^2$. Area $A = \int_0^1 (\sqrt{x} - x^2)\,dx = \left[\dfrac{2}{3}x^{3/2} - \dfrac{x^3}{3}\right]_0^1 = \dfrac{2}{3} - \dfrac{1}{3} = \dfrac{1}{3}.$