Let $R$ be the region in the first quadrant bounded below by the graph of $y=x^2$ and above by the graph of $y=\sqrt{x}$. $R$ is the base of a solid whose cross sections perpendicular to the $x$-axis are squares. What is the volume of the solid?

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To find the lower and upper bounds of integration, we need to find when the two graphs intersect:

$$x^2=\sqrt{x}$$ $$x^4=x$$ $$x^4-x=0$$ $$x(x^3-1)=0$$ $$x=0, 1$$

Each square has side lengths equal to the difference between the $y$ values of each equation.

To find the volume:

$$\int\limits_0^1 A_{\text{square}}   dx = \int\limits_0^1 s^2   dx$$ $$=\int\limits_0^1 (\sqrt{x}-x^2)^2   dx$$ $$\approx \boxed{0.129}$$