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:

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