The curves are $y=x^2, y=x$ and the cross sections are squares perpendicular to the $x-$axis such that the base of the squares is on the $xy-$plane.
My solution:
The area of the squares is given by $A=L^2$, where $L$ is the side of the square. We can find $L$ in function of $x$ computing the distance between the line and the parabola. The distance is $L=x-x^2$. Then, $A(x) = (x-x^2)^2.$
Hence, the desired volume is
$$V=\int_0^1 A(x) \mathrm{d}x = \int_0^1 (x-x^2)^2 \mathrm{d}x = \frac{1}{30}.$$
Is this correct? Is there another way to solve this kind of exercises?