# Find volume of solid generated by curves in $xy$ plane and cross sections perpendicular to $x$-axis

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?

• Alternatively $$V=\int_0^1 (\sqrt x-x)^2 \mathrm{d}x = \frac{1}{30}$$ May 14 at 2:18