The volume of a solid of revolution of the area between y = x and y = y^2 and 0<=x<=1

I have solved this problem several different ways : rotating the y = x line to the x axis (a very messy integral, but solvable), integration by cones (resembles integration by "washer" using the area of the side of cone with a side equal to $x -x^2$ ) or by the integration by displaced "washer" (using the area of the base of the aforementioned cone). It turns out, I feel \ that the last method is the simplest.

$\frac{\pi \int_0^1 \left(x^4-2 x^3+x^2\right) \, dx}{\sqrt{2}}\text{ is }\frac{\pi}{\sqrt{2}}|\frac{x^5}{5}-\frac{x^4}{2}+\frac{x^3}{3}|^1_{x=0}\text{ is }\frac{\pi}{30\sqrt{2}}(6-15+10)\text{ is }\frac{\pi}{30\sqrt{2}}$

$0.0698131700798$