[Math] Washer Method – Inner & outer radius given by the same function

Question

Consider the region enclosed by the curve y = 2x(1-x) and the x-axis. This region is to be rotated about the y-axis.
Use the washer method to write down an integral expression for the volume.

I understand how to apply to washer method, but I am confused in this case by the fact that both the inner and outer radius are defined by the same function (two solutions for each x value).

So when I take the difference of the outer radius squared and inner radius squared, they will just cancel out and give 0.

What do I do in this case?

Graph: https://i.imgur.com/XgFnges.png

Answer 1

HINT

We need the expression for $x=g(y)$ that is

$$y=2x(1-x)\iff 2x^2-2x+y=0 \iff x=\frac{2\pm \sqrt{4-8y}}{4}=\frac{1\pm \sqrt{1-2y}}{2}$$

with $y\in[0,\frac12]$. and therefore

$$V=\int_0^\frac12 2\pi y\sqrt{1-2y}\,dy$$