Volume Generated by rotating the region $y=2x-x^2$

Question

What is the volume of the solid generated by revolving the region $y=2x-x^2$, $y=1$, $x=0$, $x=2$ about $y=1$?

Edit: Here is what I have done. But I'm not sure if I'm correct What I have done So far

Answer 1

Continuing from where you left off:

$$\pi \int_0^2 (1 - (2x-x^2))^2 \ dx = \pi \int_0^2 (x-1)^4\ dx$$ $$= \pi \left[\frac{1}{5}(x-1)^5 \right]_0^2 = \frac{\pi}{5} (1^5 - (-1)^5) = \frac{2 \pi}{5} $$

If you prefer, you can use a $u$-substution but you will need to change the bounds. Intutively, since we multiply by $\frac{d}{dx} x = 1$ to differentiate $(x-1)^4$, we divide by $1$ to integrate $(x-1)^4$ as usual (the formal proof comes from the chain rule). This allows you to find $\int (2x-1)^4 \ dx$ and so on.