I don't know how to find the outer and inner radius, of the given section bound by $y=-x^2+2x$ about y=-1 , in this case, the transversal section is about $x\in[0,2]$ by washed method.
I know that $A(x)=\pi(R^2-r^2) $ and the volume $V=\pi\int_0^2(R^2-r^2)dx$
If the region revolves about x-axis $y=0$, the cross-sections formed are circles, and then : \begin{align} V(s)& = \pi\int_0^2(-x^2+2x)^2dx \\ & = \pi\int_0^2(x^4-4x^3+4x^2)dx \\ & = (...)\cfrac{32\pi}{30}\\ \end{align}
The same question i have in case of positive y-axis, any help?
Move the curve one unit up (while integrating, it's a good idea to do a basic u-substitution, $u=x-1$, and take advantage of symmetry to make things a tad simpler to calculate by hand):
$$ V=\pi\int_{0}^{2}\left[(-x^2+2x+1)^2-1^2\right]\,dx= \pi\int_{0}^{2}\left(\left[(x-1)^2-2\right]^2-1\right)\,dx=\\ \pi\int_{-1}^{1}\left(\left[u^2-2\right]^2-1\right)\,du= 2\pi\int_{0}^{1}\left(u^4-4u^2+3\right)\,du=\\ 2\pi\left[\frac{u^5}{5}-\frac{4u^3}{3}+3u\right]_{0}^{1}= 2\pi\left(\frac{1}{5}-\frac{4}{3}+3\right)=\\ 2\pi\frac{3-20+45}{15}=\frac{56\pi}{15}\ cubic\ units. $$
What you're doing is you're subtracting the volume you get by rotating a 1x2 rectangle around the x-axis from the volume you get by rotating the curve $y=-x^2+2x+1$ around that same x-axis. Here's a picture of what's going on:
Let me know if something's not right or if you don't understand something.