Find volume of solid bounded by $y=1-x$ and $ y = 2x - 2x^2$ and revolving around $x=\frac12$ The problem is:
Find the volume of the solid generated by revolving the region bounded by $y=1-x$ and $y = 2 x - 2 x^2$ around the vertical line $x=\frac{1}{2}$?
This is new to me, I tried it out, but I keep getting different answers when I check my work.

Note that $y_1=1-x$ and $y_2=2x-2x^2$ intersect at $x=\frac12$ and $x=1$. It is more convenient to use the cylinder method, integrating over $\frac12 < x< 1$,
$$V = \int_{1/2}^1 2\pi r(y_2-y_1)dx$$
$$= 2\pi\int_{1/2}^1 (x-\frac12)[(2x-2x^2)-(1-x)]dx$$
$$= 2\pi\int_{1/2}^1 (x-\frac12)(-2x^2+3x-1)dx=\frac{\pi}{48}$$
