We can "slice and dice" the region under consideration two ways:
Shell method: in this method we are going to integrate over $x$ using "very thin shells" (cylinders, approximately).
The center of each shell will be at $x = 2$, and the radius will vary from 2 (when $x = 0$) to 1 (when $x = 1$), because the radius of each shell is:
$r(x) = 2 - x$.
The surface area of each shell is:
$2\pi r(x)f(x)$
where $f(x) = x - x^2$, so our volume element that we integrate over will be:
$dV = 2\pi r(x)f(x)dx = 2\pi (2-x)(x - x^2)dx$
yielding the integral:
$\displaystyle V = 2\pi\int_0^1 (2x - 3x^2 + x^3)\ dx$
The solid of revolution itself looks like the "top half" of a doughnut, with a "hole" in the middle that extends from $x = 1$ to $x = 3$. We *could* integrate from 0 to 2, but after $x = 1$ we are "in the hole" and there's no additional volume to add, because we are actually integrating using the function:
$g(x) = \max(f(x),0)$
(past $x = 1$ the parabola is below the $x$-axis, and "outside the boundary").
Disk method: in this method we are actually integrating over $y$ (slicing parallel to the $x$-axis). This isn't as pretty, because now we have two radii, and "getting the radii right" is more of a chore.
So let's look at the parabola $y = x - x^2$ as if $y$ is the independent variable.
Solving for $x$ in terms of $y$ we get:
$x = \frac{1}{2} \pm \sqrt{\frac{1}{4} - y}$
Hopefully it's clear that the positive square root corresponds to the "top" branch of the parabola (when viewed with the $y$-axis horizontally), and the negative square root is the "lower" branch" (we have to consider these two branches, because otherwise we do not get a function of $y$).
The "lower branch" corresponds to the OUTER radius of our disk (well, we actually get a "washer" shape, not a disk, that is to say: an annulus). The "upper branch" corresponds to the inner radius.
The area of each washer is going to be:
$\pi(R(y)^2 - r(y)^2)$
where:
$R(y) = 2 - \left(\frac{1}{2} - \sqrt{\frac{1}{4} - y}\right) = \frac{3}{2} + \sqrt{\frac{1}{4} - y}$
and:
$r(y) = 2 - \left(\frac{1}{2} + \sqrt{\frac{1}{4} - y}\right) = \frac{3}{2} - \sqrt{\frac{1}{4} - y}$
so our volume element in this case will be:
$dV = \pi(R(y)^2 - r(y)^2)dy = 6\pi\sqrt{\frac{1}{4} - y}\ dy$
leaving us with the somewhat ugly integral to calculate:
$\displaystyle V = 6\pi\int_0^{1/4} \sqrt{\frac{1}{4} - y}\ dy$
If you get the same answer both ways, it's likely it is correct :)
