Suppose we have region $\mathcal{R}$ that is bounded by the parabolas $x^2$ and $2-x^2$. If we rotate $\mathcal{R}$ about line $x=1$, what is the volume?
Well, the parabolas intersect at $x= \pm 1$. I would use shell-method to find the volume of an slice to be
$$ dV = 2 \pi \underbrace{(x-1)}_{radius} \cdot \underbrace{ ( 2 - x^2 - x^2 ) }_{hieght}d x $$
Thus,
$$ \boxed{ V = \int\limits_{-1}^1 2 \pi (x-1)(2-2x^2) dx }$$
is this correct?
The region and the line can be visualized as follows:
The way to think about it is, if we have a shell formed from rotating a small rectangle at $x$ of width $dx$ and height $h(x)$, its volume is $$ dV = 2\pi \cdot r(x) \cdot h(x) \,dx = 2\pi (1-x)(2-2x^2) \, dx $$ Therefore \begin{align} V &= 4\pi \int_{-1}^1 (1-x)(1-x^2) \, dx \\ &= 4\pi \int_{-1}^1 (1 - x - x^2 + x^3) \, dx \\ &= 4\pi \int_{-1}^1 (1 - x^2) \, dx \\ &= 8\pi \int_{0}^1 (1 - x^2) \, dx \\ &= 8\pi \left(1 - \frac{1}{3} \right) \\ &= \frac{16\pi}{3} \end{align}
Note some of the integration simplification is due to odd/even functions over an interval symmetric about the origin.
Everything is correct besides the radius, it should be the other way around(i.e $1-x$) as the axis of revolution is to the right of the region $R$. A quick check would be to evaluate the radius at the left endpoint of $R$ which is $1-(-1)=2$, while at the midpoint $x=0$, we get $1-0=1$ which makes sense.