For a region R bounded above by the curve $y = e^{-x^2}$, below by the curve $y = x^2 - 1$, on the left by the curve $x = -1$, and on the right by $x = 1$, that is rotated around the vertical line $x=-3$
https://www.desmos.com/calculator/ce5rmfcejs
The volume is: $v=\int2\pi r h dr\Rightarrow\int_{-1}^{1} 2\pi (3+x)[(e^{-x^2})-(x^2-1)]dx$
Why is the radius of the shell $x+3$ and not $x+2$ since the region is 2 away from the axis of rotation?
The simple answer is that the radius of the shell is determined by the relationship to the axis of rotation to the interval on which the bounding functions are evaluated. Since the axis of rotation is $x = -3$, then for a point $(x, f(x))$ for some function $f$, the radius is $x + 3$, for $x \in [-1,1]$. So when $x = -1$, that corresponds to the innermost radius, which is $-1 + 3 = 2$, as you expect.
Just for fun: I have a little animation of this region:
