For instance, consider the following case:
The region of the $xy$-plane bounded by the $x$-axis, the line $x = 2$, and the graph of $y = x$ is rotated about the line $x = 4$ to construct a solid $S$. Applying the shells method, should I take $4-x$ or $x-4$ as the radius of my cylinder?
Do we, in Layman terms, base the $x$ values on the values of $x$ that we "have" before the revolution is applied, .ie. the left triangle below.
i.e. take $r=4-x$ because $4 >$ values of $x\in[0,1]$ on the cartesian plane:
Whereas taking $r=x-4$ implies that we are basing our values on the "right triangle below" (which we don't yet have) before the triangle area is rotated about $x=4$.
I apologise for using informal terms to explain my thoughts as I am not sure how best to describe this situation formally.
Well, if we go by your figure, it's clear that $x$ is a number between $0$ and $2$, so if we want the shell's radius to be positive, we would quantify this as $r = 4-x$. So a representative shell would have differential volume
$$dV = 2 \pi (4-x) x \, dx, \quad x \in [0,2]$$ and the total volume is
$$V = \int_{x=0}^2 2 \pi (4-x)x \, dx.$$
Now suppose we choose $x$ in such a way that we are on the "right-hand" side of the axis of rotation $x = 4$. Then $x \in [6, 8]$, and then the shell's radius would be $r = x-4$ because $x > 4$. But we now have to be very careful about the height of the cylinder: it is not simply $x$ as in the previous case, but actually $8 - x$. This is because when $x = 8$, you can see in your diagram that the height is $0$. And when $x = 6$, the cylinder has maximal height of $2$, just as in the case when $x = 2$.
Then the differential volume would be $$dV = 2 \pi (x-4) (8-x) \, dx, \quad x \in [6,8],$$ and the total volume is $$V = \int_{x=6}^8 2 \pi (x-4)(8-x) \, dx.$$
You can evaluate both integrals to verify they are equivalent.
You can also compute the volume using washers by integrating with respect to $y$. For a given height $y \in [0,2]$, the differential volume element is an annulus with inner radius $r_i = 4-2 = 2$, and outer radius $r_o = 4-y$. Therefore, its volume is $$dV = \pi(r_o^2 - r_i^2) \, dy = \pi((4-y)^2 - 2^2) \, dy, \quad y \in [0,2],$$ and the total volume is $$V = \int_{y=0}^2 \pi ((4-y)^2 - 2^2) \, dy.$$