Consider two cylinders $C_{s}$ (solid) and $C_{h}$ (hollow):
Cylinder $C_{s}$ has a radius $r_{s}$, and has the points $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ as the centers of its bases.
Cylinder $C_{h}$ has a radius $r_{h}$, and has the points $(x_{3},y_{3},z_{3})$ and $(x_{4},y_{4},z_{4})$ as the centers of its bases (no solid bases).
The hollow cylinder, $C_{h}$, acts as a cutter, where it moves a distance $d$ in the direction of the outward-pointing normal at either $(x_{3},y_{3},z_{3})$ or $(x_{4},y_{4},z_{4})$, whichever specified. So the axis is fixed.
In some cases, the cutter will never cross the solid cylinder, regardless of $d$.
In some cases, the cutter will cross the solid cylinder, but will not form a complete cut. Say when $d$ is not enough.
In some cases, when $d$ is large enough, and the axes are parallel and the base of $C_{h}$ will entirely cross the base of $C_{s}$. In this case we will have cut piece that is "exactly" cylindrical in shape.
Also there are other cases, say when the axes are perpendicular and $r_{s}>r_{h}$, so we will have a cut that is "not exactly" cylindrical in shape (it has curvy bases).
Furthermore, when the axes are neither parallel nor perpendicular, then the resultant piece may or may not have a part of the base of $C_{s}$ depending upon the direction of the cutter.
Keeping in mind the cases when $r_{s}>r_{h}$ and when $r_{s}<r_{h}$.
And there are many other cases are there!
I wish I can do some illustrations like here (NOT EXACTLY SAME). I am having two cylinders.
The simplest case when the cylinders do not cross each other, the volume is $0$.
For uncomplete cut, the volume should be considered as $0$.
Given $r_{s}, r_{h}, (x_{1},y_{1},z_{1}) ,(x_{2},y_{2},z_{2}) ,(x_{3},y_{3},z_{3}) ,(x_{4},y_{4},z_{4}), d$ and the direction, can we evaluate the volume of the cut piece, if any, without specifying the case?
Is there any explicit formula for this?