Find the volume of the region where a line segment can move I and my friends enjoy making and solving new mathematical questions. We made the following question, but we are facing difficulty. Could you show me how to solve this?
Question: Let $a, b$ be constants which satisfy $a\gt b\gt 0$. In the three dimensional space, there exist two lines $L, M$ which are positioned in skew relation. Suppose that if you take $P$ on $L$, $Q$ on $M$, then the minimum of the length of the line segment $PQ$ is $b$. When you move $P, Q$ such that $PQ\le a$, find the volume of the rigion where the line segment $PQ$ can move.
 A: A beginning:
Assume that $L$ is the $x$-axis and that $M$ is the line $$y\mapsto(my,y,b)\qquad(-\infty <y<\infty)$$
for some fixed $m\in{\mathbb R}$.
Given a sample point $X=(\xi,\eta,\zeta)\in{\mathbb R}^3$, $\>0<\zeta<b$,  there is a unique line $g_X$ through $X$ that meets both $L$ and $M$ in two points $X_L$ and $X_M$. We have to find conditions on the coordinates of $X$ that guarantee $|X_L-X_M|^2\leq a^2$.
In order to obtain $g_X$ we argue as follows: The point $X$ together with  $L$ (the $x$-axis) determines a plane with equation $\zeta y-\eta z=0$. We obtain the point $X_M$ by intersecting this plane with $M$. Computation gives
$$X_M=\left({mb\eta\over\zeta},{b\eta\over\zeta}, b\right)\ .$$
Now we know the two points $X$ and $X_M$ of $g_X$. Therefore $g_X$ has the parametric representation
$$g_X:\quad t\mapsto(1-t)X_M+tX\qquad(-\infty <t<\infty)\ .$$
Intersecting $g_X$ with the line $L$ (the $x$-axis) gives $t={b\over b-\zeta}$ and finally produces the point
$$X_L=\left({b(\xi-m\eta)\over b-\zeta},0,0\right)\ .$$
The condition $|X_L-X_M|^2\leq a^2$ now leads to
$$\left({mb\eta\over\zeta}-{b(\xi-m\eta)\over b-\zeta}\right)^2+\left({b\eta\over\zeta}\right)^2+b^2\leq a^2\ .$$
It seems that we obtain an admissible $(\xi,\eta,\zeta)$-domain $\Omega$ which is bounded by a fourth degree surface.
As Rahul Narain has remarked a plane $\zeta={\rm const.}$ intersects $\Omega$ in an ellipse $E_\zeta$ with an equation of the form $A\xi^2+2B\xi\eta+C\eta^2=R^2$. Using the formula for the area of such an ellipse we get
$${\rm area}(E_\zeta)=\pi\>{(a^2-b^2)\zeta(b-\zeta)\over b^2}\quad(0<\zeta< b)\ ,$$
and after integration we finally obtain
$${\rm vol}(\Omega)={\pi\over6}(a^2-b^2) b\ .$$
