Take a unit cube, and place it so that one of its body diagonals lies along the $z$-axis. For symmetry, assume that the vertices of the cube are at $(0,0,\pm\sqrt{3}/2)$. Then rotate it about the z-axis, forming a solid of rotation. The resulting solid is two cones joined by a "waist".
This same solid could be formed by rotating the following piecewise-defined curve in the $xz$-plane about the $z$-axis:
- A straight line from $(0,0,\sqrt{3}/2)$ to $A$, defined as $(\sqrt{2/3}, 0, \sqrt{3}/2 - 1/\sqrt{3})$.
- A curve from $A$ to $B$, defined as defined as $(\sqrt{2/3}, 0, -\sqrt{3}/2 + 1/\sqrt{3})$.
- A straight line from $B$ to $(0, 0, -\sqrt{3}/2)$.
Question: What is the equation of the curve between $A$ and $B$ so defined?
This was a question that occurred to me recently, but I don't even have a good idea how to begin solving it. I have a vague sense that this curve can be thought of as some kind of 3D envelope of the cubes as they rotate, but since a cube is not a smooth surface, I'm not sure how to apply standard techniques.
That curve is an arc of hyperbola $2x^2-4z^2=1$, the section of a hyperboloid with equation $2x^2+2y^2-4z^2=1$.
That surface is generated by the rotation of an edge of the cube, connecting a vertex in the positive $z$ half-space (for instance $P=(1/\sqrt6,1/\sqrt2,\sqrt3/6)$) with a neighbouring vertex in the negative $z$ half-space (for instance $Q=(-1/\sqrt6,1/\sqrt2,-\sqrt3/6)$). A generic point on that edge can be written as $$ M=Pt+Q(1-t)=\left({1\over\sqrt6}(1-2t),{1\over\sqrt2},{\sqrt3\over6}(2t-1)\right) $$ and the square of its distance from $z$-axis is $$ x_M^2+y_M^2={1\over6}(1-2t)^2+{1\over2}=2z_M^2+{1\over2}. $$ Point $M$ and all its rotated images lie then on the hyperboloid given above.