How to parameterize a hyperboloid in a solid of revolution

Question

The middle “hyperboloid” part of the solid of revolution is determined entirely by a single edge of the cube that does not touch one of the axis vertices - there are six such edges. Mark these on your cube. Consider one of these edges. More importantly, consider its edge segment between $(1,0,1)$ and $(0,0,1)$ as well as a separate axis segment between the two points on the planes that the axis of rotation passes through - the edge and axis segments are skew segments (i.e. not parallel, but don’t cross either . . . a phenomenon allowed by 3-d space).

I'm wondering how to:

• Parameterize the edge and axis segments – i.e. find vector-valued functions with appropriate domain restrictions to represent these segments.

•Make a change of parameter so that the domain of each of your vector-valued functions is the same. Moreover, this domain must be the same length as the length of the axis segment.

Answer 1

Hint: There are two natural coordinates: distance along the edge and rotation angle of the cube. The "hyperboloid" is two dimensional, so that is the right number of parameters. Find a parmameterization of the edge, then find where a point goes as the cube is rotated around the body diagonal.

If the coordinates of the edge were not specified, I would initially orient the cube with a body diagonal along the $z$ axis for this problem. The corners don't have such a pretty form, but finding what happens under rotation is easier.

Answer 2

I interpret this question as follows: The cube $C:=[0,1]^3$ is rotated around the axis ${\bf a}:={1\over\sqrt{3}}(1,1,1)$ through $(0,0,0)$. Thereby the edges meeting at $(0,0,0)$, resp. at $(1,1,1)$, generate two little cones, and the other six edges generate a single surface $S$ which we are told to describe mathematically. (A priori these edges would generate $6$ surfaces, but they all coincide because of symmetry.)

So let's look at the edge $$e:\quad t\mapsto(t,0,1)\qquad(0\leq t\leq 1)\ .$$ Any point of it when rotated around ${\bf a}$ will describe a circle in a plane orthogonal to ${\bf a}$. A typical such plane $\nu_h$ is given by $$\nu_h:\qquad {x+y+z\over\sqrt{3}}= h\ ,$$ where $h$ denotes the distance of $\nu$ from the origin. This plane intersects the (extended) edge $e$ at the point $$P_h=(\sqrt{3} h-1,0,1)\ .$$ The condition $P_h\in C$ implies that the variable $h$ is a priori bounded by ${1\over\sqrt{3}}\leq h\leq{2\over\sqrt{3}}$.

On the other hand the plane $\nu_h$ intersects the axis $\langle{\bf a}\rangle$ at the point $A_h={1\over\sqrt{3}}(h,h,h)$. It follows that the radius $\rho_h$ of the circle described by $P_h$ is given by $$\rho_h^2 =|P_hA_h|^2=\bigl({2h\over\sqrt{3}}-1\Bigr)^2 +{h^2\over3}+\Bigl({h\over\sqrt{3}}-1\bigr)^2=2\Bigl(h-{\sqrt{3}\over2}\Bigr)^2+{1\over2}\ .$$ It follows that the description of $S$ in its $(\rho, h)$ meridian half-planes (imagine the $h$-axis as vertical axis in these planes) is given by $$S:\quad \rho=\rho(h)=\sqrt{2\Bigl(h-{\sqrt{3}\over2}\Bigr)^2 +{1\over2}}\qquad \Bigl({1\over\sqrt{3}}\leq h\leq{2\over\sqrt{3}}\Bigr)\ .$$ This shows that the meridian curve of $S$ is a hyperbolic arc having its apex at $\rho={1\over\sqrt{2}}$ and $h={\sqrt{3}\over2}$.

In order to obtain a parametric representation of $S$ at its place in $3$-space we need two vectors ${\bf e}_1$ and ${\bf e}_2$ completing ${\bf a}$ to an orthonormal basis. The vectors ${\bf e}_1:={1\over\sqrt{2}}(1,-1,0)$ and ${\bf e}_2:={\bf a}\times{\bf e}_1={1\over\sqrt{6}}(1,1,-2)$ serve this purpose. A parametric representation of $S$ is then given by $$S: \quad(h,\phi)\mapsto h{\bf a}+\rho(h)(\cos\phi\,{\bf e}_1+\sin\phi\,{\bf e}_2)\qquad\Bigl({1\over\sqrt{3}}\leq h\leq{2\over\sqrt{3}}, \ \phi\in{\mathbb R}/(2\pi)\Bigr)\ .$$