Rotating the hyperboloid of two sheets $z^2-y^2-x^2=1$

Question

Revolving $f(x)=\frac{1}{x}$ about $f(x)=x$ yields the two-sheeted hyperboloid $2xy-2=z^2$ in $\Bbb R^3.$

I'm trying to rotate the surface so that the forward sheet lies in the first octant $(+,+,+)$ and the backward sheet lies in $(-,-,-).$

What's the equation?

Here's what I have so far. This is the graph of $2xy-2=z^2:$

It would also work to rotate the two-sheeted hyperboloid $z^2-y^2-x^2=1.$

My attempt was to use a $3$d rotation matrix that rotates points about the origin. I was able to find the right rotation matrices, but I could not reason what to multiply the rotation matrix with to get the desired result. I could probably get the final equation, with a few hints.

I believe I have to change the entries in the matrix $Q$ (as shown in the hint below by @NinadMunshi). I'm not sure what to change them to, but the new entries should rotate the hyperboloid.

Answer 1

You want to rotate the surface so that the direction $(1,0,0)$ becomes $\dfrac1{\sqrt3}(1,1,1)$. A possible rotation matrix is

$$\begin{pmatrix}\frac1{\sqrt3}&0&-\frac2{\sqrt6} \\\frac1{\sqrt3}&\frac1{\sqrt2}&\frac1{\sqrt6} \\\frac1{\sqrt3}&-\frac1{\sqrt2}&\frac1{\sqrt6}\end{pmatrix}$$

Now your equation is

$$\frac{(u+v+w)^2}3-\frac{(v-w)^2}2-\frac{(-2u+v+w)^2}6=1$$

or

$$u^2+v^2+w^2-4uv-4vw-4wu+3=0.$$

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Caution: sign and permutations errors are guaranteed, I didn't check. But the final result must not be far from correct.

https://www.wolframalpha.com/input/?i=plot+x%5E2%2By%5E2%2Bz%5E2-4xy-4yz-4zx%2B3%3D0

Answer 2

$\textbf{Hint:}$ The equation can be rewritten as

$$z^2-x^2-y^2 = 1 \implies x^TQx = 1$$

where

$$Q = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\\ \end{pmatrix}$$

Answer 3

I'm trying to rotate the surface so that the forward sheet lies in the first octant $(+,+,+)$ and the backward sheet lies in $(-,-,-).$

Let $O=(0,0,0)$ be the origin of the space $\Bbb R^3$.

By the construction, the first hyperboloid has a rotation axis $\ell_1=\Bbb R (1,1,0)$ so it is contained in its the two-sided asymptotic cone $C_1$ based at $O$, with an axis $\ell_1$ and an angle $\pi/2$ at $O$. The cone $C$ covers $\tfrac {2-\sqrt{2}}{2}$ of an area of a unit sphere $S$ centered at $O$. On the other hand, two octancts cover only $1/4$ of an area of $S$, so a required rotation with a base at $O$ is impossible.

It is easy to see that the second hyperboloid has a rotation axis $\ell_2=\Bbb R (0,0,1)$ and it is contained in its the two-sided asymptotic cone $C_2$ based at $O$, with an axis $\ell_2$ and an angle $\pi/2$ at $O$. Similarly to the first case we see that a required rotation with a base at $O$ is impossible.