Tricky 3d geometry problem

Solution 1

If we place the cube with its main diagonal from $(0,0,0)$ to $(1,1,1)$ and three edges along the axes, then we can parametrize two edges and a diagonal: $$ \begin{align} edge_1&:s\mapsto\begin{pmatrix}s\\0\\0\end{pmatrix},\quad s\in[0,1]\\ edge_2&:s\mapsto\begin{pmatrix}1\\0\\s\end{pmatrix},\quad s\in[0,1]\\ diag&:x\mapsto\frac{1}{\sqrt 3}\begin{pmatrix}x\\x\\x\end{pmatrix},\quad x\in[0,\sqrt 3] \end{align} $$ For a given $s\in[0,1]$ one can minimize the quadratic expression (just pick the vertex) $$ |diag(x)-edge_1(s)|^2 $$ with respect to $x$ to find that $s=\sqrt 3 x$ and with this the distance $f(x)$ between the point $diag(x)$ on the diagonal and the point $edge_1(s)$ on $edge_1$ is $$ f(x)=\sqrt 2 x $$ Similarly, one may deduce that for $$ |diag(x)-edge_2(s)|^2 $$ to be minimized wrt. $x$ for a fixed $s\in[0,1]$ we must have $s=\sqrt 3 x-1$ and so the distance $g(x)$ between the diagonal and $edge_2$ is $$ g(x)=\sqrt{2(x^2-\sqrt 3x+1)} $$ By symmetry, we may conclude that the curve we are rotating is $$ h(x)= \begin{cases} \sqrt 2 x&\text{ for }x\leq\tfrac13\sqrt 3\\ -\sqrt 2(x-\sqrt 3)&\text{ for }x\geq \tfrac23\sqrt 3\\ \sqrt{2(x^2-\sqrt 3x+1)}&\text{ in between} \end{cases} $$ defined on the domain $x\in[0,\sqrt 3]$ which is illustrated here:

Remark: Fixing $s$ and varying $x$ fixes a point on an edge and varies a point on the diagonal until the nearest point is found. Doing it the other way around would result in a wrong construction of fixing a point on the diagonal and finding the nearest point on the given edge, which minimizes distance orthogonal to an edge instead of orthogonal to the diagonal/axis of rotation.

To demonstrate how it fits, here is an overlay in a dynamic 3D-model of it:

The red curve is the function $h(x)$ derived above corresponding to the "union" case of the solid formed by the uncountable union of all positions of a full rotation of the cube. The purple lines describe the "intersection" case, the uncountable intersection of all positions in a full rotation of the cube.

Solution 2

Let's take the unit centered cube, with vertexes at $\pm 1$. To rotate it so that its main diagonal gets aligned with the $x$ axis (vertical axis in the figure) we can use two rotations along two axes, the first by 45 degrees, the second by $\tan^{-1}(\sqrt{1/2})=\sin^{-1}(\sqrt{1/3})$. We get then the rotation matrix:

$$ Q= \begin{pmatrix} \sqrt{\frac{1}{3}} & 0 & \sqrt{\frac{2}{3}} \\ 0 & 1 & 0 \\ -\sqrt{\frac{2}{3}} & 0 & \sqrt{\frac{1}{3}} \\ \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & \sqrt{\frac{1}{2}} & -\sqrt{\frac{1}{2}} \\ 0 & \sqrt{\frac{1}{2}} & \sqrt{\frac{1}{2}} \\ \end{pmatrix}= \begin{pmatrix} \sqrt{\frac{1}{3}} & \sqrt{\frac{1}{3}} & \sqrt{\frac{1}{3}} \\ 0 & \sqrt{\frac{1}{2}} & -\sqrt{\frac{1}{2}} \\ -2\sqrt{\frac{1}{6}} & \sqrt{\frac{1}{6}} & \sqrt{\frac{1}{6}} \\ \end{pmatrix}$$

Indeed, we can verify that the matrix is orthogonal and $Q \, (1, 1,1)'=(\sqrt{3} ,0, 0)'$

Lets consider first the upper part. That corresponds to the points spanned by the edges starting on the upper vertex, hence they correspond to the revolution of:

$$ \begin{pmatrix} x \\y \\ z \end{pmatrix} = Q \begin{pmatrix} 1 \\\alpha \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{3}}(\alpha+2) \\ \frac{1}{\sqrt{2}}(\alpha-1)\\ \frac{1}{\sqrt{6}}(\alpha-1) \end{pmatrix} $$

Then $\alpha=\sqrt{3} x-2$, in the range $\alpha \in[-1,1]$, or $x\in[1/\sqrt{3},\sqrt{3}]$.

The rotations along the $x$ axis will keep $r^2=y^2+z^2$ constant, and so

$$ r^2= \frac{2}{3}(\alpha-1)^2=\frac{2}{3}(3-\sqrt{3}x)^2$$

Or $$r = \sqrt{6}\left(1-\frac{x}{\sqrt{3}}\right)$$

For the next part, we consider another edge, starting from a neighbour vertex, say from $(1,1,-1)'$:

$$ \begin{pmatrix} x \\y \\ z \end{pmatrix} = Q \begin{pmatrix} 1 \\\alpha \\ -1 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{3}}\alpha \\ \frac{1}{\sqrt{2}}(\alpha+1)\\ \frac{1}{\sqrt{6}}(\alpha-3) \end{pmatrix} $$ with $\alpha=\sqrt{3} x$, in the range $\alpha \in[-1,1]$, or $x\in[-1/\sqrt{3},1/\sqrt{3}]$.

And here the radius is

$$ r^2= y^2+z^2=\frac{2}{3}\alpha^2+2=2x^2+2$$

Then the radius is

$$ r(x)=\begin{cases} \sqrt{6}\left(1-\frac{x}{\sqrt{3}}\right) & \mbox{if } 1/\sqrt{3} \le x \le \sqrt{3}\\ \sqrt{2x^2+2} & \mbox{if } 0 \le x \le \sqrt{1/3} \end{cases}$$

(This seems to agree with String's answer.)

So, yes the middle cross section is an hyperbola.

To compute the total volume you need to integrate : $V = 2 \int_0^\sqrt{3} \pi r(x)^2 dx $ and scale the result by multiplying it by $(L/2)^3$ (because our cube has edge length $2$, instead of $L$)

Solution 3

Hint:

This rotation-surface is called Katenoid.

It's parametrization is given by

$$x(u,v)=(a \cos (u) \cosh (v),a \sin (u) \cosh (v),a v)$$

To calculate the volume, use the function:

$$f(x)=a \cosh \left(\frac{x-A}{a}\right)+B$$.

Rotate it around the x-Axis.

Then a formulae for volume is given by:

$$V=\pi \int _a^b\ f(x)^2{dx}$$

Please, look at my comment again. Here is another rotation-surface rotation-hyperboloid.

$$x(\text{s},\text{v})\text{=}(\cos (s)-v \sin (s),v \cos (s)+\sin (s),v)$$

This one can be generated by straight-lines:

I don't know, which surface to use.

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Updated on March 18, 2020