# Question about a rotating cube?

I read an article not too long ago that posed the following problem:

What is the volume of the solid of revolution created by spinning a unit cube about an axis joining two opposing vertices?

So the shape generated will be two cones and a parabola-like curve in the "middle". I hope that makes sense. At first, I tried to find a cross section of the resulting structure and then integrating it with disks, but I think I am over-complicating it. How would I go about solving this problem if I define my unit cube to be on the first octant with $i, j,$ and $k$ (so the axis would be $r(t)=t\langle1,1,1\rangle,0 < t <1$)? Thank you.

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Wouldn't it just be 2 cones of equal size? – Graphth Mar 2 '12 at 19:59
I think you should place the solid so that the axis of revolution coincides with the $x$-axis, and decompose it into the three parts: the two cones and the middle section. The volume of the two cones is straightforward. As for the middle section, let $f(x)$ be the function describing its shape. Then the volume is $\pi \int_a^b f^2(x)dx$. – Manolito Pérez Mar 2 '12 at 20:18

## 2 Answers

Suppose the cube has vertices $\{0,1\}^3$ and the axis of revolution is from $(0,0,0)$ to $(1,1,1)$. The axis has length $\sqrt{3}$.

Using dot products, we get that three of the vertices are on a plane perpendicular to the axis at a distance of $\dfrac{2}{\sqrt{3}}$ from $(0,0,0)$ and the other three vertices are on a plane perpendicular to the axis at a distance of $\dfrac{1}{\sqrt{3}}$ from $(0,0,0)$.

Using cross products we get that each of these six vertices are at a distance of $\dfrac{\sqrt{2}}{\sqrt{3}}$ from the axis.

The vertices in each of these planes form an equilateral triangle centered on the axis and are rotated at an angle of $\dfrac{\pi}{3}$ from each other.

The lines from the ends of the axis to the vertices closest to them sweep out a cone that is $\dfrac{1}{\sqrt{3}}$ high and has a base radius of $\dfrac{\sqrt{2}}{\sqrt{3}}$. The total volume of these two cones is $$2\cdot\frac13\pi\frac{1}{\sqrt{3}}\left(\frac{\sqrt{2}}{\sqrt{3}}\right)^2=\frac{4\pi\sqrt{3}}{27}\tag{1}$$ The middle section is a bit trickier. Take a cylinder of height $1$ and radius $1$ with a line connecting corresponding points in the top and bottom. Give the top a twist of $\alpha$ with respect to the bottom (keeping the top and bottom at the same distance from each other). Projecting this line onto a plane containing the axis of the cylinder and rotating the cylinder yields the line $$y=(1-x)\cos(\theta)+x\cos(\alpha-\theta)\tag{2}$$ where x is the distance along the axis from the bottom ($x=0$) to the top ($x=1$) and y is the distance from the axis. The envelope of this family of lines is the hyperbola $$y^2=\sin^2(\alpha/2)(2x-1)^2+\cos^2(\alpha/2)\tag{3}$$ The volume of the hyperboloid of revolution is pretty simple to compute \begin{align} \int_0^1\pi y^2\mathrm{d}x &=\int_0^1\pi\left(\sin^2(\alpha/2)(2x-1)^2+\cos^2(\alpha/2)\right)\mathrm{d}x\\ &=\pi\left(\sin^2(\alpha/2)\frac12\int_{-1}^1t^2\mathrm{d}t+\cos^2(\alpha/2)\right)\\ &=\pi\left(\frac13\sin^2(\alpha/2)+\cos^2(\alpha/2)\right)\\ &=\pi\frac{2+\cos(\alpha)}{3}\tag{4} \end{align} Scaling $(4)$ for a general height and radius yields $$V=\pi r^2h\frac{2+\cos(\alpha)}{3}\tag{5}$$ In our case, $\alpha=\dfrac{\pi}{3}$ yielding $\dfrac{2+\cos(\alpha)}{3}=\dfrac56$. Therefore, the volume of the middle section is $$\pi\left(\frac{\sqrt{2}}{\sqrt{3}}\right)^2\frac{1}{\sqrt{3}}\cdot\frac56=\frac{5\pi\sqrt{3}}{27}\tag{6}$$ Adding the volumes in $(1)$ and $(6)$ we get the total volume to be $\dfrac{\pi}{\sqrt{3}}$.

$\hskip{4.5cm}$

The code for the animation above can be found here.

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That animation is just awesome. – Pedro Tamaroff Mar 4 '12 at 18:15
@robjohn: The idea of "envelopes" is one I haven't studied but I can see that it is worthy of study. I actually found the volume of the middle "part" of the spinning cube (the surface of revolution of the resulting hyperbola) by parameterizing the lines in question and defining the magnitude of the difference between the lines as the cross section of my surface. Your solution with envelopes is more elegant though, I think. Very nice animation. :) – Hautdesert Mar 8 '12 at 1:00

The "parabola-like curve" is actually not a parabola, but a hyperboloid.

Specifically, it is the curve traced by a line segment (any edge of the cube that doesn't end in either of the point on the axis) rotating about another line (the rotation axis).

For simplicity, let's say that we have a line of the form $x = a$, $y = bz$ and that we're rotating it around the $z$-axis. Then the surface swept by the line as it rotates is $x^2 + y^2 = r(z)^2$, where $r(z) = \sqrt{a^2 + (bz)^2}$ is the distance of the line from the $z$-axis in the $x,y$ plane at a given value of $z$. So, overall, the surface is given by the equation $$x^2 + y^2 = a^2 + (bz)^2$$ or, equivalently, $$\frac{1}{a^2}x^2 + \frac{1}{a^2}y^2 - \frac{b^2}{a^2}z^2 = 1$$ which defines a circular hyperboloid of one sheet.

Now all you need to do is determine $a$ and $b$ for one of the middle edges of your cube, and you're all set.

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