Question about a rotating cube?

Solution 1:

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 $\dfrac{\pi}{\sqrt{3}}$.

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The code for the animation above can be found here.

Solution 2:

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.