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.

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.
