Suppose we took a solid unit cube and balanced it on a vertex. Next, we spin the cube really quickly about the vertical axis, producing a solid of revolution:
Clearly, the top and bottom are two identical cones produced by revolving a line segment passing through the axis of rotation. What about the thing between them? It is a quadric surface (three-dimensional conic) called a hyperboloid of one sheet. Alas, this name invariably reminds me of a television commercial for a particular brand of kitchen towels:
The hyperboloid of one sheet is a doubly ruled surface, meaning that at each point we can find two straight lines drawn on the surface of the hyperboloid which pass through the point*. The only other doubly ruled surfaces are the plane and hyperbolic paraboloid.
* Proof: The hyperboloid is the surface swept out when one of the edges of the cube is revolved about the axis, so must be at least singly ruled. However, the hyperboloid possesses bilateral symmetry, so we can find a second family of lines by reflecting the first family.
One may ask what the volume of the revolved cube is. After some pretty easy stuff involving Pythagoras and vectors, we obtain the various dimensions of the cross-section of the revolved cube:
The two cones each have volume (2 sqrt(3)/27)π, and the hyperboloid has the Cartesian equation 2x²+ 2y² – 4z² = 1, where the centre of the cube is the origin. Basic integration gives a volume of (5 sqrt(3)/27)π for the hyperboloid, so the entire solid has volume π/sqrt(3). It is interesting to note that this total volume, which was obtained in a very piecewise manner, is simpler than the volumes of its conical and hyperbolic components.