A cuboid measures 10×15×30. Find the volume and the area of the surface of revolution of the cuboid about its space diagonal.
Here is a brute force computer idea (not implemented yet):
First, we do not revolve the box, but we will tip it.
The space diagonal, sd, has length L = sqrt (10^2 + 15^2 + 30^2) = 35. (That was the easy part).
One notes that if we fix one corner of the box to the origin and make sd the z-axis and then cut x,y planes perpendicular to this axis (z=const, where const goes from 0 to L) the planes intersect the surface of the box as triangles (near the top and bottom) and as rectangles in the middle.
The sd intersection point is never at the center of these triangles or rectangles but moves linearly as their sizes change with z. If one found the maximum distance from the intersection to the side of each triangle or rectangle as a function of z, this distance would serve as a radius of revolution at each z. The box's volume of revolution would be the integral of these disks (pi r(z)^2) over z.
Rather than going through the great pain of calculating r(z), we estimate it instead.
We start with the box untipped. (Base: A B C D, top: a b c d, and vertical edges Aa Bb Cc Dd, with its base flat on the z=0 plane with corner A, one end of the diagonal, fixed to the origin. AB runs along the x-axis and AD along the y-axis. With Euler angles we rotate the box first around the x-axis (so that c' d' is directly above A B (i.e. in the y=0 plane)
and then around the y-axis so that c"
is now above A. The diagonal sd = A"c" runs along the z-axis. Here, the primes represent coordinates after rotation. Note, A = A' = A", since this point is fixed.
We record a dense sampling of the initial (x,y,z) points on the surfaces of the box, and then prepare the transformed (x",y",z") points after the two rotations. We bin the new points in fine steps of z" and look for the maximum r" = sqrt(x"^2 + y"2) within each z" bin. This approximates the r"(z")
function we originally sought. We sum pi r"^2 delta_z" to obtain the volume of revolution V.
The surface area would likewise come from the integral 2 pi r" delta_z"
It's quick and dirty but might be fairly easy to accomplish.
Note, there should be symmetry in r" about its center, z" = 15/2 .
Edited on August 28, 2020, 12:24 pm
A computer approximation idea
