geometrysolid-geometry
This was just a thought I had while driving this morning, no particular application.
If you make a straight line cut through a solid object such that the resulting two pieces have the same volume, will they have the same surface area? Vice versa?
What about applications to certain shapes? Regular prisms? Logically it seems that it would be true for a sphere considering there's only one way to cut it in half to achieve the same volume.
Note – I'm only considering uniform densities
I struggled with this one, but it finally became clear. I hope this answer helps someone else along the way.
The comment around the original MathForum question that suggested that one could calculate the volume of the kinds of solids described above by taking the hyperarea of the base and multiplying it by the average of the heights, while technically on the mark, its wording sent me down the wrong road.
To find the volume of these kinds of solids one needs to multiply the volume formula for a regular simplex (also shown in the original post):
times the average of the heights of the cap.
Calculating a solid with 3 heights to the cap will need a 2 dimensional regular simplex as a base (an equilateral triangle). In this case the formula for the volume of a regular simplex will give the triangle's area, hence the confusion about hyperarea.
When calculating for heights of n dimensions, the base simplex will have n-1 dimensions.
So, a little Mathematica function will do the trick:
volSimplexWithHyperCap[heights_List] := Module[{n, simplexVolume},
n = Length[heights] - 1;
simplexVolume = Sqrt[n + 1]/(n! * Sqrt[2^(n)]);
N[simplexVolume * Mean[heights]]]
volSimplexWithHyperCap[3, {1, 1, 1}]
0.433013
This corresponds correctly to the prism volume in the question. It also, readily extends to higher dimensions:
volSimplexWithHyperCap[{1, 3, 2, 3, 5}]
0.0652186
This has the advantage over integration of being relatively simple and very fast to calculate. Not a Cholesky decomposition solution like I imagined at first, but it looks serviceable.
Maybe @MvG will think of a proof.
Thanks to everyone for their patients with all the edits.
If the angle is very small, then the area is simply $r^2 d\Omega/\cos\alpha$, where $r$ is the distance $HA$ from the vertex to the surface and $\alpha=\angle HAO$ is the angle between $HA$ and the normal to the surface $OA$.
Best Answer
Join two solids of equal volume by a small (narrow and short) tube - e.g. a sphere and a cube. Cut the tube in half.
The volumes will be equal. The surface area on each side will be very nearly the surface area of the original component solids. Since the surface area does not determine the volume (largest volume for given surface area is s sphere), there is no reason for the two surface areas to be the same.
Once you see this you will realise that equality will only occur in special cases - eg cutting along an axis of symmetry; or by mimicking the proof of the ham sandwich theorem (taking a plane along which to cut the volume in half, and rotating it continuously, and showing by using the intermediate value theorem that there is some plane which will give equal surface areas too).