Comments on: Green’s Theorem and Area of Polygons

By: Thomas Klimpel

Thomas Klimpel — Fri, 20 Jun 2014 20:23:25 +0000

The list of vertices of a polygon can also be seen as the set of oriented boundary elements (=edge segments) of the polygon. So the analogy with the 3D polyhedron would be to compute the volume based on the set of oriented boundary elements (=faces) of the 3D polyhedron. But I think one has to use Gauss' theorem (sometimes also called Divergence theorem or Ostrogradsky's theorem), not the Kelvin–Stokes theorem.

By: Jyrki Lahtonen

Jyrki Lahtonen — Sun, 08 Jun 2014 08:46:41 +0000

The problem of computing the volume of a 3D polyhedron given a list of its vertices as input is underspecified. Consider the following example. Start with a cube (8 vertices). Add a ninth vertex somewhere inside the cube. Remove from the cube a pyramid shaped part that has one of the faces of the original cube as its base and the freshly added ninth vertex as its peak. Depending on the choice of the face you get six different non-convex polyhedra, possibly all with distinct volumes (with a generically placed ninth vertex).

To describe the polyhedron you need to give its faces. Vertices alone won't be enough.

By: anorton

anorton — Sat, 07 Jun 2014 17:34:56 +0000

I considered this, but didn't make much progress. I believe that some similar formula for volume could be deduced through an application of Stoke's Theorem, but I did not work on that long enough to produce anything of note.

By: Norbert

Norbert — Sat, 07 Jun 2014 09:02:00 +0000

There is no natural order on the set of vertices of $n$ dimensional polyhedra. I think there is no good answer here.

By: Interested Reader

Interested Reader — Sat, 07 Jun 2014 05:04:59 +0000

So, how does this generalize to 3 or 4 dimensional polyhedra?

By: anorton

anorton — Fri, 06 Jun 2014 15:35:21 +0000

The first point and the last point are the same. That, is $(x_0, y_0) = (x_n, y_n)$, So, although there are $n$ vertices, there are $n+1$ terms in the summation.

You are correct regarding the typo in the last term of the summation, though--I'll change that.

By: Joel Reyes Noche

Joel Reyes Noche — Fri, 06 Jun 2014 03:09:21 +0000

Also, in your example, shouldn't the last term in the sum be (0+0)(0-2)/2?

By: Joel Reyes Noche

Joel Reyes Noche — Fri, 06 Jun 2014 03:05:03 +0000

You say that there are $n$ vertices and your $k$ is from $0$ to $n$, so wouldn't that involve $n+1$ vertices? Shouldn't the first vertex be called $(x_1,y_1)$ and $k$ be from $1$ to $n$?

By: Steven Stadnicki

Steven Stadnicki — Wed, 04 Jun 2014 23:56:34 +0000

Not only can this formula be used to find areas, centroids, etc., but it regularly is; I've used it myself for determining the "best" location to place a letter in an arbitrarily-shaped polygon.