Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Faces, Vertices, and Edges of Cylinders, Cones, and Spheres

Date: 12/28/2003 at 17:21:33
From: Cara
Subject: Characteristics of polyhedra

I need to know how many faces, vertices, and edges do cylinders,
cones, and spheres have?  Logically I would say that a sphere has 1
face, 0 vertices and 0 edges.  Problem:  a face is flat, sphere is not
flat.  Secondly this does not satisfy Euler's formula v - e + f = 2. 
I would say a cone has 2 faces, 1 edge, and 1 vertex.  Problem:  while
this does satisfy Euler, it does not satisfy the definitions.



Date: 12/28/2003 at 20:41:45
From: Doctor Peterson
Subject: Re: Characteristics of polyhedra

Hi, Cara.

To start, take a look at this page:

  Cone, Cylinder Edges?
    http://mathforum.org/library/drmath/sets/select/dm_cone_edge.html 

Properly speaking, Euler's formula does not apply to a surface, but 
to a network on a surface, which must meet certain criteria.  The 
"natural" faces and edges for these surfaces, or those determined by 
applying the definitions used for polyhedra, do not meet these 
criteria.

Just taking the natural parts of a cone, as you say, it has one 
presumed vertex, the apex; one edge, the circle at the base; and two 
faces, one flat and one curved.  (I say "presumed" because the apex is 
not really a vertex in the usual sense of a place where two or more 
edges meet, but it is a point that stands out.)  This gives

  1 - 1 + 2 = 2

So it does fit the formula; but there is no reason it should, really, 
because it doesn't fit the requirements for the theorem, namely that 
the graph should be equivalent to a polyhedron.  Each face must be 
simply connected (able to shrink to a disk, with no "holes" in it), 
and likewise each edge must be like a segment (not a circle).  One of 
our "natural" faces has a "vertex" in the middle of it, so it is not 
simply connected; and the "edge" has no ends, so it doesn't fit 
either.  These errors just happen to cancel one another out.

As another example, take a cylinder, which in its natural state has 
no vertices, two "edges", and three "faces":

  0 - 2 + 3 = 1

It doesn't work, and the theorem doesn't claim it should.

In each case you can "fix" the graph by adding one segment from top 
to bottom.  In the cone, this gives one extra vertex (on the base), 
and one extra edge, so the formula still holds.  In the cylinder, it 
gives two new vertices and one extra edge, and the formula becomes 
correct.

What do you have to do to "fix" the sphere?

Here is a deeper discussion of these ideas:

  Euler's Formula Applied to a Torus
    http://mathforum.org/library/drmath/view/51815.html 

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Polyhedra
High School Polyhedra

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/