Faces, Vertices, and Edges of Cylinders, Cones, and SpheresDate: 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/ |
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