Cylinder Area Paradox
Cylinder Area Paradox
Let be the surface of a cylinder of height and radius . ( does not include the flat circular ends of the cylinder.) This Demonstration constructs a set of triangles that tend uniformly to —yet their total area does not tend to the area of !
S
h
r
S
S
S
Divide into subcylinders (or bands) of height . Construct congruent isosceles triangles in each band with vertices at the vertices of a regular -gon inscribed in the circles at the top and bottom of each band, offset by .
S
m
h/m
2n
2n
π/n
For any point in (except the axis of the cylinder), let be the axial projection of onto . As , to say that the triangles approximate uniformly means that for any point on a triangle and any (independent of ), there is a such that for all , .
p
3
s(p)
p
S
m,n∞
S
t
ϵ>0
t
K∈
m,n>K
|t-s(t)|<ϵ
The sum of the areas of the triangles is
A(m,n)=2rnsin+
π
n
2
h
2
(mr)
2
1-cos
π
n
Depending on how the limit is taken, A(m,n) can differ. If first with held fixed and then , the limit is , the expected area of the cylinder. If first with held fixed and then , the limit is infinity. If and together so that is some positive constant , the limit can be chosen to be any number greater than .
lim
m,n∞
n∞
m
m∞
2πrh
m∞
n
n∞
m∞
n∞
m/
2
n
c
2πrh
Therefore does not have a limit.
A(m,n)
The surface is known as Schwarz's lantern, Schwarz's polyhedron, or Schwarz's cylinder.