# 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.