Cylinder Area Paradox

height h

10

radius r

5

bands m

5

arcs n

7

area of the cylinder ≈ 314.20

area of the triangles ≈ 312.90

Let

S

be the surface of a cylinder of height

h

and radius

r

. (

S

does not include the flat circular ends of the cylinder.) This Demonstration constructs a set of triangles that tend uniformly to

S

—yet their total area does not tend to the area of

S

!

Divide

S

into

m

subcylinders (or bands) of height

h/m

. Construct

2n

congruent isosceles triangles in each band with vertices at the vertices of a regular

2n

-gon inscribed in the circles at the top and bottom of each band, offset by

π/n

.

For any point

p

in

3



(except the axis of the cylinder), let

s(p)

be the axial projection of

p

onto

S

. As

m,n∞

, to say that the triangles approximate

S

uniformly means that for any point

t

on a triangle and any

ϵ>0

(independent of

t

), there is a

K∈

such that for all

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,

lim

m,n∞

A(m,n)

can differ. If first

n∞

with

m

held fixed and then

m∞

, the limit is

2πrh

, the expected area of the cylinder. If first

m∞

with

n

held fixed and then

n∞

, the limit is infinity. If

m∞

and

n∞

together so that

m/

2

n

is some positive constant

c

, the limit can be chosen to be any number greater than

2πrh

.

Therefore

A(m,n)

does not have a limit.

The surface is known as Schwarz's lantern, Schwarz's polyhedron, or Schwarz's cylinder.