GABRIEL'S HORN
… OR THE INFINITE PAINT
CAN
Consider the region R in the first quadrant, bounded
above by y = 1/x and by x = 1 on the left.
The region is revolved around the x-axis to form a solid
of infinite length. This object is famous in mathematics and
is called Gabriel's Horn (because it looks like an angelic trumpet).
It has also been called the Infinite Paint Can because the volume
is some finite value but if you paint the exterior it would need
an infinite volume of paint.
(a) Find the volume of the solid.
This is an easy task and I leave the work for you.
(b) Show that the surface area is infinite.
This is a considerably more difficult task. The basic formula
for the surface area of a solid of revolution is fairly straightforward,
but the actual integral is quite tricky.
At this point I am stuck. I know of no techniques that
will allow me to integrate further. Thus, we need a new plan
of attack. First, let us see if numerical approximations help.
Above are the approximations of the surface area for the object
on intervals [1, 10], [1, 1000], and [1, 1 000 000]. Though a
million is an incredibly small number on the way to infinity
(actually, all numbers are infinitely far from infinity), we
do not see the surface area approaching a limit. This seems paradoxical
since we already know the Horn has a finite volume. What would
the Horn look like in the neighborhood of x = 1,000,000?
The radius at that point is 0.0000001, thus the "Horn"
will be incredibly thin. How is area still piling up? In that
neighborhood, the "tube" is nearly a straight line
with nearly no hollow center.
The key word is "nearly." Infinity is a tricky topic
and we cannot be too cavalier in tossing away what seems to be
nothing. Recall the summation:
This seemed like it would eventually reach a limit since the
terms in the series are so incredibly small for large values
of n. Alas, what we expect does not always turn out to
be so.
Can the integral for the surface area be evaluated at all?
Yes! I do not recall the correct method, but the TI-89 can solve
it.
Evaluating this as a definite integral yields infinite surface
area, as shown below.
I am certainly not happy with this definite integral because
I do not know how to evaluate the indefinite integral. I have
no idea if the math is correct or if it is a bunch of rubbish.
Fortunately, there is another method that allows me to know with
complete certainty that the surface area is infinite.
The trick is to compare the messy integral with one where
the definite integral can be determined easily and precisely
but also that is definitely smaller than the integral we are
trying to determine. If the smaller integral is infinite, then
the more difficult, larger integral must also be infinite.
Thus, compare 
with 
.
I claim S1 must be larger than (or equal
to) S2 for all values of b that
are greater than or equal to 1. In , the
radius of the surface of revolution is 
,
while in S2 the radius is 1/x.
Since 
for all x, then 
for all x greater than or equal to
1. Thus, the surface of revolution for S1
has a greater radius for all x greater than or equal to
1. Given equal lengths for the solids, the solid with the greater
radius for all x will have a greater surface area.
Hence, 
. Our next question is 
?
Yes, as is shown below.
Therefore, since S1 > S2
and S2 has infinite surface area, then
S1 must also have infinite surface area.
This proves that the mathematical object known as Gabriel's
Horn has (as you have shown) a volume of p
cubic units but has infinite surface area. Thus, theoretically,
if it were painted with golden paint it could hold a total volume
of approximately 3.14 cubic units of paint, but it would need
infinitely many gallons of paint to paint the exterior. Now,
that really is miraculous!
This page was created on June 12, 2004.
Last Updated on August 25, 2004.
