The paradox

known as "Gabriel's

Horn" deals with the

solid formed when the

graph y=

^{1}/_{x} is

rotated around the x-axis, considered when x≥1. It

looks a bit like a horn (hence the name) with its wide

opening at x=1, facing the origin, and

narrowing toward the

x-axis as x→

∞.

The volume of any solid formed by rotation of a graph over a≤x≤b with y=r around the x-axis is:

b
∫ πr^{2}*d*x
a

Therefore, volume of the horn solid is the

improper integral ∞ ∞
π ∫ (^{1}/_{x})^{2}*d*x=-π^{1}/_{x} ] = **π**
1 0

It turns out that the volume is exactly equal to π.

The surface area of any solid formed by rotation of a graph over a≤x≤b with y=r around the x-axis is:

b
∫ 2πr√(1+(*d*y)^{2})*d*x
a

The surface area of this solid is another

improper integral:

∞
∫ 2π^{1}/_{x}√(1+(*d*/*d*x ^{1}/_{x})^{2})*d*x=**∞**
1

The

proof of this is a bit on the

lengthy side, but you can

trust me that it's

true. Or, if you have a

worthwhile calculator like the

TI-89, you can just type it in and prove it to

yourself.

So, this solid has a finite volume (V=π), but an infinite surface area! This means that if the horn was a **paint can**, it would not be able to hold enough paint to cover its own surface. However, this is seemingly impossible; think of it this way. Since the can would be full, the inside surface of the can (which, of course is the same as the outside surface, since the thickness of the can is 0) would be covered by paint. However, this area is so large that no amount of paint could possible cover it. Somehow, our π units of paint are covering an infinite surface area, and still filling in the space between the walls. Therefore, **π>∞**??

There's a good 3D picture of the Horn, plus a humorous one of the "Paint-Can Paradox" at "http://www.geocities.com/Eureka/Plaza/4033/gabriel_index.en.html". You can play with an excellent 3D model of the Horn at "http://mathworld.wolfram.com/GabrielsHorn.html".