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Consider the curve $\frac{1}{x}$ where $x \geq 1$. Rotate this curve around the x-axis.

One Dimension - Clearly this structure is infinitely long.

Two Dimensions - Surface Area = $2\pi\int_∞^1\frac{1}{x}dx = 2\pi(\ln ∞ - \ln 1) = ∞$

Three Dimensions - Volume = $\pi\int_∞^1{x}^{-2}dx = \pi(-\frac{1}{∞} + \frac{1}{1}) = \pi$

So this structure has infinite length and infinite surface area. However it has finite volume, which just does not make sense.

Even more interesting, the "walls" of this structure are infinitely thin. Since the volume is finite, we could fill this structure with a finite amount of paint. To fill the structure the paint would need to cover the complete surface area of the inside of this structure. Since the "walls" are infinitely thin, why would a finite amount of paint not be able to cover the outside of the "walls" too?

Please help me make sense of this whole thing.

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The infinite is often counterintuitive. You cannot actually fill it with paint, though: Planck would get in your way (eventually, the horn is thinner than atoms). –  Arturo Magidin Dec 17 '10 at 3:04
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This is known as Gabriel's Horn: mathworld.wolfram.com/GabrielsHorn.html –  Douglas S. Stones Dec 17 '10 at 3:12
    
Related, I suppose: en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox –  Aryabhata Dec 17 '10 at 3:14
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A simpler $2$-dim example: one obtains finite area and infinite perimeter by appending the rectangles $\rm R_n$ of height $\rm 1/2^n $ and width $1,$ for all $\rm\ i\in \mathbb N$ –  Bill Dubuque Dec 17 '10 at 3:35
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Does the existence of infinitely long regions of the plane with finite area, such as the area under $y = exp(-x)$ in the first quadrant, involve any sort of paradox? The finite volume of rotation described by pacman is similarly an instance of how a finite value can be the limit of an infinite series. –  hardmath Dec 17 '10 at 4:22
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5 Answers

It seems you already have made sense of this whole thing from a mathematical point of view (it has infinite surface area and finite volume -- there's no contradiction here). The paradox is that infinity, at times, does not match well with our day-to-day experiences.

Here's a related example (and hopefully a little bit easier to picture): Take a unit square. Cut it in half vertically down the middle -- forming two pieces. The area clearly remains the same, but the total perimeter (counting both pieces) has increased by 2. Keep cutting the right-most piece vertically down the middle forever. The area remains unchanged throughout, whereas the perimeter is $4+\lim_{n \rightarrow \infty} \sum_{i=1}^n 2=\infty$.

alt text

Now, if you wanted to make it look more like a horn, you could move the pieces around, for example:

alt text

which still has area 1 and perimeter infinity. If you wanted to make it have volume 1 and surface area infinity, replace "square" with "cube".

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+1 Very nice example indeed. –  Adrián Barquero Dec 17 '10 at 3:51
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+1 'cause you're so close to 2,000 –  uncle brad Dec 17 '10 at 4:12
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The same concept is contained in the Dirac delta function. It has infinite height & perimeter but always has area equal to 1. –  Tpofofn Dec 17 '10 at 12:45
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The answer to the paint question is as follows:

To define terms, let's say you are rotating about the $x$ axis and the volume is between the surface and the $y-z$ plane. If we take a small (but finite) patch of the $y-z$ plane of $\Delta A$ area then the area above it is roughly $\Delta A$ as $\sqrt{y^2 + z^2} \rightarrow \infty$ and the volume is roughly $\frac{\Delta A}{\sqrt{y^2 + z^2}}$. So as we move away from the $x$ axis, the trapped volume goes to zero but the surface area does not.

Now back to the paint analogy. It takes infinite paint because a fixed volume of paint covers a fixed surface area (i.e. one gallon / 400 sq.ft.). This is true even as $\sqrt{y^2 + z^2} \rightarrow \infty$, so we find ourselves having to paint a surface area equivalent to the $y-z$ plane. Now if we tried to fill the volume with the paint, we are no longer required to cover at a given rate of 1 gal. / 400 sq.ft. In other words our film thickness is $\frac{1}{\sqrt{y^2 + z^2}}$ which tends to zero.

So in summary, your assumption regarding the walls being infinitely thin is true, but you relate volume and surface area using the metaphor of paint. Paint always has a finite spread rate no matter how thin the film. This film does not get thinner as we move away from the $x$ axis. The volume between the curve and the $y-z$ plane however does.

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The mathematics has already been explained; when the mathematics contradicts your intuition the answer is to ignore your old intuition and find new intuition.

As for the physics, Gabriel's horn is in no meaningful sense a physically realistic object. One way to summarize the lesson here is that the assumption that very thin objects have zero thickness is only reasonable if this causes a reasonable amount of error in one's computations, but in the construction of Gabriel's horn the thickness of the membrane comes to dominate the computation and it is no sense reasonable to ignore this error. Any time a mathematical model fails to accurately reflect reality one should question the assumptions that go into building that model, and to the extent that this is a mathematical model of anything, the assumption that thickness is negligible is the one that should be thrown out. If you try to repeat the construction of Gabriel's horn with a fixed finite thickness you will, of course, find that the resulting object has infinite volume.

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I'd like to add something to the discussion made above.

Let $f:]0,+\infty[ \to [0,+\infty[$ a measurable function, $N\in \mathbb{N}$, $x\in \mathbb{R}^N$ and $t\in \mathbb{R}$. Consider the set:

$$E_f:=\{ (x,t)\in \mathbb{R}^{N+1}:\ |x|\leq |f(t)|\}$$

which is the body of revolution generated by $f$ with respect to the $t$ axis in $\mathbb{R}^{N+1}$. Using cylindrical coordinates, one finds for the Lebesgue measure $\mathcal{L}^{N+1}(E_f)$ the following expression:

$$\mathcal{L}^{N+1} (E_f)=\omega_N \int_0^{+\infty} f^N (t)\ \text{d} t\; ,$$

where $\omega_N$ is the volume of the unit ball in $\mathbb{R}^N$ (i.e. $\omega_N:=\pi^{\frac{N}{2}}/ \Gamma (\frac{N}{2} +1)$), hence $\mathcal{L}^{N+1}(E_f)=\omega_N \| f\|_{L^N}^N$ and $\mathcal{L}^{N+1} (E_f)$ is finite iff $f\in L^N(]0,+\infty[)$.

On the other hand, if $f$ is also Lipschitz continuous, it is easy to compute the surface area (or De Giorgi perimeter) $\mathcal{P} (E_f)$ of $E_f$: using cylindrical coordinates one finds:

$$\mathcal{P} (E_f)=N\omega_N \int_0^{+\infty} f^{N-1}(t)\ \sqrt{1+|f^\prime (t)|^2} \text{d} t\; ,$$

thus:

$$N\omega_N \| f\|_{L^{N-1}}^{N-1} \leq \mathcal{P} (E_f) \leq N\omega_N\sqrt{1+\| f^\prime \|_{L^\infty}^2}\ \| f\|_{L^{N-1}}^{N-1}$$

and $\mathcal{P} (E_f)$ is finite iff $\| f\|_{L^{N-1}}^{N-1}$ does.

Let $\mathcal{S} (E_f)$ be the measure of the sections of $E_f$ obtained cutting the set with any hyperplane $\Pi:=\{ (x,t)|\ \langle a,x\rangle =0\}$ ($|a|=1$) containing the axis of revolution; then:

$$\mathcal{S} (E_f) =\omega_{N-1} \int_{0}^{+\infty} f^{N-1}(t)\ \text{d} t\; ,$$

and also $\mathcal{S} (E_f)$ is finite iff $f\in L^{N-1} (E_f)$.

Since $f$ is defined in the interval $[0,+\infty[$, one in general doesn't have $f\in L^{N}\Rightarrow f\in L^{N-1}$ not even if $f$ is Lipschitz (e.g., $f(x):=\chi_{[0,1[}(x)+x^{1-N}\chi_{[1,+\infty[} (x)$ is $L^N$ but not $L^{N-1}$): thus in general it is always possible to pick a Lipschitz function $f$ such that $\mathcal{L}^{N+1} (E_f)$ is finite and $\mathcal{P} (E_f)$, $\mathcal{S} (E_f)$ are not.

N.B.: Instead of the De Giorgi perimeter, one can use the Hausdorff measure $\mathcal{H}^{N}$ or the Minkowski content $\mathcal{M}$ as well: in fact there is equality among $\mathcal{P} (E_f)$, $\mathcal{H}^{N} (E_f)$ and $\mathcal{M} (E_f)$ because the boundary $\partial E_f$ is sufficiently regular when $f$ is Lipschitz.

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I found this explanation illuminating. It's from Jerome Keisler's Elementary Calculus.

Suppose you have a clay cylinder with radius 1 and length 1. Its cross-sectional area is $\pi$, so its total volume is $\pi$. The surface area, not counting the ends, is $2\pi$.

Now roll the cylinder with your hands so that it is only half as thick as it was. Its radius is now $\frac12$, so its cross-sectional area is only $\frac14\pi$. The volume must remain the same, so the cylinder's length is now $4$, and its surface area, again not counting the ends, has doubled, to $4\pi$.

You can repeat this process, rolling it into a snake with radius $\frac14$. Its length then increases to $16$, and its surface area doubles again to $8\pi$.

At each step as you roll the cylinder into a thinner and thinner snake, the volume always remains constant, but the surface area increases without bound. A snake with a radius of $r$ has a length of $\frac1{r^2}$ and a surface area of $\frac{2\pi}r$, both of which go to infinity as $r$ goes to zero. None of this should seem surprising.

Now instead of rolling the whole snake, just roll the right-hand half of the snake at each step. The volume is still constant, as it must be, and the surface area still goes to infinity. The shape you get is quite similar to Gabriel's horn.

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