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Infinite Curve Problem

  1. Oct 25, 2014 #1
    NB. first time using Latex so apologies if something came out wrong, I've done my best to double check it.

    Consider the curve [itex]y = \frac{1}{x}[/itex] from [itex]x=1[/itex] to [itex]x=\infty[/itex]. Rotate this curve around the x axis to create a funnel-like surface of revolution. By slicing up the funnel into disks with [itex]r=\frac {1}{x}[/itex] and thickness [itex]dx[/itex] (and hence volume ([itex]\pi r^2 dx[/itex])) stacked side by side, the volume of the funnel is
    [itex]V = \int_1^\infty \frac{\pi}{x^2} dx=- \frac{\pi}{x}\mid\int_1^\infty = \pi,[/itex]

    which is finite. The surface area however involves the circumferential of the disks, which is [itex](2\pi r)dx[/itex] multiplied by a [itex] \sqrt{1+y'2}[/itex] factor accountng for the tilt of the area. The surface of the funnel is therefore

    [itex]A = \int_1^\infty \frac{2\pi\sqrt{1+y'2}}{x} dx > \int_1^\infty \frac{2\pi}{x}dx[/itex]
    which is infinite. As the volume is finite but the area is infinite, the funnel can be filled with paint but you can't paint it. Which appears to be a paradox since one should be painting the inside surface when filling up the funnel. But the inside surface=outside surface given the funnel has no thickness.


    So the question asks me to basically make sense of this paradox, I've done a lot of headscratching over it but thinking logically I simply can't find find our whats going on here, since , having checked over the equations theoretically the statements posed by the question are logical and I see no disconnect, so seems possible but impossible in reality given the paint should be at a constant (visible) thickness which I assume.
     
    Last edited: Oct 25, 2014
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  3. Oct 25, 2014 #2

    PeroK

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    First, you can't fill something mathematically infinite with paint nor paint it.

    But, thinking mathematically, you can paint it: you just need a diminishing thickness of paint. Essentialy that is what is happening with the volume. You start with the equivalent of thick paint (filling the funnel) but as you go further from the origin the thickness of the funnel reduces without limit.

    You could paint it on the outside - with a finite volume of paint - by copying what you do on the inside.
     
  4. Oct 25, 2014 #3

    epenguin

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    Intriguing. But they have made it seem unfamiliar by taking a 3-dimensional case. If you think of it, in all the integrals
    y dx that you have ever done that were finite, the curve in the x, y plane had infinite length.
    So it's not a special case and the fact is actually familiar, though still needing some explaining perhaps.

    ( the ∫ from Σ does not always come out very well I see!)
     
    Last edited: Oct 25, 2014
  5. Oct 26, 2014 #4

    Mentallic

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    It's interesting that given paint is made up of atomic particles, it hence has a certain size and so at some point, no paint would be able to fit further down the funnel. Whatever depth this is at, you would have only painted a finite surface area, and so you still have an infinite area that you physically can't paint.
     
  6. Oct 26, 2014 #5
    If you want to google this, it's called Gabriel's Horn.
     
  7. Oct 26, 2014 #6

    LCKurtz

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    Here's my take. Let's talk about painting the xy plane. I will suppose we have an infinitely non viscous paint, by which I mean, if a coat of it has any positive thickness, it could be spread over a greater area.

    Definition: The plane is painted if given any ##r>0## the disk centered at the origin of radius ##r## is painted.

    Theorem: If you have ##m>0## cubic units of paint (the good stuff defined above), that is sufficient to paint the plane.

    Proof. Suppose ##r>0##. Apply your ##m## units of paint to the disk of radius ##r## centered at the origin. This can be done since the thickness of the coat is ##\frac m {\pi r^2}>0##. So there is no contradiction, in principle, about painting an infinite surface with a finite volume of paint.

    So you could just pour your paint on the origin and let it spread.
     
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