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Math Help - Conceptualising integration/vol. of revolution

  1. #1
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    Conceptualising integration/vol. of revolution

    The area under the curve y= x^(-2/3) from x=1 to infinity is undefined. Yet if the region is rotated about the x-axis the volume of revolution IS defined as 3pie.
    I've done the math but can't explain how it is so conceptually. The best I can come up with is that the area is not necessarily infinite just indeterminable. But I think that this may be b.s.
    Any help would be greatly appreciated.

    Woollybull.
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  2. #2
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    Why not take a shot at the surface area, while you are at it?

    "Conceptually"? What does that mean? The "concept" you should be contemplating is "convergence". It converges or it doesn't.

    Perhaps the answer lies in your use of the word "pie". Pie is for eating. Pi is for mathematics and Greeks. Are you distracted by lunch and not really thinking about mathematics?

    Also, your use of the word "infinite" is not useful. There may be a day, in Geometry or Complex Analysis, where "infinity" will be a place to stop and visit, but if you abandon this for now and consider simply that the value of the integral increases without bound as x increases, where's the question?

    The physical concept of NONconvergence is very simple:

    1) Pick a value.
    2) You will exceed it eventually.

    You see, it is not a matter of being "indeterminable". It is a matter of being able to demonstrate that there is no such value to which the integral settles.
    Last edited by TKHunny; May 15th 2008 at 10:23 AM.
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  3. #3
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    I am unable to visualise how the area doesnt settle but the volume does settle. Can you help me with that??

    Thanks.
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  4. #4
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    Surface area is infinite, but volume is finite.

    2{\pi}\int_{1}^{\infty}(\frac{1}{x^{\frac{2}{3}}})  \sqrt{1+\frac{4}{9x^{\frac{10}{3}}}}dx

    But, \sqrt{1+\frac{4}{9x^{\frac{10}{3}}}}\geq{1}, \;\ if \;\ x\geq{1}

    So, \frac{2{\pi}}{x^{\frac{2}{3}}}\sqrt{1+\frac{4}{9x^  {\frac{10}{3}}}}\geq{\frac{2{\pi}}{x}}

    Therefore, \lim_{L\rightarrow{\infty}}\frac{2{\pi}}{x}dx={\in  fty}

    Think of it as filling the solid with paint and letting it seep through the surface.
    Then you could paint an infinite surface area with a finite amount of paint.
    Something to scratch ones head about, huh?
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  5. #5
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    What Galactus said is a very interesting thing which is hard to accept.
    That you can fill the object with paint but you cannot paint it. I know it sounds strange. But you can read more here. I consider the reverse question, i.e. whether there is a shape which cannot be filled with paint but can still be painted more interesting. Because I believe that such a shape cannot exists as a consequence of the isoperimetric inequality. Though I am not completely sure if one can apply this inequality to a problem like this.
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  6. #6
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    I appreciate your help guys.
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