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Math Help - [READ]Unusual geometrical shapes

  1. #1
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    [READ]Unusual geometrical shapes

    1)In fractal geometry there exist shapes which have INFINITE perimeter and FINITE area (Koch Snowflake). Which I find amazing, however I was wondering what if the positions were reversed? i.e. INFINITE area but FINITE perimeter. I doubt that this is true but I cannot proof it. For one thing there is no well-defined term for a "fractal" we can just give examples.

    2)This question is related to the previous. There are solids called "supersolids" these are solids which have INFINITE surface area but FINITE volume (Gabriel's Horn-you can fill it with paint but not paint it with paint). Now reverse the positions, INFINITE volume but INFINITE surface area. However, this problem I think is easier because we can give a well-defined term to a supersolid. We can define it like this: If a function is countinous and positive for x>a, then show that if the improper integral for surface area converges if and only if the improper integral for volume converges! This might actually be true which I am going to try to prove.
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  2. #2
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    "If the improper integral for surface area converges if and only if the improper integral for volume converges! This might actually be true which I am going to try to prove."

    Greetings PerfectHacker Sir:

    May I assume your assertion stated in blue above to correctly read: "The improper integral for surface area converges if and only if the improper integral for volume converges" --irrespective of truth value, of course? That is, is the opening word "If" a type-O? If that be the case, what are your thoughts regarding the convergent integral y=e^-x, x in [0,inf) and bound by, say, z=0 and z=1 for instance? Here we have finite volume bound by infinite surface area. Does this example not serve to counter your assertion? Or have I failed to comprehend some aspect of the claim?

    Season's Greetings to you.

    Rich B.
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  3. #3
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    Indeed! I have made a mistake!
    "If the suface integral converges then the volume integral converges"
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  4. #4
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    Hi P.Hkr:

    ...a simple matter of inadvertent inclusion of the bi-conditional terminology, iff.

    Regards,

    Rich B.
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    Since I learned Latex math this is what my question is:
    Given that \int_{a}^\infty f(x)\sqrt{1+[f'(x)]^2}dx converges then: \int_{a}^\infty [f(x)]^2dx must converge.
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  6. #6
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    Quote Originally Posted by ThePerfectHacker
    1)In fractal geometry there exist shapes which have INFINITE perimeter and FINITE area (Koch Snowflake). Which I find amazing, however I was wondering what if the positions were reversed? i.e. INFINITE area but FINITE perimeter. I doubt that this is true but I cannot proof it. For one thing there is no well-defined term for a "fractal" we can just give examples.
    This will be incredibly heuristic, but take a unit circle. Now stretch the origin infinitely far as if the surface was made of rubber, and you have a finite perimeter with infinite area, in some kind of non-euclidean geometry.
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  7. #7
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    Give it mathematical equation which violates my hypothesis then I would be happy.
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