• Dec 21st 2005, 01:56 PM
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.

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.
• Dec 22nd 2005, 07:09 AM
Rich B.
"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.
• Dec 22nd 2005, 01:14 PM
ThePerfectHacker
Indeed! I have made a mistake!
"If the suface integral converges then the volume integral converges"
• Dec 23rd 2005, 12:51 AM
Rich B.
Hi P.Hkr:

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

Regards,

Rich B.
• Dec 25th 2005, 12:51 PM
ThePerfectHacker
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.
• Dec 25th 2005, 11:22 PM