Special Curves

~~Kribz~~

Posers and Puzzles

25 Nov 04

~~Kribz~~

Bannedtown, TX

Joined: 11 Nov 04
Moves: 77

25 Nov 04

Can you construct a function f(x) such that when rotated around the x-axis, it yields a three-dimensional curve whose surface area is finite yet whose volume is infinite?

What about such a function that yields a shape of finite volume but infinite surface area?

Kribz

~~Kribz~~

Bannedtown, TX

Joined: 11 Nov 04
Moves: 77

25 Nov 04

Time saving hint: don't try to construct them from scratch. Either you know the answers or you don't. They are intended to be answered Yes or No, not by the construction of f(x).

Kribz

THUDandBLUNDER

Joined: 29 Feb 04
Moves: 22

25 Nov 04

1 edit

"What about such a function that yields a shape of finite volume but infinite surface area?"

Gabriel's Trumpet.

TheMaster37

Kupikupopo!

Out of my mind

Joined: 25 Oct 02
Moves: 20443

25 Nov 04

Originally posted by Kribz
Can you construct a function f(x) such that when rotated around the x-axis, it yields a three-dimensional curve whose surface area is finite yet whose volume is infinite?

What about such a function that yields a shape of finite volume but infinite surface area?

Kribz

Yes & Yes

Acolyte

Now With Added BA

Loughborough

Joined: 04 Jul 02
Moves: 3790

25 Nov 04

1 edit

Originally posted by TheMaster37
Yes & Yes

I get No & Yes. Finite volume, infinite area is fine, but not the other way around. For revolution purposes we may assume that f(x) is non-negative. Then:

Area = S(2pi*f(x)*sqrt(1+f'(x)^2))dx (where S means integrate over the domain of f)
>= 2pi*S(|f(x)|)dx

So for finite area, f must be 1-integrable.

Volume = pi*S(f(x)^2)dx

So for infinite volume, f must not be 2-integrable.

However, all 1-integrable functions are 2-integrable by Hoelder's inequality in the case where the exponents are both 1/2:

S|fg| =< sqrt(S|f|)sqrt(S|g|)

Hence there is no function for which a finite-area surface and an and infinite-volume solid of revolution exist.

Nemesio

Ursulakantor

Pittsburgh, PA

Joined: 05 Mar 02
Moves: 34824

25 Nov 04

Originally posted by THUDandBLUNDER
"What about such a function that yields a shape of finite volume but infinite surface area?"

Gabriel's Trumpet.

http://curvebank.calstatela.edu/torricelli/torricelli.htm

This math is so neat and so inaccessible to me that it blows my mind.

Cool stuff!

THUDandBLUNDER

Joined: 29 Feb 04
Moves: 22

26 Nov 04

Originally posted by THUDandBLUNDER
"What about such a function that yields a shape of finite volume but infinite surface area?"

Gabriel's Trumpet.

I meant Gabriel's Horn.

AThousandYoung

1st Dan TKD Kukkiwon

tinyurl.com/2te6yzdu

Joined: 23 Aug 04
Moves: 26757

26 Nov 04

I think yes and yes.

THUDandBLUNDER

Joined: 29 Feb 04
Moves: 22

26 Nov 04

Originally posted by AThousandYoung
I think yes and yes.

As for objects with infinite volume but finite surface area, I don't think this is possible. One well-known property of a sphere is that it maximizes volume for a given surface area. If infinite volume in finite surface area were possible, then you could improve on a sphere.

~~Kribz~~

Bannedtown, TX

Joined: 11 Nov 04
Moves: 77

27 Nov 04

1 edit

Originally posted by THUDandBLUNDER
One well-known property of a sphere is that it maximizes volume for a given surface area. If infinite volume in finite surface area were possible, then you could improve on a sphere.

Good observation. I think this closes this problem, as we have a proof by construction for second question, and a proof by contradiction against the first question.

cosmic voice

Joined: 19 Jun 04
Moves: 2930

03 Dec 04

Originally posted by Kribz
Can you construct a function f(x) such that when rotated around the x-axis, it yields a three-dimensional curve whose surface area is finite yet whose volume is infinite?

What about such a function that yields a shape of finite volume but infinite surface area?

Kribz

Finite volume with infinite surface area is possible.
The simplest example is the Kronecker delta function, rotated about the
x-axis.
The other way round is not possible.