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unJon
unJon 
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February 21st, 2021 at 10:22:05 AM permalink
Quote: chevy



I guess I have always disagreed with this as a "paradox". The problem I always had was with equating surface area to volume. Gabriel's Horn hides this in a calculation where we concentrate on one integral being finite but another is infinite. But consider the x-y plane. The surface area of that is infinite, but the volume of paint required to "paint" it is 0. In the case of Gabriel's Horn, you go out and buy your can of paint with volume, pi "units", you paint the inside (or outside) using up 0 "units", and you still have pi "units" of paint left to fill the horn with.

As to Teliot's original question, I agree with the calculations Teliot and ThatDonGuy discussed...but if I go back to comparing surface area to volume, does equating them have some physical or geometric meaning I am missing?



These are good points. But there is a natural relationship between surface area and volume. And, for me, there is some impact to the paradox. Like this: here is an infinite amount of paper for you to take and use all of it but create a shape with a finite amount of volume. You could not take that paper and make a sphere with finite volume. But you could make a Gabriel’s Horn. To your point you could also just lie it flat and make a plane (though no volume really “enclosed” there).
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ThatDonGuy
ThatDonGuy
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February 21st, 2021 at 10:47:09 AM permalink
Quote: teliot

Quote: ThatDonGuy


The only value where they are equal is t = 1 (i.e. both volume and surface area are zero), and even this assumes that the disc of radius one at t = 1 is not included in the area.

The surface area = INTEGRAL[1, t] (2 PI / t) dt = 2 PI INTEGRAL[1, t] (1/t) dt = 2 PI (ln t - ln 1) = 2 PI ln t.
The volume = INTEGRAL[1, t] (PI / t^2) dt = PI INTEGRAL[1, t] (1/t^2) dt = PI (-1/t + 1/1) = PI (1 - 1/t).
These are equal when 2 ln t = 1 - 1/t.
2 ln 1 = 1 - 1/1, so they are equal when t = 1.
If t > 1, then they are equal when 2 ln t + 1/t - 1 = 0.
Let f(t) = 2 ln t + 1/t - 1
df/dt = 2/t + 1/t^2; since this is positive for all t > 1, f(t) is strictly increasing for all t > 1, which means f(t) > f(1).

If the disc is included, then the surface area = 2 ln t, and they are equal when ln t = -1/t, but for t > 1, ln t > 0 and -1/t < 0.


Your solution was my original thought on this. But now, I disagree about the surface area calculation you've given ... thinking out loud, if you replace your formula with "The Surface Area > ..." and the rest of your argument works.



I can see where the surface area in "if the disc is included" is wrong; it should be PI (2 ln t + 1). Is that what you were referring to?
chevy
chevy 
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February 21st, 2021 at 10:55:24 AM permalink
unJon:

Interesting comparison of using infinite paper to create a finite volume Gabriel's Horn, but not a sphere. To that end I concede, there are intrinsic relationships between surface area and volume.

I think I like your implied style "Can you make an object with infinite surface area and finite (non-zero) volume?" as a better wording than the video's "paint it" paradox.

My objection was 2D vs 3D, and not really an infinity issue. My objection still holds for a 1x1x1 cube....takes zero volume of paint to coat the inside(surface area), but 1 volume unit of paint to fill it. We wouldn't compare 6 "units" to coat it with 1 "unit" to fill it.
teliot
teliot
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February 21st, 2021 at 11:02:56 AM permalink
Quote: ThatDonGuy

Quote: teliot

Quote: ThatDonGuy


The only value where they are equal is t = 1 (i.e. both volume and surface area are zero), and even this assumes that the disc of radius one at t = 1 is not included in the area.

The surface area = INTEGRAL[1, t] (2 PI / t) dt = 2 PI INTEGRAL[1, t] (1/t) dt = 2 PI (ln t - ln 1) = 2 PI ln t.
The volume = INTEGRAL[1, t] (PI / t^2) dt = PI INTEGRAL[1, t] (1/t^2) dt = PI (-1/t + 1/1) = PI (1 - 1/t).
These are equal when 2 ln t = 1 - 1/t.
2 ln 1 = 1 - 1/1, so they are equal when t = 1.
If t > 1, then they are equal when 2 ln t + 1/t - 1 = 0.
Let f(t) = 2 ln t + 1/t - 1
df/dt = 2/t + 1/t^2; since this is positive for all t > 1, f(t) is strictly increasing for all t > 1, which means f(t) > f(1).

If the disc is included, then the surface area = 2 ln t, and they are equal when ln t = -1/t, but for t > 1, ln t > 0 and -1/t < 0.


Your solution was my original thought on this. But now, I disagree about the surface area calculation you've given ... thinking out loud, if you replace your formula with "The Surface Area > ..." and the rest of your argument works.



I can see where the surface area in "if the disc is included" is wrong; it should be PI (2 ln t + 1). Is that what you were referring to?

if you want an exact value for the surface area over a finite interval, you cannot disregard the term in the surface area integral that is disregarded in the video in the infinite upper limit case.

For example, find the limit x = t that gives a surface area equal to twice the volume. I have no idea how to do this.
Personal website: www.ijmp.org
unJon
unJon 
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February 21st, 2021 at 11:15:58 AM permalink
Quote: chevy


My objection was 2D vs 3D, and not really an infinity issue. My objection still holds for a 1x1x1 cube....takes zero volume of paint to coat the inside(surface area), but 1 volume unit of paint to fill it. We wouldn't compare 6 "units" to coat it with 1 "unit" to fill it.



Yes agree with your point here.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
teliot
teliot
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February 21st, 2021 at 11:16:43 AM permalink
Quote: chevy

unJon:

Interesting comparison of using infinite paper to create a finite volume Gabriel's Horn, but not a sphere. To that end I concede, there are intrinsic relationships between surface area and volume.

I think I like your implied style "Can you make an object with infinite surface area and finite (non-zero) volume?" as a better wording than the video's "paint it" paradox.

My objection was 2D vs 3D, and not really an infinity issue. My objection still holds for a 1x1x1 cube....takes zero volume of paint to coat the inside(surface area), but 1 volume unit of paint to fill it. We wouldn't compare 6 "units" to coat it with 1 "unit" to fill it.

Then space filling curves should be equally objectionable. A one dimensional line filling two-dimensional space.
Personal website: www.ijmp.org
ThatDonGuy
ThatDonGuy
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February 21st, 2021 at 12:53:35 PM permalink
Quote: teliot

if you want an exact value for the surface area over a finite interval, you cannot disregard the term in the surface area integral that is disregarded in the video in the infinite upper limit case.

For example, find the limit x = t that gives a surface area equal to twice the volume. I have no idea how to do this.


Maybe I am calculating it wrong, but I thought the surface area = the integral that sums the circumferences of the circles from x = 1 to x = t.
The circumference of the circle at x is 2 PI (1/x), whose integral with respect to x is 2 PI ln (x), so the integral over x = 1 to t is 2 PI ln t - 2 PI ln 1 = 2 PI (ln t - 0) = 2 PI ln t.
Then again, that method doesn't seem to work when I try it with calculating the surface area of a sphere...
teliot
teliot
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February 21st, 2021 at 1:51:16 PM permalink
Quote: ThatDonGuy

Maybe I am calculating it wrong, but I thought the surface area = the integral that sums the circumferences of the circles from x = 1 to x = t.
The circumference of the circle at x is 2 PI (1/x), whose integral with respect to x is 2 PI ln (x), so the integral over x = 1 to t is 2 PI ln t - 2 PI ln 1 = 2 PI (ln t - 0) = 2 PI ln t.
Then again, that method doesn't seem to work when I try it with calculating the surface area of a sphere...

watch the video @13:30 and you will see what's going wrong with your calculation method.
Personal website: www.ijmp.org
charliepatrick
charliepatrick
Joined: Jun 17, 2011
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February 21st, 2021 at 2:20:24 PM permalink
Occasionally The [London] Times publishes a maths puzzle. This one is NOT easy (although easier than some they have had) and usually you need to find the first key and just work your way through (rather like suduko). Using programming/excel is optional, but I know people who can solve it just using a calculator. (Typically they have clues where you have primes, squares etc and know some of the digits. By pure fluke there's only one answer that fits. Luckily this one isn't quite that devious.)

I have found this link with a PDF

(If this doesn't work the link is here https://www.crosswordsolver.org/forum/799536/listener-4647/ but the page might have spoilers lower down as it's a forum.)
DogHand
DogHand
Joined: Sep 24, 2011
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February 21st, 2021 at 2:47:54 PM permalink
Quote: teliot

watch the video @13:30 and you will see what's going wrong with your calculation method.



The surface area for a Gabriel's Horn of length t is the integral from 1 to t of 2*pi*sqrt(1-1/x^4)/x.

If for simplicity we let a == sqrt(1+1/t^4) and b == sqrt(2) then this integral is

Area = (pi/2)*(2*(b-a) + ln((1+a)*(b-1)/((-1+a)*(b+1))))

As shown in the video,

Volume = pi*(1-1/t)

Set Area =Volume and calculate t.

As ThatDonGuy indicated earlier in the thread, the only value of t that makes them equal is t=1

Dog Hand

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