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Topic: Infinite, or finite area? (Read 487 times) |
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Benoit_Mandelbrot
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Infinite, or finite area?
« on: Jan 16th, 2004, 8:37am » |
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We have the function ln(x). Is there a finite area above the curve and below the x-axis from 0 to 1, and what is the arc length? How about revolving it around the axises? Is there finite volume, and how about surface area?
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| « Last Edit: Jan 19th, 2004, 6:01am by Benoit_Mandelbrot » |
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ThudnBlunder
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Re: Infinite, or finite area?
« Reply #1 on: Jan 16th, 2004, 9:15am » |
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See also Gabriel's Horn.
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| « Last Edit: Jan 16th, 2004, 9:16am by ThudnBlunder » |
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Benoit_Mandelbrot
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Re: Infinite, or finite area?
« Reply #2 on: Jan 16th, 2004, 9:34am » |
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That has to do with 1/x though, not ln(x), so...
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| « Last Edit: Jan 16th, 2004, 10:00am by Benoit_Mandelbrot » |
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Sameer
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Re: Infinite, or finite area?
« Reply #3 on: Jan 16th, 2004, 9:44am » |
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I wonder who is going to post the answer so that we can start poking our noses (well i don't consider myself in league of this Uber Puzzlers) but this is too easy for me too  ... so will wait until the author gives a nod 
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Sameer
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Re: Infinite, or finite area?
« Reply #4 on: Jan 16th, 2004, 9:45am » |
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wow that was my 100th post yippeee 
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John_Gaughan
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Re: Infinite, or finite area?
« Reply #5 on: Jan 16th, 2004, 4:11pm » |
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Edit: I fixed some of the math. The gerbil wasn't running full speed that day and I erroneously assumed ex crossed the y-axis at e instead of 1. /Edit
on Jan 16th, 2004, 8:37am, Benoit_Mandelbrot wrote:| We have the function ln(x). Is there a finite area above the curve and below the x-axis from 0 to 1, and what is the arc length? How about revolving it around the axises? Is there finite volume, and how about surface area? |
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Area
:: Figuring this out as a logarithm is a bit tough for me, so I chose to flip it around and find the area of ex from -[infty] to 0.
A = lim (a -> -[infty]) [int] (from a to 0) of ex dx
A = lim (a -> -[infty]) (e0 - ea)
A = 1 - 0
The area is 1. ::
Arc Length
:: Rather than figure this one out, as I don't remember the answer off the top of my head, my guess is that the arc length is infinite. Even though the domain has a lower bound, the range is unbounded. ::
Volume
:: This is similar to the area problem, except we need to revolve it around the X axis. Since there are no odd shapes, I choose the disk method. The radius of any disk is ex. The area is [pi]r2. The area is then [pi]e2x. When you integrate this over the length of the function, you get volume.
V = (1/2) [pi] lim (a -> -[infty]) [int] (from a to 0) of e2x 2dx
V = lim (a -> -[infty]) ([pi]/2) (e0 - e2a)
V = ([pi]/2) (1 - 0)
V = [pi]/2
Surface Area
:: [hide]Rather than explain, since it is similar to the volume, here it goes:
A = 2[pi] lim (a -> -[infty]) [int] (from a to 0) of ex dx
A = 2[pi] lim (a -> -[infty]) (e0 - ea)
A = 2[pi] (1 - 0)
A = 2[pi]
::
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| « Last Edit: Jan 29th, 2004, 6:00am by John_Gaughan » |
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Sameer
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Re: Infinite, or finite area?
« Reply #6 on: Jan 16th, 2004, 4:32pm » |
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agh... u can integrate from a to 0 as lnx and e^x are inverses of each other giving area to be 1 and not 1 + e
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John_Gaughan
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Re: Infinite, or finite area?
« Reply #7 on: Jan 16th, 2004, 5:28pm » |
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I didn't remember how to do that and couldn't find it when I looked it up. I know it is possible I just didn't remember the rule.
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John_Gaughan
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Re: Infinite, or finite area?
« Reply #8 on: Jan 29th, 2004, 6:07am » |
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on Jan 16th, 2004, 4:32pm, Sameer wrote:| agh... u can integrate from a to 0 as lnx and e^x are inverses of each other giving area to be 1 and not 1 + e |
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You're right, I fixed the solution. For some reason that day I forgot that logarithms/exponentials cross the axis at 1, not the base. Anyway, I don't remember how to integrate a logarithm at x=0, although I suppose I could do (int) ey dy from -[infty] to 0, which is the exact same thing in terms of the result. So my answer still stands 
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