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 Post subject: Calculus (16) A Multiple IntegralPosted: Tue, 30 Dec 2008 10:12:29 UTC
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Evaluate:

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 Post subject: Re: Calculus (16) A Multiple IntegralPosted: Tue, 30 Dec 2008 19:52:17 UTC
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commutative wrote:
Evaluate:

Spoiler:
Easy to prove

and so it is just a matter of fiddling about... to give

where H_n=1+(1/2)+(1/3)+...+(1/n) is the n-th Harmonic number.

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 Post subject: Re: Calculus (16) A Multiple IntegralPosted: Wed, 31 Dec 2008 00:57:41 UTC
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outermeasure wrote:
commutative wrote:
Evaluate:

Spoiler:
Easy to prove

and so it is just a matter of fiddling about... to give

where H_n=1+(1/2)+(1/3)+...+(1/n) is the n-th Harmonic number.

your answer is correct. it can be proved by induction that:

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 Post subject: Re: Calculus (16) A Multiple IntegralPosted: Wed, 31 Dec 2008 08:57:40 UTC
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Oops, it should be f(x) rather than f(y).

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 Post subject: Posted: Wed, 31 Dec 2008 19:30:16 UTC
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Joined: Fri, 9 Nov 2007 00:04:38 UTC
Posts: 128
Here's an interesting little variation on this I found while doing the Putnam exam (the above problem was about half of the 2008 B2) (sorry if this is hi-jacking, but it's basically the same problem but requires a bit more thinking):
Evaluate

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"The reason that every major university maintains a department of mathematics is that it is cheaper to do this than to institutionalize all those people."

Some people live their lives by the KISS principle (Keep It Simple Stupid), but I prefer a slightly different one:
Make It Fun And Complicate It Further, or MIFACIF. Ever try and find the volume of a cube in spherical coordinates?

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 Post subject: Posted: Wed, 31 Dec 2008 22:45:50 UTC
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Joined: Fri, 9 Nov 2007 00:04:38 UTC
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cody wrote:
As for your integral,

I get:

Yep, that's it. I'm curious if you used the Cauchy formula for repeated integration or did the calculations from scratch?

ps: A really fun one is finding the length of y=1/x in the coordinate system (note that it's not an orthogonal coordinate system, i.e. it's metric tensor has off-diagonal entries)

_________________
"The reason that every major university maintains a department of mathematics is that it is cheaper to do this than to institutionalize all those people."

Some people live their lives by the KISS principle (Keep It Simple Stupid), but I prefer a slightly different one:
Make It Fun And Complicate It Further, or MIFACIF. Ever try and find the volume of a cube in spherical coordinates?

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 Post subject: Posted: Thu, 1 Jan 2009 19:21:52 UTC
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cody wrote:
I hope no one gets peeved about this, but regarding simplethinkers signature.

I have derived the volume of a unit cube in spherical coordinates.

I don't believe this. The range for phi looks wrong (going only up to arctan(sqrt(2)) maximum....). I believe you will need to split into more than 1 triple integrals if you want nice limits, e.g. doing rho last you will have basically 3 parts:- one for 0<rho<1 (an eighth of sphere), one for 1<rho<sqrt(2) (spherical non-geodesic hexagon, with 3 sides geodesic), one for sqrt(2)<rho<sqrt(3) (spherical non-geodesic triangles). Those spherical integrals can be done with Gauss-Bonnet.

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 Post subject: Posted: Thu, 1 Jan 2009 19:40:22 UTC
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cody wrote:
Hey simplethinker. I just lumbered through it from scratch. Similar to the method in

the first problem. May I ask you to demonstrate the Cauchy method you mentioned. I

was unfamiliar with it until you mentioned it.

The Cauchy Integral formula is a complex analysis means of doing things like integration.

It states that

Where is any closed, simple, rectifiable curve in the complex numbers which contains the point .

For more information, I recommend you to the wikipedia article on the matter.

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