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Topic: prove ln function (Read 361 times) |
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fatball
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prove ln function
« on: Dec 10th, 2007, 10:48am » |
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I = lnlnln ( x^2/a^2 + y^2/b^2 + z^2/c^2 ) where x^2 + y^2 + z^2 <=1 , and a,b,c are constant.
(a)show I = lnlnln ( x^2/b^2+y^2/c^2+z^2/a^2 ) = lnlnln ( x^2/c^2+y^2/a^2+z^2/b^2 )
(b)hence find I by spherical coordinates........
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Obob
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Re: prove ln function
« Reply #1 on: Dec 10th, 2007, 1:49pm » |
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Assuming that you mean to integrate ln ln ln (x^2/a^2+y^2/b^2+z^2/c^2) over the region x^2+y^2+z^2 <= 1, this problem doesn't make a whole lot of sense. For x, y, z all close to zero, x^2/a^2+y^2/b^2+z^2/c^2 is close to zero, so that its logarithm is negative. Then we can't take the logarithm again.
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towr
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Re: prove ln function
« Reply #2 on: Dec 10th, 2007, 2:03pm » |
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Maybe it's a bad transcription, and those ln's are suppose to be in(t)?
Integrating over x^2/a^2 + y^2/b^2 + z^2/c^2 within the sphere x^2 + y^2 + z^2 <= 1 might qualify as easy. And certainly a) would certainly be simple enough (we can just switch axes).
Personally I wouldn't opt to use spherical coordinates to do the integration, I'd rather use the known volume of a sphere and go from there (unless that's disallowed).
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| « Last Edit: Dec 10th, 2007, 2:05pm by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Obob
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Re: prove ln function
« Reply #3 on: Dec 10th, 2007, 4:45pm » |
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Ah yes, that makes much more sense towr.
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