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Guten Tag allerseits,
zur Kenntnis eine Korrespondenz mit support.
Mit freundlichen Grüssen
Hans Dolhaine
_________________________________
VTR-TS
Phone: +49-211-797-4809
Fax: +49-211-798-1853
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E-Mail: Hans.Dolhaine@XXXXXXX.com
----- Weitergeleitet von Hans Dolhaine/KGaA/HENKEL am 22.10.2003 14:23
-----
support@wolfram.c
om An: Hans.Dolhaine@XXXXXXX.com
16.10.2003 05:30 Kopie:
Entscheidung Thema: [TS 23099]--Re:integrals
erforderlich ?
|--------|
| [ ] ja |
|--------|
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From: Hans.Dolhaine@XXXXXXX.com
Date: Wed, 15 Oct 2003 11:28:20 +0100
Subject: integrals
To: support@XXXXXXX.com
hello to you,
I encountered a problem which may be a bug, but I'm not quite sure. Gauss
states, that integrals over surfaces may be transfomed to integrals over
volumes according to
surfaceintegral [ a . dS ] = volumeintegral [ div . a dV ]
Please have a look at the attached notebook. I have a vector aa whose
divergence simplifies to 1/r, so the volumeintegral (over the whole sphere)
is rather simple). Trying to do the surfaceintegral yields a result which
is quite different. When I try numerical integration I get a warning that
the integrand(s) is (are) singlular at some points and no evaluation takes
place. That is ok. But the symbolic integration, if the sequence of doing
the integrals is chosen appropriately, gives without any complaints some
values. Generally I trust these, but this seems to be a bug: no reference
to the possible singularities is given and the final result is totally
different from the volumeintegral.
Is this a bug or have I overlooked something?
Kind regards
Hans Dolhaine
(See attached file: support.nb)
_________________________________
VTR-TS
Phone: +49-211-797-4809
Fax: +49-211-798-1853
Mobile: 0171 97 17 049
E-Mail: Hans.Dolhaine@XXXXXXX.com
=======================================================
Hello,
Thank you for taking the time to send us this report.
You are correct that some of the results from Integrate
in your example are wrong. In particular, the result
from each of the last three Integrate examples in the
notebook that was included with your message should be
2Pi/3. This can be verified using numerical integration.
In[]:= 2 Pi NIntegrate[Evaluate[i1[[1]] /. R -> 1],
{t, 0, Pi/2, Pi}, SingularityDepth -> 999]
Out[]= 2.0944
In[]:= NIntegrate[Evaluate[i1[[2]] /. R -> 1],
{p, 0, Pi, 2 Pi}, {t, 0, Pi/2, Pi}]
Out[]= 2.0944
In[]:= NIntegrate[Evaluate[i1[[3]] /. R -> 1],
{p, 0, 3, 2 Pi}, {t, 0, Pi/2, Pi}]
Out[]= 2.0944
The behavior of these examples in the current version
of Mathematica (Version 5.0) has changed, but remains
incorrect. I have filed these examples so that this
behavior can be investigated.
I have not reviewed the mathematics or the analysis
that you described, and do not know if that analysis
is correct.
Unfortunately, there are no known workarounds for this
error in the Integrate function. We apologize for any
difficulties caused by this error.
Technical Support
Wolfram Research
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<<attachment: support.nb>>
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