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Re: ContourPlot3D and NIntegrate

  • To: mathgroup at smc.vnet.net
  • Subject: [mg100989] Re: ContourPlot3D and NIntegrate
  • From: antononcube <antononcube at gmail.com>
  • Date: Fri, 19 Jun 2009 20:47:02 -0400 (EDT)
  • References: <h0bu15$8lu$1@smc.vnet.net>

Hi,

If you use the following defintion

VV[x_?NumericQ, y_?NumericQ, z_?NumericQ] :=
 Module[{u, phi},
  NIntegrate[(x^2 + y^2 + z^2)^(-3)*u*phi, {u, 0, 1}, {phi, 0, 2 Pi},
   Method -> {Automatic, SymbolicProcessing -> 0}, PrecisionGoal ->
6]]

then the plot computation

ContourPlot3D[  VV[x, y, z], {x, 1, 2}, {y, 1, 2}, {z, 1, 2}] //
AbsoluteTiming

takes 44 seconds on my 2 year old MacBook Pro

Note that I have put the symbolic processing time to zero. You can
further see if using smaller precision gives satisfactory and faster
results for you.


Anton Antonov



On Jun 5, 4:09 pm, Henning Heiberg-Andersen
<henning.heibergander... at gmail.com> wrote:
> Hi,
>
> I don't understand what goes wrong in this sequence:
>
> *
>
> In[1]:=VV[x_,y_,z_]:=
>
> Module[{u,phi},NIntegrate[(x^2+y^2+z^2)^(-3)*u*phi,{u,0,1},{phi,0,2Pi}]
>
> ]
>
> In[2]:=
>
> ContourPlot3D[VV[x,y,z],{x,1,2},{y,1,2},{z,1,2}],
>
> AND THEN:
>
> NIntegrate::inumr: The integrand \[NoBreak](phi$12637167
> u$12637167)/(x^2+y^2+z^2)3\[NoBreak] has evaluated to non-numerical value=
s
> for all sampling points in the region with boundaries
> \[NoBreak]{{0,1},{0,6.28319}}\[NoBreak].
> =87<http://reference.wolfram.com/mathematica/ref/message/NIntegrate/inu=
mr...>
>
> Can anybody help?
>
> Sincerely,
>
> Henning Heiberg-Andersen



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