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MathGroup Archive 2006

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Re: Noob question about definite integration

  • To: mathgroup at smc.vnet.net
  • Subject: [mg72220] Re: Noob question about definite integration
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Fri, 15 Dec 2006 07:05:49 -0500 (EST)
  • References: <eloret$o8e$1@smc.vnet.net>

Mathematica is right!

f[r_] := (h^2 - r^2)^3

symb = Integrate[(4/3)*Pi*((h^2 - r^2)^3)^3, {r, 0, h}]
(262144*h^19*Pi)/692835

In order to justify Mathematica's analytic result compare the following
tables

TableForm[(N[#1, 20] & )[(symb /. h -> #1 & ) /@ Range[10]]]

TableForm[(NIntegrate[(4/3)*Pi*((#1^2 - r^2)^3)^3, {r, 0, #1},
WorkingPrecision -> 40, PrecisionGoal -> 20] & ) /@ Range[10]]

Regards
Dimitris


Ï/Ç some guy named Dave Ýãñáøå:
> Hey all,
>
> I have a function that I want to integrate. It's basically a smoothing
> function over a vector space, and I want to normalize it. The function itself
> looks like this:
>
>    f(r) = (h^2 - r^2)^3
>
> where r is the scalar length ||x|| of some displacement vector x, and h is a
> constant. We are really only interested in values of r in [0,h], and for 2D
> vectors. However, I have some research literature involving 3D vectors that I
> want to compare against, just for sanity.
>
> So, if x is a 3D vector, then the surface of the sphere at radius r should be
> (4/3)(f(r))^3.  We should then be able to get the definite integral from
> Mathematica by doing something like:
>
>    Integrate[ 4/3 * Pi( (h^2 - r^2)^3 )^3, {r, 0, h}]
>
> When I do so, I get back
>
>    262144 pi h^19 \ 692835
>
> However, the literature I have says that the integral should yield
>
>    64 pi h^9 / 315
>
> So, have I screwed something up? Is the literature wrong, perhaps?
> 
> Thanks.
> 
> Dave


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