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