Re: Volume integration pb

*To*: mathgroup at smc.vnet.net*Subject*: [mg104092] Re: Volume integration pb*From*: Leonid Shifrin <lshifr at gmail.com>*Date*: Mon, 19 Oct 2009 07:10:27 -0400 (EDT)*References*: <200910180924.FAA17110@smc.vnet.net>

Fred, your integral is ill-defined. In the first of the methods you list, this shows up when you look at the particular point z = 0 - you get a logarithmic divergence. In the second method, this shows up for a particular point r = 0 - you get a quadratic divergence in the term -2 z^2/(3 (r^2 + z^2)^(5/2). Most explicitly, you see the problem when you go from cylindrical to polar coordinates where you expression becomes dtheta dphi d rho*rho^2*rho^2*Sin[theta]*(Sin[theta]^2 - 2 Cos[theta]^2)/(3 rho^5), with rho = Sqrt[z^2+r^2], and theta being the usual polar angle (such that r = rho*Sin[theta], h = rho*Cos[theta]). Integrating over theta in the small neibourhood of the origin from 0 to Pi gives zero, while integration over rho is logarithmically divergent (at the origin) integral d rho/rho. Thus, we have a 0*Infinity situation. What this means is that the value of the integral will depend on the way it is computed - that is, the procedure which we use to compute the integral becomes a part of the definition. One may say that the integral must be somehow "regularized" at the origin, then computed, then regularization removed. Since it diverges only logarithmically, removing the regularization gives a finite answer, but different in each case. Your two methods of computing this integral can be considered two different regularizations, each giving a different meaning to the integral. The bottom line: neither result is "correct" or "wrong", because the integral is ill - defined. Similar situation happens in computing various anomalies in Quantum Field Theory, for example. In that case, the way you regularize the integral is a part of the definition of the model (determines the physics of the problem). Which regularization is correct in your case must be decided by you. Regards, Leonid On Sun, Oct 18, 2009 at 2:24 AM, Fred Bartoli <@news.free.fr> wrote: > Integrating (r^2 - 2 z^2)/(3 (r^2 + z^2)^(5/2)) over a cylinder of R > radius and -h/2 to +h/2 height. > > (* Integrating first over radius then over height : *) > Integrate[2\[Pi] r (r^2-2 z^2)/(3 > (r^2+z^2)^(5/2)),{r,0,R},Assumptions->R>0&&z>0]; > int1=Integrate[%,{z,-h/2,h/2},Assumptions->R>0&&h>0] > > > (* Then integrating first over height then over radius : *) > Integrate[(r^2 - 2 z^2)/(3 (r^2 + z^2)^(5/2)), {z, -h/2, h/2}, > Assumptions -> h > 0 && r > 0] > int2 = Integrate[2 \[Pi] r %, {r, 0, R}, > Assumptions -> h > 0 && R > 0] > > int2-int1//Simplify > > -> 4 Pi/3 > > Am I missing something? > > > -- > Thanks, > Fred. > >

**References**:**Volume integration pb***From:*Fred Bartoli <""@news.free.fr>