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Volume integration pb

  • To: mathgroup at smc.vnet.net
  • Subject: [mg104083] Volume integration pb
  • From: Fred Bartoli <""@news.free.fr>
  • Date: Sun, 18 Oct 2009 05:24:29 -0400 (EDT)
  • Reply-to: myname_with_a_dot_inbetween at free.fr

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.


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