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 05:28:46 -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.