Re: help with integration
- To: mathgroup at smc.vnet.net
- Subject: [mg121652] Re: help with integration
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Sat, 24 Sep 2011 22:32:13 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- Reply-to: hanlonr at cox.net
Integrate[
r^2*ArcCos[x/r] - x*Sqrt[r^2 - x^2], {x, Sqrt[3] y,
Sqrt[r^2 - y^2]}, Assumptions -> {y > 0, r > 2 y}] // Simplify
y^3/3 + (2/3)*r^2*Sqrt[r^2 - 3*y^2] + y^2*Sqrt[r^2 - 3*y^2] -
r^2*y*(1 + Sqrt[3]*ArcCos[(Sqrt[3]*y)/r]) + r^2*Sqrt[r^2 - y^2]*
ArcSec[r/Sqrt[r^2 - y^2]]
Assuming[{y > 0, r > 2 y},
Integrate[
r^2*ArcCos[x/r] - x*Sqrt[r^2 - x^2], {x, Sqrt[3] y,
Sqrt[r^2 - y^2]}] // Simplify]
y^3/3 + (2/3)*r^2*Sqrt[r^2 - 3*y^2] + y^2*Sqrt[r^2 - 3*y^2] -
r^2*y*(1 + Sqrt[3]*ArcCos[(Sqrt[3]*y)/r]) + r^2*Sqrt[r^2 - y^2]*
ArcSec[r/Sqrt[r^2 - y^2]]
% === %%
True
Bob Hanlon
---- Salman Durrani <dsalman96 at yahoo.com> wrote:
=============
I am trying to do the following integration:
Integrate[r^2*ArcCos[x/r] - x*Sqrt[r^2 - x^2], {x, Sqrt[3] y, Sqrt[r^2 - y^2]} ]
When I try, Mathematica comes back to be with a very long conditional expression.
How do I tell mathematica that r is positive, real in the above equation not complex.
Thanks
Kahless