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