Re: help with double integration

*To*: mathgroup at smc.vnet.net*Subject*: [mg121777] Re: help with double integration*From*: Peter Pein <petsie at dordos.net>*Date*: Sat, 1 Oct 2011 03:09:15 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <j63t52$6dv$1@smc.vnet.net>

Am 30.09.2011 10:03, schrieb Salman Durrani: > Hello > > I am trying to use mathematica to do the following double integration: > > Integrate[ > Integrate[ > r^2*ArcCos[x/r] - x*Sqrt[r^2 - x^2], {x, Sqrt[3] y, > Sqrt[r^2 - y^2]}], {y, 0, r/2}] > > The correct answer is: > r^4(pi^2/144 + pi/(24Sqrt{3}) - 1/32) > > However I am unable to get mathematica to produce the correct result. I have tried splitting into two integrations and using the assumptions option (suggested by newsgroup): > > 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 > > The above produces the result for the inner integration but then > > Integrate[ > 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]], {y, 0, > r/2}] // Simplify > > does not produce any result. > > Am I missing something fundamental here. Any help would be appreciated. > > Thanks > > Kahless > Hi, if you express ArcCos in terms of Log it works well: In[1]:= Integrate[TrigToExp[r^2*ArcCos[x/r] - x*Sqrt[r^2 - x^2]], {y, 0, r/2}, {x, Sqrt[3]*y, Sqrt[r^2 - y^2]}, Assumptions -> r > 2*y > 0] Out[1]= (1/288)*(-9 + 4*Sqrt[3]*Pi + 2*Pi^2)*r^4 and In[2]:= Expand[(1/288)*(-9 + 4*Sqrt[3]*Pi + 2*Pi^2)*r^4 - r^4*(Pi^2/144 + Pi/(24*Sqrt[3]) - 1/32)] Out[2]= 0 shows that this is the expected result. hth, Peter