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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



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