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Re: help with double integration

  • To: mathgroup at smc.vnet.net
  • Subject: [mg121779] Re: help with double integration
  • From: "Dr. Wolfgang Hintze" <weh at snafu.de>
  • Date: Sat, 1 Oct 2011 03:09:37 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <j63t52$6dv$1@smc.vnet.net>

"Salman Durrani" <dsalman96 at yahoo.com> schrieb im Newsbeitrag 
news:j63t52$6dv$1 at smc.vnet.net...
> 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
>
Hello,

sometimes Mathematica needs some manual help. Then it works fine as we 
shall see.

Let us start with your last integral which Mathemacia could not solve 
directly, and let's split in into three parts

1)

Integrate[
y^3/3 + 2/3 r^2 Sqrt[r^2 - 3 y^2] + y^2 Sqrt[r^2 - 3 y^2] , {y, 0,
r/2},Assumptions->r>0]

((9 + 4*Sqrt[3]*Pi)*r^4)/96

no problem!

2)

- Integrate[
r^2 y (1 + Sqrt[3] ArcCos[(Sqrt[3] y)/r]) , {y, 0,
r/2}] // Simplify // InputForm

-((9 + 7*Sqrt[3]*Pi)*r^4)/144

no problem!

3)
Integrate[
r^2 Sqrt[r^2 - y^2] ArcSec[r/Sqrt[r^2 - y^2]], {y, 0,
r/2}] // Simplify

Integrate[r^2*Sqrt[r^2 - y^2]*
ArcSec[r/Sqrt[r^2 - y^2]], {y, 0, r/2}]

Problem!
In order to solve 3) we change the variable of integration from y to 
y/r
(notice the extra factor r in front of the expression. This takes care 
of the change in dy)

3.1)
Integrate[
r (r^2 Sqrt[r^2 - y^2] ArcSec[r/Sqrt[r^2 - y^2]]) /. y -> r t, {t, 0, 
1/2}] //
Simplify

Integrate[r^2*Sqrt[-(r^2*(-1 + t^2))]*
ArcSec[r/Sqrt[-(r^2*(-1 + t^2))]], {t, 0, 1/2}]

The problem persists!
Now tell Mathematica that r>0

3.2)
Integrate[
r (r^2 Sqrt[r^2 - y^2] ArcSec[r/Sqrt[r^2 - y^2]]) /. y -> r t, {t, 0, 
1/2},
Assumptions -> r > 0] // Simplify

((-9 + 3*Sqrt[3]*Pi + Pi^2)*r^3)/144

Hurray! Now adding the three terms gives the desired result.

3.3) Unfortunately, simply imposing r>0 in 3) is not sufficient.

Hope this helps.

Regards,
Wolfgang




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