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Mathematics Calculus Wolfram Language

Solving a triple integral with assumptions

Sinval Santos

Posted 3 years ago

In a physics problem, I came across the following triple integral: exp = (r Sin[\[Alpha]] (Cos[\[Theta]] Cos[\[Phi]] Sin[\[Alpha]] - Cos[\[Alpha]] Sin[\[Theta]]))/( 4 (-2 + Sqrt[2]) \[Pi]^2 (p^2 + r^2 - 2 p r Cos[\[Alpha]] Cos[\[Theta]] - 2 p r Cos[\[Phi]] Sin[\[Alpha]] Sin[\[Theta]])^(3/2)); Assuming[0 < r < 1 && 0 < \[Alpha] < \[Pi]/4 && \[Alpha] != \[Theta] && 0 < \[Theta] < \[Pi]/4 && 0 < \[Phi] < 2 \[Pi], \!\( \SubsuperscriptBox[\(\[Integral]\), \(0\), \(2 \[Pi]\)]\( \SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Pi]/4\)]\( \SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]exp\ \SuperscriptBox[\(p\), \(2\)] Sin[\[Alpha]]\ \[DifferentialD]p \ \[DifferentialD]\[Alpha] \[DifferentialD]\[Phi]\)\)\)] I tried to perform each integration separately, but got no result. Is there any transformation or procedure, that I am not aware of, to accomplish this task? There is also an attached file. Grateful, Sinval Attachments: Triple integral.nb

In a physics problem, I came across the following triple integral:

exp = (r Sin[\[Alpha]] (Cos[\[Theta]] Cos[\[Phi]] Sin[\[Alpha]] - 
     Cos[\[Alpha]] Sin[\[Theta]]))/(
  4 (-2 + Sqrt[2]) \[Pi]^2 (p^2 + r^2 - 
     2 p r Cos[\[Alpha]] Cos[\[Theta]] - 
     2 p r Cos[\[Phi]] Sin[\[Alpha]] Sin[\[Theta]])^(3/2));
Assuming[0 < r < 1 && 0 < \[Alpha] < \[Pi]/4 && \[Alpha] != \[Theta] &&
   0 < \[Theta] < \[Pi]/4 && 0 < \[Phi] < 2 \[Pi], \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(2  \[Pi]\)]\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Pi]/4\)]\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]exp\ 
\*SuperscriptBox[\(p\), \(2\)] Sin[\[Alpha]]\ \[DifferentialD]p \
\[DifferentialD]\[Alpha] \[DifferentialD]\[Phi]\)\)\)]

I tried to perform each integration separately, but got no result. Is there any transformation or procedure, that I am not aware of, to accomplish this task? There is also an attached file. Grateful, Sinval

POSTED BY: Sinval Santos

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Sinval Santos

Posted 2 years ago

Thanks again for your attention Hans. I also tried it with numerical integration and I got a lot of error messages. Sincerely, sinval

POSTED BY: Sinval Santos

Hans Dolhaine

Hans Dolhaine, retired

Posted 2 years ago

POSTED BY: Hans Dolhaine

Sinval Santos

Posted 2 years ago

Hello, really this problem is almost impossible, and thank you for your contribution Hans. Regards, Sinval

POSTED BY: Sinval Santos

Hans Dolhaine

Hans Dolhaine, retired

Posted 2 years ago

Hello, I am afraid your problem is next to impossible. Here is a remark, which is unfortunately essentially useless, as you do not integrate over theta. Your integrand exp = (r Sin[\[Alpha]] (Cos[\[Theta]] Cos[\[Phi]] Sin[\[Alpha]] - Cos[\[Alpha]] Sin[\[Theta]]))/( 4 (-2 + Sqrt[2]) \[Pi]^2 (p^2 + r^2 - 2 p r Cos[\[Alpha]] Cos[\[Theta]] - 2 p r Cos[\[Phi]] Sin[\[Alpha]] Sin[\[Theta]])^(3/2)); a bit modified e1 = 4 Pi^2 (Sqrt[2] - 2) exp p^2 Sin[\[Alpha]] Now let hh = p^2 + r^2 - 2 p r Cos[\[Alpha]] Cos[\[Theta]] - 2 p r Cos[\[Phi]] Sin[\[Alpha]] Sin[\[Theta]] j1 = Sin[\[Alpha]]/(4 (Sqrt[2] - 2) p Pi^2 ) D[hh^(-1/2), \[Theta]] // Simplify j1 is your integrand as derivative of theta. exp - j1 // Simplify But as already mentioned integration over theta is no issue and this is not a step to the solution of your problem.

Hello, I am afraid your problem is next to impossible.

Here is a remark, which is unfortunately essentially useless, as you do not integrate over theta.

Your integrand

exp = (r Sin[\[Alpha]] (Cos[\[Theta]] Cos[\[Phi]] Sin[\[Alpha]] - 
            Cos[\[Alpha]] Sin[\[Theta]]))/(
      4 (-2 + Sqrt[2]) \[Pi]^2 (p^2 + r^2 - 
             2 p r Cos[\[Alpha]] Cos[\[Theta]] - 
             2 p r Cos[\[Phi]] Sin[\[Alpha]] Sin[\[Theta]])^(3/2));

a bit modified

e1 = 4 Pi^2 (Sqrt[2] - 2) exp p^2 Sin[\[Alpha]]

Now let

hh = p^2 + r^2 - 2 p r Cos[\[Alpha]] Cos[\[Theta]] -  2 p r Cos[\[Phi]] Sin[\[Alpha]] Sin[\[Theta]]

j1 = Sin[\[Alpha]]/(4 (Sqrt[2] - 2) p Pi^2  ) D[hh^(-1/2), \[Theta]] // Simplify

j1 is your integrand as derivative of theta.

exp - j1 // Simplify

But as already mentioned integration over theta is no issue and this is not a step to the solution of your problem.

POSTED BY: Hans Dolhaine

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