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How can I compute this triple integral?

enter image description here

I tried this but it has been running for a long time and not return result yet.

Integrate[
 Boole[0 <= x^2 + y^2 + z^2 <= 4]*((y - 1)^3/(x^2 + y^2 + z^2 + 1)), {x, 0, 2}, {y, -2, 
  2}, {z, -2, 2}]
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    $\begingroup$ NIntegrate will give a numerical value (-16.755...) $\endgroup$
    – mikado
    Apr 16 at 20:26
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    $\begingroup$ @mikado that seems right. Someone got -16π/3, why is it hard for Mathematica to get the exact result? $\endgroup$
    – hana
    Apr 16 at 20:29
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    $\begingroup$ @hana If I had to guess (and I would be), because "work over this prism and then test membership in a hemisphere" is harder to work with than "work over this sphere and then test membership in a hemisphere", and that the latter likely makes Mathematica more likely to try certain methods (e.g. it knows a sphere is relevant from the domain specification, so will try spherical coordinates etc.). The domain as a prism doesn't lead it to realize a sphere is present, as that's been coded into the function rather than the domain, so it doesn't try those methods. But, again: just a guess. $\endgroup$
    – Zibadawa
    Apr 17 at 11:53

2 Answers 2

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Integrate[
 Boole[x >= 0]*((y - 1)^3/(x^2 + y^2 + z^2 + 1)), {x, y, 
   z} ∈ Ball[{0, 0, 0}, 2]]

-((16 π)/3)

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Changing to spherical polar coordinates gives an answer that agrees numerically

π Integrate[(r^2 Cos[θ] (r Sin[θ] - 1)^3)/(r^2 + 1), {r, 0, 2}, {θ, -π/2, π/2}]
(* -((16 π)/3) *)

But doesn't explain why Mathematica struggles with the Cartesian formulation.

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