2
$\begingroup$

$F_{z}=\iiint_{\Omega} \frac{z-a}{\left[x^{2}+y^{2}+(z-a)^{2}\right]^{\frac{3}{2}}} d v$

$\Omega$: ImplicitRegion[x^2 + y^2 + z^2 <= r^2, {x, y, z}]

MMA code:

Clear["Global`*"];
reg = ImplicitRegion[x^2 + y^2 + z^2 <= r^2, {x, y, z}];
f[x_, y_, z_] := 
  \!\(TraditionalForm\`\((z\  - \ 
        a)\)/\((x^2\  + \ y^2\  + \ \((z\  - \ a)\)^2)\)^\((3/2)\)\);
Integrate[f[x, y, z], Element[{x, y, z}, reg], 
 Assumptions -> r > 0 && a > 0 ]

($Aborted)

Or,

Clear["Global`*"];
reg = Ball[{0, 0, 0}, r];
f[x_, y_, z_] := (z - a)/(x^2 + y^2 + (z - a)^2)^(3/2);
Integrate[f[x, y, z], Element[{x, y, z}, reg], 
 Assumptions -> r > 0 && a > 0]

($Aborted)

Both of them can't get the result:

$- \cdot \frac{4 \pi R^{3}}{3} \cdot \frac{1}{a^{2}}$

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2
  • 3
    $\begingroup$ Recommendation: Eliminate useless factors that have nothing to do with the problem, here $G$ and $\rho_0$. $\endgroup$ Mar 2, 2022 at 6:35
  • $\begingroup$ Thanks, done. @David G. Stork $\endgroup$
    – lotus2019
    Mar 3, 2022 at 2:37

1 Answer 1

3
$\begingroup$

After switching to the spherical coordinates, Mathematica is able to crack it:

Clear["Global`*"];
Integrate[t^2*Sin[\[Theta]]*(z - a)/(x^2 + y^2 + (z - a)^2)^(3/2) /. {x -> 
t*Cos[\[Phi]]*Sin[\[Theta]], y -> t*Sin[\[Phi]]*Sin[\[Theta]], 
z -> t*Cos[\[Theta]]}, {t, 0, r}, {\[Phi], 0, 2*Pi}, {\[Theta], 0, Pi}, 
Assumptions -> r > 0 && a >= -r && a <= r]

ConditionalExpression[-((4*a*Pi)/3), a > 0]

Integrate[t^2*Sin[\[Theta]]*(z - a)/(x^2 + y^2 + (z - a)^2)^(3/2) /. {x -> 
t*Cos[\[Phi]]*Sin[\[Theta]], y -> t*Sin[\[Phi]]*Sin[\[Theta]], 
z -> t*Cos[\[Theta]]}, {t, 0, r}, {\[Phi], 0, 2*Pi}, {\[Theta],0,Pi}, Assumptions->r>0&&a > r]

-((4 \[Pi] r^3)/(3 a^2))

Maple 2021 confirms the former and the latter.

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1
  • $\begingroup$ Thanks a lot!! @user64494 $\endgroup$
    – lotus2019
    Mar 3, 2022 at 1:22

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