So I'm trying to evaluate the triple integral $$\displaystyle \iiint \limits_{R} \displaystyle \frac{1}{((x-a)^2+y^2+z^2)^{1/2}} \mathrm dV$$ for $a>1$ over the solid sphere $0 \leq x^2 + y^2 + z^2 \leq 1$.

Apparently, there's an interpretation that I should be able to draw from this to. Not too sure what it is.

So the first thing that came to mind when I saw the integral was to apply spherical coordinates, but this doesn't make the denominator of the integrand any less messy.

Using spherical coordinates, the integrand becomes

$$\displaystyle \iiint \limits_{R} \displaystyle\frac{1}{(\rho^2-2a\rho\sin\phi\cos\theta+a^2)^{1/2} } \rho^2\sin\phi \space\mathrm d\rho\mathrm d\phi\mathrm d\theta$$ (I haven't bothered to add the bounds yet), which doesn't look that much more friendly.

Any support for this question would be appreciated.

  • 2
    $\begingroup$ i think using cylindrical coordinates $(x,r,\theta)$ brings this down into a 2d integral with an extra factor of $2\pi$ out front $\endgroup$ – gt6989b Mar 26 '18 at 21:00
  • 1
    $\begingroup$ I don't remember this stuff well at all, so beware, but you're looking at (a constant times) the mean electric potential of a charge at $(a,0,0)$ over the unit sphere. The potential will be a harmonic function over the unit sphere, because the charge is outside it ($a>0$), and the mean value property of harmonic functions would imply that the integral is proportional to the value of the potential at the center of the sphere. $\endgroup$ – stochasticboy321 Mar 26 '18 at 21:00
  • $\begingroup$ @Kaynex it's the denominator that doesn't look friendly $\endgroup$ – user98937 Mar 26 '18 at 21:05
  • $\begingroup$ My apologies, misunderstood $\endgroup$ – Kaynex Mar 26 '18 at 21:10
  • 1
    $\begingroup$ @user98937 you represent the $(y,z)$ as $(r,\theta)$ $\endgroup$ – gt6989b Mar 26 '18 at 21:19

Orienting ourselves better will simplify things.

ie. make $\phi = 0$ align with the x-axis.

$\iiint \frac {\rho^2\sin\phi}{(\rho^2 -2a\rho\cos\phi + a^2 )^\frac 12}\ d\rho\ d\phi\ d\theta$

But I think that cylindrical will be easier.

$y = r\cos\theta\\ z = r\sin\theta\\ x = x$

$\iiint \frac {r}{(r^2 + (x-a)^2)^\frac 12}\ dr\ dx\ d\theta$

$\int_0^{2\pi}\int_{-1}^{1}\int_0^{\sqrt {1-x^2}} \frac {r}{(r^2 + (x-a)^2)^\frac 12}\ dr\ dx\ d\theta$

$\int_0^{2\pi}\int_{-1}^{1} ((r^2 + (x-a)^2)^\frac 12|_0^{\sqrt {1-x^2}}\ dx\ d\theta\\ \int_0^{2\pi}\int_{-1}^{1} (1 - 2ax + a^2)^\frac 12 - |x-a| \ dx\ d\theta$


  • $\begingroup$ I didn't get the $1/4$ constant when I integrated and evaluated $\displaystyle \frac {r}{(r^2+(x-a)^2)^{1/2}}$ with respect to $r$. I got a constant of 1. $\endgroup$ – user98937 Mar 27 '18 at 21:03
  • $\begingroup$ My mistake, thanks... $\endgroup$ – Doug M Mar 27 '18 at 21:08
  • $\begingroup$ no worries...also, how would I go about integrating $\left|{x-a}\right|$? $\endgroup$ – user98937 Mar 27 '18 at 21:09
  • $\begingroup$ It is given that $a>1$ and $-1\le x\le 1$ so $|x-a| = a-x$ $\endgroup$ – Doug M Mar 27 '18 at 21:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.