# Moment, spheroid, charge redistribution

Let $$I_k:= c \int_{\mathbb R^3} (3x_k'^2-r'^2) \,\,\,d^3 x'$$ where ${r'}^2={x'}_1^2+{x'}_2^2+{x'}_3^2$ and $c$ is a constant = density of charge (uniform) in the body.

Suppose this integral is evaluated for a solid spheroid $${x^2\over a^2}+{y^2\over a^2}+{z^2\over b^2}\le 1$$

Now suppose we hollow out the spheroid and place all the charge uniformly over the shell. Is there a good way of seeing whether $I_k$ remains constant or changes?

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As a rule "output values" change when the "input" is changed. If in a particular setup they don't we have a "theorem". - In the case at hand the question would heavily profit from a thorough editing. –  Christian Blatter Oct 22 '12 at 11:25
Thank you, @ChristianBlatter . I am not entirely sure what you mean... –  Gregory Oct 22 '12 at 11:27

We have $I_x=I_y=-I_z/2=\langle x^2-z^2\rangle$. Moving the charge further out increases the average value of the squared coordinates. In the case of an oblate spheroid, $\langle x^2-z^2\rangle\gt0$, and in the limit $a\gg b$ we have $I_x=I_y=-I_z/2\approx\langle x^2\rangle$, so moving the charge to the surface increases the absolute value of all three integrals. In the case of a prolate spheroid, $\langle x^2-z^2\rangle\lt0$, and in the limit $b\gg a$ we have $I_x=I_y=-I_z/2\approx-\langle z^2\rangle$, so again moving the charge to the surface increases the absolute value of all three integrals, but with the opposite sign. The border case is the sphere, for which $I_x=I_y=I_z=0$.

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I'd reformulate the question as follows:

Consider a constant charge density $c$ on the rotationally symmetric ellipsoid $$B:\quad {x^2+y^2\over a^2}+ {z^2\over b^2}\leq 1\ .$$ We are interested in the two integrals (edited after seeing Joriki's answer) $$J:=c\int\nolimits_B \bigl(3x^2-(x^2+y^2+z^2)\bigr)\ {\rm d}(x,y,z)=c\int\nolimits_B (x^2-z^2)\ {\rm d}(x,y,z)$$ and $$J':=c\int\nolimits_B \bigl(3z^2-(x^2+y^2+z^2)\bigr)\ {\rm d}(x,y,z)=-2 J\ .$$ How do their values change when the same total charge is uniformly (with respect to surface measure ${\rm d}\omega$) distributed over the surface $\partial B$?

The answer is not at all obvious and may depend in a "nonelementary" way on the ratio $a/b$.

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Perhaps you could parameterize the volume of integration with some parameter (call it $\lambda$). Then take $dI_k/{d \lambda}$, and apply Leibniz's rule on the integrand and the limits of integration. If its zero, well, there you go. The only issue I can think of is that an thin hollow shell (but still with some thickness) and putting the charge entirely on the skin of the surface may not be equivalent situations.

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Do you disagree with my answer? It seems you think it's possible that the derivative might be zero? If so, please point out where you think the argument fails. –  joriki Oct 22 '12 at 12:46
I haven't worked it out in detail. I was just answering his original question - "how can I tell if this changes upon doing something?" –  clustro Oct 22 '12 at 13:07