Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

3 Answers 3

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 $I_x=I_y=I_z=0$.

share|improve this answer
add comment

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 $a/b$.

share|improve this answer
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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