# Differential Spherical Shells - Triple Integrals

1. Jan 16, 2013

### thetasaurus

1. The problem statement, all variables and given/known data

Despite the fact that this started as an extended AP Physics C problem, I turned it into a calc problem because I (sort of) can. If it needs to be moved please do so.

There is a hollow solid sphere with inner radius b, outer radius a, and mass M. A particle of mass m travels outward from the center. Find the force of gravity between b and a.

Disclaimer: I know that there are other/simpler ways to do this. I just want to know at what stage my thinking/math went wrong.

2. Relevant equations

$F_{g}(r)\frac{GmM(r)}{r^{2}}$

$\rho=\frac{M}{\frac{4\pi}{3}(a^3-b^3)}$

I am trying to find M(r), the mass internal to the particle at radius r (aka the mass contributing to the force of gravity due to the shell theorem).

The correct function should be

$F_{g}(r)=\frac{Gm(r^3-b^3)}{r^{2}(a^3-b^3)}$

3. The attempt at a solution

I'm going to sum up the volume spherical shells from radius b to r, then multiply by ρ to get the mass.

The internal radius of this shell is r>b, and the external radius is r+dr

$dV=\frac{4\pi}{3}((r+dr)^3-r^3)$

$dV=\frac{4\pi}{3}(3r^2dr+3rdr^2+dr^3)$

Once expanded I realized I would have to use multiple integrals to integrate the exponentiated dr's

$V=\frac{4\pi}{3}\left(3\int_b^{r}{r^2dr}+3\iint_b^{r}{rdrdr}+\iiint_b^{r}{drdrdr}\right)$

Which, when carried through, yields the result

$M(r)=\frac{M}{a^3-b^3}\left(5/3r^3-1/2b^3-b^2r-1/2br^2\right)$

It's close, but it's not

$M(r)=\frac{M}{a^3-b^3}\left(r^3-b^3\right)$

I believe I did the integration right, so where did I go wrong? I thought it was weird finding a volume by integrating a volume, but it seemed like it should work. Can anyone who can shed light on why this is wrong? I don't need you to tell me I should have done another method, I need to know theoretically why I can't do this. Thanks in advance.

P.S... If anyone could help me with the code for my integral expression it'd be much appreciated because those tiny sumas look terrible.

Last edited: Jan 16, 2013
2. Jan 17, 2013

### haruspex

In the limit (which is what calculus is about), the higher powers of dr become irrelevant. Just throw them away. There is no multiple integral here.

3. Jan 17, 2013

### thetasaurus

As you said, I get the correct solution when I get rid of the second and third dr's. That error is probably due to the fact that I don't actually know what I'm doing, but how can I mathematically justify omitting them? When is it ever okay to just get rid of stuff?

4. Jan 17, 2013

### haruspex

Calculus, whether integration or differentiation, is a limit process. At some point, you let the dx's tend to zero. If you do it right, you get a leading term that does not, in general, go to zero.
In this case, you are summing terms like r2dr, rdr2, dr3. The number of such terms, over some finite range A, varies roughly as A/dr. The terms will therefore sum, separately, to things like AR2m ARdr, Adr2, where R is a bound on the value of r across the range. In the limit, the second and third of those vanish.