Differential Spherical Shells - Triple Integrals

[thetasaurus]

Homework Statement

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.

Homework 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


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

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.

 

[haruspex]

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

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.

 

[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?

 

[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.