# Differential Spherical Shells - Triple Integrals

• thetasaurus
In summary: Therefore, all you care about is the first, which is what you got.In summary, the conversation discusses a problem involving a hollow solid sphere and a particle traveling outward from the center. The conversation also mentions using calculus to find the force of gravity between the inner and outer radius of the sphere. The conversation concludes with a discussion about the limit process and justifying omitting certain terms in the calculation. Ultimately, the correct solution involves getting rid of some of the higher powers of dr.
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.

Last edited:
thetasaurus said:
$dV=\frac{4\pi}{3}(3r^2dr+3rdr^2+dr^3)$
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.

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?

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.

## 1. What is a differential spherical shell?

A differential spherical shell is a 3-dimensional shape that is formed by rotating a circle about a fixed axis in 3-dimensional space. It is called a "differential" shell because it is made up of infinitely thin layers, each with a different radius.

## 2. How are triple integrals used in differential spherical shells?

Triple integrals are used to find the volume of a differential spherical shell. The integration is done with respect to three variables - the radius, the angle of rotation, and the height of the shell. This allows us to calculate the volume of the shell by adding up the infinitesimal volumes of each layer.

## 3. What is the formula for calculating the volume of a differential spherical shell?

The formula for calculating the volume of a differential spherical shell is V = ∫∫∫ r^2 sinθ dr dθ dϕ, where r is the radius, θ is the angle of rotation, and ϕ is the height of the shell.

## 4. How does the orientation of the differential spherical shell affect the triple integral?

The orientation of the differential spherical shell affects the limits of integration for the triple integral. Depending on the orientation, the limits for each variable may change, resulting in a different calculation for the volume of the shell.

## 5. What are some real-world applications of differential spherical shells and triple integrals?

Differential spherical shells and triple integrals are used in various fields such as physics, engineering, and mathematics. They are used to calculate the volume of objects such as spheres, water tanks, and even the Earth's atmosphere. They are also used in the design of bridges, tunnels, and other structures that require precise calculations of volume and mass.

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