# What would a calculus author have to say on ##\int r^2dm##?

• I
• DrBanana
DrBanana
TL;DR Summary
Trying to learn about the right path to understanding 'differentials'
So I've been searching around for rigorous explanations for things like ##dx## in physics, I'm not looking to fully commit myself to reading the relevant literature at the moment but just want to know what I'll have to do in order to understand. Perhaps I'll make a separate thread about that.

In most introductory physics textbooks, the method for calculating the moment of inertia of a rigid body is like this: you consider it as a collection of really small particles, so the approximation for the moment of inertia is ##I = \sum m_i r_i ^2 ##
As the number of particles go to infinity, and consequently each one's mass goes to zero, ##I=\int r^2 dm##.

So, some questions:
1. This isn't limited to moment of inertia but other derivations in physics in general: I was only taught to thing of integrals as areas under curves, but what about the cases (like in moment of inertia) where we are adding up small things, and not directly computing the area under some curve? What is the rigorous way to do this?

2. In a single variable calculus textbook, the 'x' in ##\int f dx## just refers to the integral's variable. However in the above integral expression that can't be the case as the radial distance r doesn't directly depend on m. To reconcile this, I found this answer on stackexchange: https://physics.stackexchange.com/a/550227/259006

So my question for this point is, if I confronted the author of a single variable calculus textbook (e.g. Courant), what would he have to say? That is, is it possible to give a satisfactory meaning to ##\int r^2 dm## using what one learns in standard single variable calculus (no hyper reals, no differential forms etc), or does one have to learn advanced methods (by the way Courant does actually calculate moment of inertia in his book but only in a special case where the mass is confined to a plane region bounded by a curve).

3. On wikipedia I find this expression for the moment of inertia:
Relevant section: https://en.wikipedia.org/wiki/Moment_of_inertia#Motion_in_a_fixed_plane
Even though I don't know what triple integrals are yet, I can get a feel if this and it's more satisfying than simply ##\int r^2 dm##. But, where does it come from? I don't know of any book that expresses it like this.

P.S. I now realise that there is a second part to Courant's book that might deal with stuff like this, but I would still like to know the answers to my questions.

Honestly? There are so many things that should be addressed in your post that I could hold an entire lecture. But I cannot hold an entire lecture here. You have three questions, and according to our rules, these should be three different threads. Unfortunately, every single one of your questions is by itself an invitation to hold a lecture. You cannot learn physics from Wikipedia, from PF, or from SE. The notation ##dx## is highly dependent on context. Physicists use it quite sloppy, sometimes as something infinitesimally small, sometimes as a coordinate, sometimes as differential, and sometimes as pure information about which variable is to integrate. Or is it primarily all about inertia?

My recommendation: narrow it all down to what you really want to understand, start a separate thread, or read a summary of a subject in one of our INSIGHTS BLOGs.

pinball1970, bhobba, SammyS and 4 others
fresh_42 said:
Honestly? There are so many things that should be addressed in your post that I could hold an entire lecture. But I cannot hold an entire lecture here. You have three questions, and according to our rules, these should be three different threads. Unfortunately, every single one of your questions is by itself an invitation to hold a lecture. You cannot learn physics from Wikipedia, from PF, or from SE. The notation ##dx## is highly dependent on context. Physicists use it quite sloppy, sometimes as something infinitesimally small, sometimes as a coordinate, sometimes as differential, and sometimes as pure information about which variable is to integrate. Or is it primarily all about inertia?

My recommendation: narrow it all down to what you really want to understand, start a separate thread, or read a summary of a subject in one of our INSIGHTS BLOGs.
Sorry I didn't know about that particular rule; I'll make a separate thread later. I gave three insights blogs a cursory glance, the one about hyper real numbers, the one where a derivative is looked at ten different ways (https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/) and the one about a gauge integral.

Alright,

in the equation

$$\int r^2 \,dm$$

it is true that ##r## is not a function of ##m## BUT ##dm## absolutely depends on ##r## so the integral is ultimately in terms of ##r##.

A single integral is over a line (or a curve)
A double integral is over an area
A triple integral is over a volume

So let's think about what each of these cases would look like

for a line (say a thin rod)
$$\int r^2 \,dm = \int r^2 \lambda \left(r \right) \, dr$$ where ##\lambda \left( r \right)## is a linear mass density function

for an area (maybe a thin plate)

$$\int \int r^2 \,dm = \int \int r^2 \sigma \left(r \right) \, dA$$ where ##\sigma \left( r \right)## is an area mass density function.

for a volume

$$\int \int \int r^2 \,dm = \int \int \int r^2 \rho \left(r \right) \, dV$$ where ##\rho \left( r \right)## is an volume mass density function.

A differential , in a general sense, is the approximate change along the fitting object dimension -wise. In one dimension, given## y=f(x) ##differentiable, the differential## dy## is the approximate change of## y ## along the tangent line, so that## dy=f'(x)dx##. Then dy is not the actual change ##\Delta f(x)=\Delta y=f(x+h)-f(x)##, but rather the change along the tangent line. Similar for Real-valued functions of several variables, where we approximate with tangent planes or , in general, with hyperplanes.

WWGD said:
A differential , in a general sense, is the approximate change along the fitting object dimension -wise. In one dimension, given## y=f(x) ##differentiable, the differential## dy## is the approximate change of## y ## along the tangent line, so that## dy=f'(x)dx##. Then dy is not the actual change ##\Delta f(x)=\Delta y=f(x+h)-f(x)##, but rather the change along the tangent line. Similar for Real-valued functions of several variables, where we approximate with tangent planes or , in general, with hyperplanes.
Is this only true for spatial dimensions?

DrBanana said:
Is this only true for spatial dimensions?
Not sure what you mean. What other setting do you have in mind?

DrBanana said:
Is this only true for spatial dimensions?
No, for all dimensions, often time.

bhobba
WWGD said:
A differential , in a general sense, is the approximate change along the fitting object dimension -wise. In one dimension, given## y=f(x) ##differentiable, the differential## dy## is the approximate change of## y ## along the tangent line, so that## dy=f'(x)dx##. Then dy is not the actual change ##\Delta f(x)=\Delta y=f(x+h)-f(x)##, but rather the change along the tangent line. Similar for Real-valued functions of several variables, where we approximate with tangent planes or , in general, with hyperplanes.
So, even without this definition, it is possible to show that ##\int dy= \int f'(x)dx##, where dy and dx just tell you with respect to which variable we're integrating. My question is, is it possible to think of the ##dy## in the integral as a linear approximation along a tangent curve like you said, or do the two necessarily have to be different? I only ask because when solving a separable differential equation, you initially have dy's and dx's, which can be thought of differentials as you stated, but then you integrate both sides (or more formally just use the fundamental theorem of calculus, so it just seems you're integrating both sides), in that case do the meanings of the dy's change? Of course, I know that when solving ##\frac{dy}{dx}=h(x)p(y)##, we can just do the following:
##\frac{d}{dx}\int \frac{1}{p(y)} \frac{dy}{dx} dx= h(x) ##
## \int \frac{1}{p(y)} \frac{dy}{dx} dx= \int h(x) dx##
## \int \frac{1}{p(y)} dy = \int h(x) dx## (by the substitution rule)

But I would really like to think of doing this in terms of the differentials (approximations using tangents) you mentioned.

fresh_42 said:
No, for all dimensions, often time.

Differentials can have two meanings in modern calculus.

First, as what are called infinitesimals. That approach is explained here:

Elementary Calculus – An infinitesimal approach by Keisler. Freely available here: https://www.math.wisc.edu/~keisler/calc.html

The other is purely definitional based on notation. The usual notation dy/dx = f(x) suggests the following definition

dy = f(x)dx

It is a notation that is sometimes useful. For example:

∫dy = ∫f(x)dx.

Change of variable is easy in this notation:

dy/dx = y'(x). dy = y'(x)dx. dy = (dy/dx)dx. So the change of variable formula is then easy ∫f(y)dy = ∫f(y(x))dy = ∫f(y(x))(dy/dx)dx.

I would explain it this way in a beginning calculus course, leaving the full proof to a real analysis course.

Thanks
Bill

I don't remember the details now, but , IIRC, the Inverse function theorem was used to justify the manipulation of differentials as fraction. Will try to remember.

• Calculus
Replies
10
Views
196
• Introductory Physics Homework Help
Replies
13
Views
2K
• Mechanics
Replies
10
Views
1K
• Calculus
Replies
4
Views
879
• Calculus
Replies
22
Views
3K
• Calculus
Replies
3
Views
1K
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
325
• Calculus and Beyond Homework Help
Replies
7
Views
902