6
$\begingroup$

After years, I still find myself having trouble really internalizing the meaning of various differentials in integrals—specifically, when they come about via reasoning regarding physical phenomena. When I come back for review, I fall prey to the same problems I had when originally learning the material. It's not that I don't necessarily understand the correct solution, but I more often than not don't underestand why the incorrect one is incorrect.

Example 1: Consider the process of deriving the moment of inertia of a thin disk of mass $M$ and radius $R$. My immediate thought is "I'd like to derive this by summing the moment of inertia of concentric rings of various radii."

$$I = \sum m_i r_i^2$$ ... where $m_i$ is the mass of a particle on the ring and $r_i$ will be the distance of the particle from the center of the ring (or, the ring's radius).

The moment of inertia of one point on the circumference of the ring of radius $r$ is:

$$dI_{ring} = dm ~ r^2 = \left(\frac{m}{\pi R^2}\right)r^2$$

... but shouldn't there be a $dr$ somewhere? It's about here where I flounder around trying to figure out why I don't have a $dr$, what $dr$ really means (so that I can insert it into the appropriate place), whether or not I should actually have $dr$ or $dm$, etc. Then, I ask myself "... what am I really summing over—what would the bounds of integration be?" (From $0$ to $2\pi r$, because I'm summing over tiny piece of the circumference? Am I confusing myself by using $dr$ and not, say, $dS$?)

Example 2: Let's say I get past all that, and find the moment of inertia of a thin hoop of radius $r$ to be $I = \left( \frac{2mr}{R^2}\right)r^2$. I'd now like to sum these hoops to form a disk. So...

$$dI_{disk} = \left( \frac{2mr}{R^2}\right)r^2$$

... where I'd like $r$ to vary from $0$ to $R$. Again, what about $dr$ (or $d\text{[whatever]}$)? Well, I know I want $r$ to vary, so... my integral should look something like...

$$\int_0^R dI$$

... right?

$$\int_0^R \left( \frac{2mr}{R^2}\right)r^2 = \int_0^R \frac{2mr^3}{R^2}$$ ... $dr$?

This will go on for a long time, until I inevitably make a post on Stack Exchange asking for help.


I've read through many examples, and have walked myself through many derivations that fall into this category—and I understand them fully when I do. The problem is, the knowledge that I gain from doing this doesn't seem to generalize. I can't seem to intuit a kind of general rule of thumb for these types of problems, and it's particularly frustrating.

Will someone please help elucidate what this god damned mysterious differential is in such a way that, perhaps, provides a general rule of thumb?

$\endgroup$
  • 1
    $\begingroup$ Maybe this helps: dont think of dm as a point. Think of it as a small volume element. This volume element you approximate by the (for now) finite dr times r dphi. The 'shrinking to zero' gets done automaticaöly when u write down the integral. $\endgroup$ – lalala Apr 14 '17 at 19:11
7
$\begingroup$

The issue is that while formally integrals like $$I = \int dI = \int r^2 dm$$ make sense, they're useless for actual calculations because we don't have a parametrization for $I$ or $m$ -- we can't integrate over these variables. What we do almost all the time is to introduce a parametrization, and then integrate with respect to the parameter.

For example, we might define $I(r)$ as the moment of inertia, only counting masses with radius $r$ or less. In that case, we really want $I(R)$, where $R$ is the radius of the object, and $$I = \int dI = \int_0^R \frac{dI}{dr} \, dr.$$ Here, $dI/dr$ is the rate of change of the moment of inertia as we count larger and larger radii, so that $dI/dr\, dr = dI$ is the contribution of the moment of inertia due to a thin slice of thickness $dr$.

That means that in your case, the right expression for $dI$ is $$dI = r^2 dm = r^2 (\rho \cdot 2 \pi r dr) = \frac{2Mr^3}{R^2} dr$$ where $\rho$ is the density of the object. In the second expression, we're implicitly thinking of $dm$ as the amount of mass in a little radius $dr$. If you're uncomfortable with the second equality, think of $m(r)$ as a function that gives the amount of mass inside a radius $r$ (like we defined $I(r)$). Then we're simply replacing $dm$ with $(dm/dr)dr$ by the chain rule.

$\endgroup$
  • $\begingroup$ This is exactly what I was looking for, man. $\endgroup$ – AmagicalFishy Apr 14 '17 at 23:07
  • $\begingroup$ I never saw the value of the "$\mathrm{d}m$" constructions, especially in introductory courses in which the student has never seen such a thing. The only thing he or she has ever integrated over is a coordinate (and possibly a parametrization of a path). $\endgroup$ – garyp Jan 31 '18 at 14:27
  • $\begingroup$ @garyp I think it's a perfectly fine notation to use -- if you used something else, it would almost certainly take up more space, and it nicely suggests $dm = (dm/dr)dr$ by 'fractions'. But I agree that it needs to be explained explicitly when introduced. $\endgroup$ – knzhou Jan 31 '18 at 14:54
0
$\begingroup$

Here's how I do it, maybe it will work for you.

Integral is essentially a continuous summation. I need to consider what set of elements I am summing actually represent, and how they vary.

There's (as ever) more than one way to solve a problem, and in terms of your example it would be equally possible to solve it either
a) summing/integrating all elements from a particular direction then summing all of those in every direction OR
b) summing/integrating all elements at a particular radius in every direction and then summing all of those for every radius.

These schemes are chosen to ensure all relevant elements are included in the integration.

The easiest summation/integration is usually the one for which the summed elements are all similar. That would be choice b) here as the size of all elements on a defined radius is constant.

$\endgroup$
0
$\begingroup$

I don't know if I'll be able to fully answer your question, but for the example you've posted, this is how I've usually approached it.

When you want to calculate $dm$ of a small circular ring (where we will then integrate over the whole disk), we want to find the small portion of mass of that ring. This is given by

$$dm = \rho(r) * 2 \pi r *dr$$

The reason this is the formula is because we have the mass density of the disk ($\rho$), and then the "area" that is created from a small disk is the circumference of the disk ($2\pi r$) multiplied by the thickness of the disk ($r$). You can think of it like unfurling the small ring into a rectangle with sides $2\pi r$ and $dr$. From there, you can then integrate from $0$ to $R$.

In general, I find that, if you're going to replace on differential by another, you will usually have some other differential tagged on. What I mean by this is that, when I replaced $dm$ in my expression, I get another differential $dr$ at the end.

In the case of the disk, the way I think about what the differential should be is through thinking about what I am summing over. Here, we're using infinitesimally thin rings, and they all have the same "thickness" $dr$. However, the radius $r$ for each ring certainly is changing, which is why we also have a factor of $r$ in there as well.

I hope this helps (at least, for this example). My suggestion would be to keep on going through examples, and after a while the differentials part will become clearer.

$\endgroup$

protected by Qmechanic Apr 14 '17 at 19:49

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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