Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,853 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

How do I determine the measure for a volume integral?

+ 2 like - 0 dislike
9 views

If $I = \int r^2 dm$, how do I set up an integral over the volume of any object? I can't use any assumptions about symmetry or shortcuts because the goal is to rotate around an arbitrary axis.

$m = \rho v$ so $I = \rho\int r^2 dv$, but for a cube $v = xyz$ so $dv = yz dx + zx dy + xy dz$. I guess?
How do I go from that to $I = \rho\int\int\int r^2 dx dy dz$?

What is actually going on? Why don't I replace $dv$ with $yz dx + zx dy + xy dz$ and get $I = \rho\int yzr^2 dz + \rho\int zxr^2 dy + \rho\int xyr^2 dz$?

Or for a cylinder, $v = \pi(r_o^2 - r_1^2)h$ and $dr = \pi(r_o^2 - r_1^2)dh + 2\pi hr_o dr_o - 2\pi hr_i dr_i$. How do I set up the volume integral with this?

To clarify: I'm asking for the general principle. When I think of a shape, such as an arbitrarily rotated cylinder, I need to know what to do to set up the volume integral.

How does this work?

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user jnm2
asked Nov 5, 2011 in Mathematics by jnm2 (20 points) [ no revision ]
Downvote not appreciated. I'm asking for help here. What should I have done differently?

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user jnm2
Not my downvote, but I would point out that calculating rotational inertia is a bit of a tenuous connection for what is essentially just a question about calculus.

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user wsc
Oh, so this is in the wrong place?

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user jnm2
Added a +1 to remove the negative vote.

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user Antillar Maximus
I'm going to try migrating this to math.SE, since it is more of a mathematical question.

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user David Z

1 Answer

+ 3 like - 0 dislike

It is not true that ${\rm d}V$ in the volume integral $\int {\rm d}V$ means ${\rm d}(xyz)$. Instead, it means $\int{\rm d}x\,{\rm d}y\,{\rm d}z$: the infinitesimal volume ${\rm d}V$ is the same thing as the product of the three infinitesimal "linear factors": it makes absolutely no sense to go from the infinitesimal ${\rm d}V$ to the "whole" $V$ and then "differentiate it back".

A triple integral is just a sequence of three integrations in a row. You may first integrate over $z$, then over $y$, then over $x$. Alternatively, you may often use more convenient coordinates – axial, spherical, or others – and make the calculation more tractable. Many of those triple integrals are exactly solvable, others are not. It's a purely mathematical question which of them may be expressed in terms of simple functions.

In these integrals, while calculating the moment of inertia, you may write down the general formula $$ I = \int {\rm d} V\,\rho\,r^2 $$ where $\rho$ is a mass density at the given point (where the small volume ${\rm d}V$ is located). If $\rho$ is equal to zero except for an interval, you may replace the integral above, which was assumed to be from $-\infty$ to $+\infty$ so that the whole space is covered, by the integral over the interval where $\rho$ is nonzero.

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user Luboš Motl
answered Nov 5, 2011 by Luboš Motl (10,248 points) [ no revision ]
Most voted comments show all comments
I know what it means. I'm asking why. It's fine to be handed a magic transformation $I = \int \rho r^2 dV$ -> $I = \int\int\int \rho r^2 dx dy dz$, but that doesn't help me learn anything. What do I do to get there on my own? How I can set up a triple integral for the volume of a cylinder, or sphere, or parabolic mirror?

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user jnm2
To clarify: I'm asking for the general principle. When I think of a shape, such as an arbitrarily rotated cylinder, I need to know what to do to set up the volume integral.

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user jnm2
Dear @jnm2, the first equation in your comment above isn't a "magic transformation": it's a totally trivial identity. $\int$ is the integral sign but it's still a sum of infinitesimal pieces. The infinitesimal pieces are proportional to $dV$ and one may choose cubic shapes of $dV$ to make it clear that it may be written as $dx\,dy\,dz$. I don't understand what's your problem. If you can't understand this simple thing about the volume integral, you should give up studying physics quantitatively. Also, I wrote you how you can calculate the triple integral for any shape you mentioned.

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user Luboš Motl
Also, your question as well as your comments below my answer make it spectacularly self-evident that you don't understand what either of these symbols mean, in a striking contradiction with your assertion.

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user Luboš Motl
Let me calculate the moment of inertia of a ball of radius $R$ here. I choose spherical coordinatets so $dV=r^2\,dr\,d\Omega$. The integral over $d\Omega$ gives $4\pi$, the integral of $r^4dr$ (there was $r^2$ from the moment) from $0$ to $R$ gives $R^5/5$: I can factorize them here. So the result is $4\pi \rho R^5/5=3/5 MR^2$. Less symmetric shapes are more complicated but it's pure maths.

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user Luboš Motl
Most recent comments show all comments
@jnm2 Let's say we want to get $I_z$ for a circular cylinder of height $H$ and radius $R$ that is symmetric around the z-axis: $\int_0^H dz \int_{-R}^{R} dy \int_{-\sqrt{R^2-y^2}}^{+\sqrt{R^2-y^2}} dx\,\rho\,(x^2 + y^2)$

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user mmc
@mmc, that was the biggest help. The light bulb goes on...

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user jnm2

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$y$\varnothing$icsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...