# How do I determine the measure for a volume integral?

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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
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

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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 (10,248 points)
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
Sorry, this was the integral of $r^2=x^2+y^2+z^2$. If I calculate the correct $r_z^2$, the squared distance from the $z$-axis, it's only $x^2+y^2$ so I only get two terms out of three. Due to the spherical symmetry, it's $2/3$ of the result above, so that the moment of inertia is $(2/5)MR^2$ for a solid ball.

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user Luboš Motl
You did it again: how did you get $dV = r^2drd\Omega$? That's kind of my entire question. If I don't understand this very trivial thing, that's most likely why I'm asking. Thank you very much for your evaluation of my competence. :-) That aside, even if you aren't able or willing to explain the thing you think is very simple that I've been asking about, I would appreciate being pointed in the right direction. How to find the correct substitution for the $dV$ of an arbitrary coordinate system with an arbitrary axis of rotation. Like a cylinder tilted at $\frac{\pi}{3}$ radians.

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user jnm2
@jnm2 You can calculate the moment of inertia of any object around any given axis using any coordinate system. The calculations can be simpler in some coordinate systems for some problems, but using a specific coordinate system is only a matter of convenience.

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user mmc
@mmc If it's only a matter of convenience, how could I use the Cartesian system for a cylinder? Not that you'd need to, but I just want to understand this.

This post imported from StackExchange Mathematics at 2014-06-02 20:29 (UCT), posted by SE-user jnm2
@jnm2 The volume elements in different coordinate systems are related by a change of integration variables. It's a generalization of the usual integration by substitution.

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

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