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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?

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Downvote not appreciated. I'm asking for help here. What should I have done differently? –  jnm2 Nov 5 '11 at 16:13
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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. –  wsc Nov 5 '11 at 16:29
    
Oh, so this is in the wrong place? –  jnm2 Nov 5 '11 at 16:46
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Added a +1 to remove the negative vote. –  Antillar Maximus Nov 5 '11 at 18:45
    
I'm going to try migrating this to math.SE, since it is more of a mathematical question. –  David Z Nov 6 '11 at 0:36
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migrated from physics.stackexchange.com Nov 6 '11 at 0:37

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1 Answer

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.

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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? –  jnm2 Nov 5 '11 at 16:18
    
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. –  jnm2 Nov 5 '11 at 16:26
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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. –  Luboš Motl Nov 5 '11 at 18:50
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@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. –  mmc Nov 5 '11 at 20:50
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@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: $\int_0^H dz \int_{-R}^{R} dy \int_{-\sqrt{R^2-y^2}}^{+\sqrt{R^2-y^2}} dx\,\rho\,(x^2 + y^2)$ –  mmc Nov 5 '11 at 20:55
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