How do I determine the measure for a volume integral?

Question

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

Comments

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

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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.

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

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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

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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

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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

Answer 1

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

Comments

@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

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@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

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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.

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

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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: $\int_0^H dz \int_{-R}^{R} dy \int_{-\sqrt{R^2-y^2}}^{+\sqrt{R^2-y^2}} dx\,\rho\,(x^2 + y^2)$

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

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@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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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

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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