1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Moment of inertia with triple integrals

  1. Jun 18, 2010 #1
    so i have to find the moment of inertia of a solid cone given by the equations z = ar and z = b by using a triple integral. The density of the cone is assumed to be 1. so the integral looks like ∫ ∫ ∫ r^2 dV. so first i did it with dV = rdrdθdz with limits r (from 0 to z/a), θ (from 0 to 2pi), and z (from 0 to b) and i got the answer (pi)b^5 / (10a^4) which seems to be right. now i am trying to do the triple integral instead with dV = rdzdrdθ. i am having trouble deciding the limits now. for z, i know that r and θ are to be kept constant so my limits for z are from ar to b. then my limits for r are from 0 to z/a and my limits for θ are from 0 to 2pi. however after doing the integrations, my answer turned out messy involving a's, b's, and z's which shouldn't be in the final answer. how do i set up the limits correctly when integrating with respect to z first, then r and theta afterwards?
     
  2. jcsd
  3. Jun 18, 2010 #2
    Ok, your first try r:0~z/a, θ:0~2pi, z:0~b. And first integrate r then z(or last).
    why z:0~b, not z:ar~b?
    The reason is we first integrate r, and r:0~z/a, or not?
    Now, your secend try z:ar~b, θ:0~2pi, and r:0~z/a.
    We first integrate z, why r:0~z/a appears z again?
    z/a isn't a constant.
     
  4. Jun 18, 2010 #3
    Woah, you call this an introductory physics problem!?!?

    OK, so I don't know the equation for the moment of inertia of a cone off the top of my head. But if you do your math right, then your answer from the cylindrical coordinates integral will be right, but your answer from the spherical coordinates integral will be wrong. You've got this funny equation in physics which says,

    [tex]I = \int r^2dm[/tex]

    But what they don't tell you is that the "r" is the perpendicular distance from the axis of rotation (which you usually set to be the z axis). If you do it in spherical coordinates, you can't just assume that the r in the above integrand is the same as the r from spherical coordinates. Instead, you need to use:

    [tex]I = \int [rsin(\theta)]r^2sin(\theta)drd\theta d\phi[/tex]

    Multiplying r by [itex]sin(\theta)[/itex] will take the component of [itex]\vec{r}[/itex] in the xy plane, which is what you want.

    As for the limits of integration to use in spherical coordinates...I'm going to have to think about that one for a second. Spherical coordinates are a very unnatural way to describe a cone.
     
  5. Jun 18, 2010 #4
    To 3#
    For this problem used cylindrical coordinate maybe better.
     
  6. Jun 18, 2010 #5
    oh i got it now. the limits on r should be from 0 to b/a. setting that limit gave me the same answer as i got before. thanks for the help everyone! :smile:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook