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: Areal moment of inertia of a disk integral question

  1. Jul 31, 2008 #1
    Hi, I know the following calculation is incorrect but have been unable to find the error.

    The calculation is to integrate

    [tex]I = \int z^2 \sqrt{r^2-z^2}dz.[/tex]

    Trying u-substitution:

    [tex]u = \sqrt{r^2 - z^2}[/tex]

    [tex]z^2 = r^2 - u^2[/tex]

    [tex]2z dz = - 2udu[/tex]

    [tex]dz = \frac{-udu}{z} = \frac{-udu}{\sqrt{r^2 - u^2}}, [/tex]

    the integral becomes

    [tex]I = \int (r^2 - u^2) u \left(\frac{-udu}{\sqrt{r^2 - u^2}}\right) = -\int \sqrt{r^2 - u^2}~u^2 du = - I~?[/tex]

    [tex]I=-I \Rightarrow I = 0[/tex] but this is the areal moment of inertia integral for a disk (if boundaries of integration were added from 0 to R) and by converting it to an integral in terms of [tex]\theta[/tex] it's easy to show that it equals [tex]\pi R^4/4[/tex]. Would be much obliged if someone would point out the error for me.

    Thank you,
  2. jcsd
  3. Jul 31, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    I believe the error comes from ignoring you the bounds of integration. For example, if you integrate from z = 0...r then you would integrate from u = r...0, which takes care of the negative sign.
  4. Jul 31, 2008 #3
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook