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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


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    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
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