Areal moment of inertia of a disk integral question

[musemonkey]

Hi, I know the following calculation is incorrect but have been unable to find the error.

The calculation is to integrate

I = \int z^2 \sqrt{r^2-z^2}dz.

Trying u-substitution:

u = \sqrt{r^2 - z^2}

z^2 = r^2 - u^2

2z dz = - 2udu

dz = \frac{-udu}{z} = \frac{-udu}{\sqrt{r^2 - u^2}},


the integral becomes


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~?

I=-I \Rightarrow I = 0 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 \theta it's easy to show that it equals \pi R^4/4. Would be much obliged if someone would point out the error for me.

Thank you,
Musemonkey

 

[nicksauce]

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.

 

[musemonkey]

thanks!