# Areal moment of inertia of a disk integral question

1. Jul 31, 2008

### 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

2. Jul 31, 2008

### 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.

3. Jul 31, 2008

thanks!