- #1

- 129

- 0

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

[tex]m = \lambda L[/tex]

[tex]dm = \lambda dL[/tex]

[tex]I_{ring} = \int_{0}^{L}\lambda r^2dL[/tex]

And that would give the requiside mr

^{2}. My question is, why can't I just integrate THAT up to get the I of a disk. I mean something like this:

[tex]I_{disk} = \int_{0}^{A}\int_{0}^{L}\lambda r^2dLdA[/tex]

Where A is the area of the disk. Doesn't figuring out the I for a ring essentially skip the first step for determining the I for a disk?