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## Homework Statement

A hollow cylinder has mass

*m*, and outside radius [tex]r_{2}[/tex], and an inside radius [tex]r_{1}[/tex]. Use intergration to show that the moment of inertia about its axis is give by [tex]I=\frac{1}{2}m(r^2_2+r_1^2).[/tex]

Uniform composition is assumed.

## Homework Equations

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

[tex]dm=\sigma dA[/tex]

and

[tex]A=\pi r^2[/tex]

(note, later on I constrict the lower and upper bounds of integration to account for the now entire area of the circular side, which sounds as though it is assumed here)

## The Attempt at a Solution

Note: Using equations above.

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

[tex]I=\int r^2\sigma dA[/tex]

[tex]dA=2\pi rdr[/tex]

[tex]I=\sigma 2\pi\int r^3dr[/tex]

(here I insert my upper and lower bounds)

[tex]I=\sigma 2\pi\int_{r_1}^{r_2} r^3dr[/tex]

[tex]I=\sigma 2\pi \left(\frac{r^4_2}{4}-\frac{r^4_1}{4} \right )[/tex]

[tex]I=\sigma \pi \left(\frac{r^4_2}{2}-\frac{r^4_1}{2} \right )[/tex]

[tex]I=\frac {1}{2}\sigma \pi \left(r^4_2-r^4_1 \right )[/tex]

[tex]\sigma=\frac{m}{A}[/tex]

and since

[tex]A=\pi r^2_1-\pi r^2_2[/tex]

then

[tex]I=\frac{m}{2(\pi r^2_1-\pi r^2_2)} \pi \left(r^4_2-r^4_1 \right )[/tex]

[tex]I=\frac{\left(r^4_2-r^4_1 \right )m}{2 (r^2_1-r^2_2)}[/tex]

This is as far as I venture into the problem before I realize that I've done something wrong if I'm supposed to get the answer they assume at the beginning of the problem. Where did I go wrong?

Thank you very much for the help, it's greatly appreciated.