# Moment of inertia of a sphere

## Homework Statement

calculate the moment of inertia of a sphere of mass M and radius R by integrating over thin shells

Ishell=(2/3)mR2

## The Attempt at a Solution

this is what i have so far
the sphere is decomposed into infinitesimal shells with surface area 4$$\pi$$r2
the mass of each shell is dm=$$\rho$$(4$$\pi$$r2)dr
after expanding rho and canceling terms I get
dm=(3m/r)dr

I=$$\int$$(2/3)(dm)r2dr from 0 to R.
I=2mrdr from 0 to R
this gives me I=mr2

does anyone see what I did wrong?
PS. those "pi" are not supposed to be powers. sorry, i'm not sure how to change them

## Answers and Replies

kuruman
Science Advisor
Homework Helper
Gold Member
You did not replace dm with ρ(4πr2) under the integral sign. Instead, you integrated dm into m and then integrated over dr once more.

i'm sorry, i didn't write it all out

I=$$\int$$(2/3)dmr2 dr
from here I replaced dm with (3m/r)dr so i got

I=$$\int$$(2/3)(3m/r)r2 dr
then i got
I=$$\int$$2mr dr from 0 to R
which gave me mr2

kuruman
Science Advisor
Homework Helper
Gold Member
... from here I replaced dm with (3m/r)dr...
And why is that correct? What expression did you use for the density?