# Moment of inertia of spherical shell

1. Oct 29, 2007

### karnten07

1. The problem statement, all variables and given/known data
Show that the moment of inertia of a uniform thin spherical shell of mass m and radius a about a diameter is 2/3 ma^2

2. Relevant equations
Moment of inertia = Sum m x R^2
volume of sphere = 4/3 pi r^3

3. The attempt at a solution
I think i have to integrate the moment of inertia euation for this spherical shell, but im not sure im understanding my notes here.
The inertia is equal along all axes, so by adding the inertia along each axis gives me the total inertia?

i have some rough sketchy notes i took in a class, so i will write them here to see if anyone can decipher it.

x^2 + y^2 + z^2 = a^2 where a is the radius of the sphere

Any ideas would be much appreciated, thankyou

Last edited: Oct 29, 2007
2. Oct 30, 2007

### learningphysics

The challenge is setting up the integral... I recommend using spherical or cylindrical coordinates... You need to choose an axis through the center... any one is fine... z-axis.

$$I = \int\int\int \sigma*r^2dV$$

$$\sigma$$ is the density of the sphere.

$$r = \sqrt{x^2 + y^2}$$

so

$$I = \int\int\int \sigma*(x^2+y^2)dV$$

try to set up this integral over the volume of the sphere using spherical or cylindrical coordinates... you can use cartesian too but it's a little more tedious...

3. Oct 30, 2007

### Mindscrape

I know lp is trying to pedagogical, but I strongly advise against any coordinates but spherical. You don't integrate cubes in spherical coordinates, and likewise you don't integrate spheres in cartesian coordinates. (Another Hint: you will need to rederive, or look-up, the cartesian to spherical formulas.)