Did I do this right or get lucky? Deriving spherical shell moment of inertia

[HHHisthegame]

Homework Statement

Basically just to find moment of inertia of a spherical shell

Homework Equations

Moment of Inertia = Int(r^2 dm)

The Attempt at a Solution

I am in 12th grade and don't know much about integrals? Did I get lucky or am I right?

I = Int(r^2dM)

dM= (phi)(dA)

I = Int(r^2 phi dA)

Pull out phi, which equals M/4piR^2

I started by looking at half a sphere so I made that M/2piR^2. To get dA I figured you think of a bunch of hoops all going up to one point with a circumferance of 2piR, as you get higher up the radius changes however. A triangle says that R sin theta will get you the new radiuses, so for dA I wrote the integral of 2 pi R sin theta, dTheta. I put that in as dA, this R is a constant so now I have (M 2 pi R)/(2 pi R ^2) on the outside which cancels to M/R

Now I have

(M/R) Int(r^2 sin Theta, dTheta, dR) the R goes from 0 to R the theta from 0 to pi/2. The integral of sin Theta over that integral is 1, so it ends up

(MR^3)/(3R)

R's cancel

MR^2/3

That's half a sphere so multiply by 2

2/3 M R^2

Was that luck or does that work?

 

[Andrew Mason]

I would use Cartesian co-ordinates. Put the centre of the sphere at the origin. Divide the sphere into thin rings of area $dA = 2\pi y dz$ and express y as a function of z.

See this calculation for a solid sphere: http://hyperphysics.phy-astr.gsu.edu/hbase/isph.html#sph

AM