Determining Moment of Inertia of a sphere.

[aFk-Al]

Determining Moment of Inertia of a Sphere

I'm having some troubles determining the moment of inertia of a sphere about it's central axis. My original question was to calculate it for a cylinder, which I've done, but I'd like to know how to find it for a sphere.
Here is the problem solved for a cylinder:

Problem:
A uniform solid cylinder has radius R, mass M, and length L. Calculate its moment of inertia about its central axis (the z axis).

Solution:
I divided the cylinder into infinitesimally small layers because I knew that $$dV = (2\pi*r*dr)*L$$. From here I calculated the integral $$I = \int \rho*r^2 dV = \int_{0}^R \rho*r^2*(2\pi*r*L)dr = 2*\pi*\rho*L*R^4$$
I substituted $$\frac {M}{\pi*R^2*L}$$ (or $$\frac{M}{V}$$) for $$\rho$$ into the equation to get $$I = \frac{1}{2}*\pi*(\frac {M}{\pi*R^2*L})*L*R^4 = \frac{1}{2}*M*R^2$$

I understand this, but when I tried to get it as a sphere I ended up getting the wrong answer. Could anyone please show me how to start the problem with a sphere?

 

[berkeman]

Do it in spherical coordinates, being careful to take the sin(theta) for each little volume piece (since the spherical r value is from the origin, not from the axis of rotation). And be careful to use the correct value for dV in spherical coordinates.

 

[Ja4Coltrane]

For the solid spere, I like to add up the moment of inertia of a bunch of disks form -R to R.

 

[cristo]

How did you try to perform your integral for the sphere? Think of your choice of coordinate system--hint: spherical polar coordinates.

damn: really should have refreshed quicker!

 

[aFk-Al]

Couldn't I divide the sphere into a hemisphere and multiply it by 2? E.G. $$2*\pi*\int_{0}^R (\sqrt{R^2-r^2}*r^2)*r^2 dr$$ Where R is the radius of the sphere. So essentially I'm adding a bunch of $$\pi*r^2$$ (circles) to get a half of a sphere. Then multiplying the whole thing times two.

 

[Ja4Coltrane]

hmm, why don't you show a little more work so that it is easier to follow.
one thing though, you might want to reconsider muliplying it by two--think about it.

 

[aFk-Al]

I have no idea how to integrate that, so I don't have much work to show. The idea makes sense in my head, but I don't know how to follow through with it.