# Products of Inertia

1. Feb 2, 2013

### unscientific

1. The problem statement, all variables and given/known data

http://i45.tinypic.com/hwcsy0.png

3. The attempt at a solution

I'm not sure how to find the rest; Ixy, Iyz and Izx...
Usually for integrals such as moments of inertia you will be able to reduce it to only one variable. However, there are 2 variables here; xy, yz and zx. How do i reduce it to only one?

Last edited by a moderator: Feb 3, 2013
2. Feb 2, 2013

### vela

Staff Emeritus
You don't. You evaluate the triple integrals.

3. Feb 3, 2013

### unscientific

Are my Ixx, Iyy and Izz are correct?

Oh, so it's dM = dx dy dz,

then the range for dx is from -√(a2-y2-z2) to √(a2-y2-z2)

for dy it's from -√(a2-z2) to √(a2-z2)

for dz it's from 0 to a

Are my ranges of integration right?

4. Feb 3, 2013

### vela

Staff Emeritus
No, they're not. You can't say $y^2+z^2=a^2-x^2$ because that's true only on the spherical part of the surface. Inside the hemisphere, it doesn't hold.

That's dV, not dM. You should write dM = k dV = k dx dy dz.

Yes.