Inertia Products: Solve Ixy, Iyz, Izx

[unscientific]

Homework Statement

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

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?

 

[vela]

You don't. You evaluate the triple integrals.

 

[unscientific]

You don't. You evaluate the triple integrals.

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?

 

[vela]

Are my Ixx, Iyy and Izz are correct?

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.

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

That's dV, not dM. You should write dM = k dV = k 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?

Yes.