Inertia Tensor of a Hollow Sphere and of a Slender Rod

[Bomberman334]

Homework Statement

I need to find the Inertia Tensor of a Hollow Sphere and of a Slender Rod with center of mass set at the origin for my calculus 2 final project. I know how to do the triple integrals I am just having trouble figuring out what the limits should be for each of these shapes.

Attached is the my assignment, the ones I am referencing here are questions Three and Four.

Homework Equations

The components of the inertia tensor are
I_xx= ∭ (y^2+z^2 ) ρdv
I_yy= ∭ (x^2+z^2 ) ρdv
I_zz= ∭(x^2+y^2 ) ρdv
I_xy= I_yx= ∭xy ρdv
I_xz= I_zx= ∭xz ρdv

The Attempt at a Solution

I can't really start on the work until I know the limits...

However i know the limits of a filled sphere are
X= -R to R
Y = sqrt(R^2 -X^2)
Z = sqrt(R^2 -X^2-Y^2)

 

[SteamKing]

For the hollow sphere, rather than trying to set up a triple integral for the inside and out side radii, why not determine the inertia tensor for a solid sphere with radius = Ro and the inertia tensor for another solid sphere with radius = Ri. Since both spheres have the same center point and axes references, the inertia tensor for the spherical shell can be easily determined. (Note: you have omitted Iyz from your formulas in the OP)

 

[Bomberman334]

So are you saying I should do triple integrals for the sphere with radius Ro and Radius Ri, or is there some other method?

Sorry I'm being forced to do this for my Calculus Project with absolutely no knowledge of Inertia Tensors, I've pretty much tried to learn this in two weeks with nothing but my project as a reference.

 

[Bomberman334]

I figured out how to do the rod one using the formula for a solid cylinder, but I am still stuck on the hollow sphere. Could I just take the integral of a sphere radius r and then compare it to an integral where I use the term r-1 instead of r in the limits?

I.E.

X = -sqrt((r^2)-(z^2)-(y^2)) to sqrt((r^2)-(z^2)-(y^2))
Y = -sqrt((r^2)-(z^2)) to sqrt((r^2)-(z^2))
Z =-R to R

and

X = -sqrt(((r-1)^2)-(z^2)-(y^2)) to sqrt(((r-1)^2)-(z^2)-(y^2))
Y = -sqrt(((r-1)^2)-(z^2)) to sqrt(((r-1)^2)-(z^2))
Z =-(R-1) to R-1

 

[Bomberman334]

Anyone have any ideas?

 

[SteamKing]

Hint: If you find Ixxo for a sphere with radius Ro and Ixxi for a sphere with radius Ri, then for a hollow sphere with outer radius Ro and inner radius Ri, Ixx = Ixxo - Ixxi. Similar relationships exist for the other elements of the inertia tensor. This is a fundamental property of definite integrals: integral|0 to b - integral|0 to a = integral|a to b.

For symmetrical bodies, Ixy = Iyz = Ixz = constant. Care to guess what this constant is?

Let's see some calculations from you.

 

[Bomberman334]

I figured out a way to do it without the triple integral but I want to let you know that I feel like you didn't understand what I was saying.

I couldn't show any calculations since all of my calculations would rely on the having upper and lower bounds to work with. Since I couldn't find said bounds I was lost.

Thanks for the help though!