Moment of inertia around displaced and rotated axis

[kontejnjer]

Homework Statement

A homogenous disk with radius R and mass m lies in the xy plane so its center matches the origin O. Point O' is on the z axis at a distance s from point O. Axis y' passes through point O' at an angle \theta with the z axis. Find the moment of inertia around axis y'.

Homework Equations

I_{ij}=\int dm(\delta_{ij}r^2-x_{i}x_{j})

The Attempt at a Solution

The new coordinates x',y',z' are connected with the old x,y,z through the equations:

x'=x
y'=y\text{sin}\theta+(z-s)\text{cos}\theta
z'=-y\text{cos}\theta+(z-s)\text{sin}\theta

Since the body has radial symmetry and is 2 dimensional, we can use polar coordinates:

x=rcos\varphi
y=rsin\varphi

and the third coordinate z=0 since its in the xy plane, hence:

x'=r\text{cos}\varphi
y'=r\text{sin}\varphi \text{sin}\theta-s\:\text{cos}\theta
z'=-r\text{sin}\varphi \text{cos}\theta-s\:\text{sin}\theta

The body is homogenous so:

\sigma=\frac{dm}{dA}=const.\rightarrow dm=\sigma dA
\sigma=\frac{m}{A}=\frac{m}{\pi\:R^{2}}

In polar coordinates:

dA=r\:drd\varphi

with limits:

0\leq r\leq R
0\leq \varphi\leq 2\pi

However, I'm not sure whether the "moment of inertia" would in this case mean the I_{y'\:y'} component from the inertia tensor. If that's the case, then:
I_{y'\:y'}=\int dm(x'^2+z'^2)=\sigma \int \int r\:drd\varphi[(r\text{cos}\varphi)^2+(-r\text{sin}\varphi \text{cos}\theta-s\:\text{sin}\theta)^2]
which Wolfram evaluates as:
I_{y'\:y'}=\frac{1}{4} m \left(R^2+R^2 \text{cos}^2\theta+4 s^2 \text{sin}^2\theta\right)

Would this be the correct procedure or not?

 

[SteamKing]

Is there a picture associated with the problem description?

 

[kontejnjer]

Not really, but from the description I'm guessing it looks like the attached pic. Also, y' in the pic should lie in the yz plane, and x',y',z' are orthogonal axes (again, that's what I'm assuming as no picture was given).