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).