In your attachment, it is clearly stated that you are to find the inertia tensor.
Hence, it is insufficient to find the moments of inertia, you must also find the products of inertia.
I will assume henceforth, that we are to find the inertial tensor with respect to the origin.
1. Spin and inertial tensor:
Let a point particle with mass \delta{m} have distance \vec{r} to the origin, and let the particle's velocity be \vec{v}
The origin is at rest, so that we have:
\vec{v}=\vec{\omega}\times\vec{r}
Hence, the spin of the particle about the origin is given by:
\delta\vec{S}=\vec{r}\times\delta{m}\vec{v}=<br />
\vec{r}\times\delta{m}(\vec{\omega}\times\vec{r})=\delta{m}((\vec{r}\cdot\vec{r})\vec{\omega}-(\vec{r}\cdot\vec{\omega})\vec{r})
by a common vector identity.
With index notation, we have:
\delta{S}_{i}=\delta{I}_{ij}\omega_{j},\delta{I}_{ij}=\delta{m}(r_{k}^{2}\delta_{ij}-r_{i}r_{j})
where \delta_{ij} is the Kronecker delta, 1\leq{i,j,k}\leq3, and Einstein's summing convention has been adopted.
\delta{I}_{ij} is the inertial tensor associated with the given particle.
Since we have a rigid body, the body's inertial tensor with respect to the origin, I_{ij} is simply the sum of the point particles' associated
inertial tensors, i.e.,
I_{ij}=\int\delta{I}_{ij}
Note in particular, that the diagonal elements i=j in the inertia tensor are simply the moments of inertia about the axes \vec{i}_{1},\vec{i}_{2},\vec{i}_{3}
It is seen that the inertial tensor is symmetric; the off-diagonal elements are called products of inertia, and are, for example, important in wobbling/instability phenomena.
Calculations follow in the next post.
