Inertia tensor integral

Question

$$I_{xx}=\sum{m_\alpha}(y_\alpha^2+z_\alpha^2)$$ $$ I_{xy}=-\sum m_\alpha x_\alpha y_\alpha $$

How can these two definitions of the inertia tensor be combined into one definition to get $$ I_{ij}=\int \rho (r^2\delta_{ij}-r_i r_j)dV? $$

Answer 1

$$I_{xx}=\sum{m_\alpha}(y_\alpha^2+z_\alpha^2)$$ $$ I_{xy}=-\sum m_\alpha x_\alpha y_\alpha $$ How can these two definitions of the inertia tensor be combined into one?

Those expressions only cover two of the six independent elements of the inertia tensor of a collection of point masses. The expression for $I_{xy}$ obviously generalizes to $$I_{ij}=-\sum m_\alpha r_{\alpha,i}r_{\alpha,j}\quad(i \ne j) \tag{1}$$ A way to generalize the first expression to describe the three diagonal elements of the inertia tensor is to first recognize that $y_\alpha^2 + z_\alpha^2$ is equal to $r_\alpha^2 - x_\alpha^2$, leading to $$I_{xx}=\sum{m_\alpha}(r_\alpha^2-x_\alpha^2)$$ This generalizes to $$I_{ii}=\sum{m_\alpha}(r_\alpha^2-r_{\alpha,i}^2)\tag{2}$$ Combining expressions (1) and (2) yields $$ I_{ij} = \begin{cases} \sum m_\alpha (r_\alpha^2-r_{\alpha,i}r_{\alpha,j}) & j=i \\ \sum m_\alpha (\phantom{r_\alpha^2}-r_{\alpha,i}r_{\alpha,j}) & j \ne i \end{cases}\tag{3} $$ Equation (3) can be succinctly written using the Kronecker delta as $$I_{ij}=\sum{m_\alpha}(r_\alpha^2\delta_{ij}-r_{\alpha,i}r_{\alpha,j})$$ For a mass distribution, the above becomes, with a bit of abuse of notation, $$I_{ij}=\int(r^2\delta_{ij}-r_i r_j)\,dm$$ Recognizing that $dm = \rho dV$, this becomes $$I_{ij}=\int \rho (r^2\delta_{ij}-r_i r_j)\,dV$$