What does it mean to take a mass moment of inertia about a single point? This website here: https://www.chegg.com/learn/calculus/calculus/moment-of-inertia-about-the-origin
Shows the following: 
How is it possible to define a mass moment of inertia about the origin which is a point and not an axis. What is the direction of rotation referred to?
What situation would this represent?
 A: The original problem seems to ask about "the axis at the origin perpendicular to the plane", but still, we do have a definition of inertia about a point instead of an axis
consider total angular momentum of a single particle as a vector, denoted by $\vec{L}$
, by definition $$
\vec{L} \equiv \vec{r} \times \vec{P}
$$
componentwise:
$$
\begin{aligned}
L^i & =(r \times P)^i \\
& =\varepsilon^{i j k} r_j P_k
\end{aligned}
$$
but $
\vec{P}=(\vec{\omega} \times \vec{r})m
$ , so its $k$th component is:
$$
P_k=(\varepsilon_{klm}\omega^l r^m)m
$$
so we have:
$$
\begin{aligned}
L^i & =\varepsilon^{i j k} \varepsilon_{k l m} r_j r^m \omega^l m \\
& =\left(\delta_l^i \delta_m^j-\delta_m^i \delta_l^j\right) r_j r^m \omega^l m \\
& =\left(r_j r^j \delta_l^i-r^i r_l\right) \omega^l m \\
& =\left(|r|^2 \delta_l^i-r^i r_l\right) m \omega^l
\end{aligned}
$$
so it's natural to introduce a tensor(called inertia tensor) by:
$$
I_j^i=\left(|r|^2 \delta_j^i-r^i r_j\right) m
$$
then we can express angular momentum in a tensorial form:
$$
L^i=I_j^i \omega^j
$$
for continuous distribution, simply replace m by $\rho d^3x$ and do integral
