Why do we talk about inertia tensor? When we talk about the inertia of a rigid body, in calculating the angular momentum as a function of the moment of inertia and angular velocity, the inertia tensor is introduced. But why is it a tensor?
The moment of inertia is defined as $$I=\int_V \! r^2 \, \mathrm{d}m.$$ where $r^2$ is the squared distance of the $dV$ from the axis of rotation and $m$ is the mass, why is $I$ a tensor?
According to "Manifolds, Tensor Analysis and Applications - Marsden, Ratiu and Abraham" a tensor over a Banach space $E$ is a multilinear map defined on the cartesian product of $r$ dual space $E*$ and $p$ space $E$ which takes values on $\mathbb{R}$. I can't see this definition on $I$.
$I$ seems to me a linear functional from $\mathbb{R}^3$ to $\mathbb{R}^+$ because it takes a function $r^2=x^2+y^2+z^2$ and gives out a scalar.
The question is:
Why we talk about a tensor of inertia and not a linear operator of inertia or a linear functional of inertia?
 A: The moment of inertia you mentioned is only for a single, given axis of rotation. It's used to compute the angular momentum related to his axis:
$$L=I\omega$$
If you want to generalize the result in 3D space to compute the vector angular momentum as a function of the rotation vector, then:
$$\vec{L}=I\vec{\omega}$$
$I$ needs to become a $3\times 3$ matrix, which is called the inertia tensor.
If the rotation happens around a single axis $z$, you can choose a basis where one of the base vectors is alongisde this axis. Then the expression simplifies to:
$$L_z
=\vec{L}.\vec{e}_z=
\begin{pmatrix}
a & b & c\\
e & f & g\\
i & j & k
\end{pmatrix}
\begin{pmatrix}
0\\
0\\
\omega
\end{pmatrix}.\vec{e}_z
=k\omega$$
which is "single axis" result that you mentioned.
The basis in which $I$ is diagonal defines a specific set of axes that is sometimes useful is study the dynamics of the solid. In this basis, the (scalar) definition of $I$ that you gave can be used for each axis to compute three moments of inertie that are the eigenvalues of $I$.
A: A rank 2 tensor is something that relates two vectors. In this case, the MMOI tensor relates the rotational velocity vector to the angular momentum vector.
Given a solid whose internal particles are designated with $\vec{r}$ in relation to the center of mass, you have the following volume integral to find the angular momentum of the body.
$$ \vec{L} = \int \vec{r} \times (\vec{v}\, {\rm d}m) = \int  \vec{r} \times (\vec{v} \rho \,{\rm d} V). $$
Here $\vec{v} = \vec{\omega} \times \vec{r}$ is the motion of each particle, so
$$ \vec{L} = \int  [\vec{r} \times  ( \vec{\omega} \times \vec{r})] \rho\, {\rm d} V.$$
With the vector identity $\vec{a}\times ( \vec{b} \times \vec{c}) = \vec{b} ( \vec{a} \cdot \vec{c}) - \vec{c} ( \vec{a}\cdot \vec{b}) = ( \vec{a}\cdot \vec{c}) \vec{b} - (\vec{c} \odot \vec{a}) \vec{b} $, where $\cdot$ is the inner (dot) product, and $\odot$ is the outer product,
$$ \vec{L}=  \int  \left(  \vec{r} \cdot \vec{r} - \vec{r} \odot \vec{r} \right) \vec{\omega}\, \rho\,{\rm d} V.$$
And the integral is factored by the mass moment of inertia tensor
$$ \vec{L} = \mathrm{I}\, \vec{\omega} $$
$$ \mathrm{I} \equiv \int  \left(  \vec{r} \cdot \vec{r} - \vec{r} \odot \vec{r} \right)  \rho\, {\rm d} V.$$

If $\vec{r} = \pmatrix{x \\y \\ z}$ then the integral is
$$ \mathrm{I} \equiv \int  \begin{bmatrix} y^2+z^2 & -x y & -x z \\ -x y & x^2+z^2 & - y z \\ -x z & - y z & x^2+y^2 \end{bmatrix}  \rho\, {\rm d} V, $$
from which you are familiar with the 2D version
$$ \mathrm{I}_{zz} \equiv \int  (x^2+y^2)  \rho\, {\rm d} V.$$
