Integral formula for inertia tensor Writing down the balance of angular momentum, we introduce the inertia tensor by the formula
\begin{equation}
J(t)a \cdot b = \int_{S(t)} \rho (t,x)\left( a \times \left( x - X(t) \right)\right)\cdot \left(b \times \left( x - X(t) \right) \right) dx
\end{equation}
for some vectors $a,b$, some body $S(t)$ at time $t$, the density $\rho$ and the centre of mass $X(t)$.
Now, what confuses me, is that later we use the expressions
\begin{equation}
J(t)a \ \ \ \text{and} \ \ \ J(t)a \times a,
\end{equation}
which (as far as I can see) are not immediately clear from the above formula. 
For the first one I would expect something like:
\begin{equation}
J(t)a = \int_{S(t)} \rho (t,x)\left( a \times \left( x - X(t) \right)\right)dx,
\end{equation} 
and for the second one:
\begin{equation}
J(t)a \times a = \int_{S(t)} \rho (t,x)\left( a \times \left( x - X(t) \right)\right)\times \left(a \times \left( x - X(t) \right) \right) dx
\end{equation}
But those are just guesses. Can someone tell me the precise definition of these terms? Thank you in advance.
 A: Welcome to Physics SE! 
The definition of the inertia tensor that you seem to be using can also be written, as on the Wikipedia Moment of Inertia page
$$
\mathbf{J} = -\int d\mathbf{x} \, \rho(\mathbf{x}) \, 
\big[ \Delta\mathbf{x} \big] \cdot \big[ \Delta\mathbf{x} \big]
$$
where $\Delta \mathbf{x} = \mathbf{x}-\mathbf{X}$
and $[\ldots]$ is short for a $3\times3$ skew-symmetric matrix constructed
from the vector $\Delta \mathbf{x}\equiv (\Delta x_1, \Delta x_2, \Delta x_3)$. 
When one of these matrices multiplies a vector, the result can be represented as a vector cross product:
\begin{align*}
\big[ \Delta\mathbf{x} \big] \cdot \mathbf{b} 
&= 
\begin{pmatrix} 0 & -\Delta x_3 & \Delta x_2 \\
\Delta x_3 & 0 & -\Delta x_1 \\
-\Delta x_2 & \Delta x_1 & 0 \end{pmatrix} 
\begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}
\\
&= \begin{pmatrix} \Delta x_2 \, b_3 - \Delta x_3 \, b_2 \\ 
\Delta x_3 \, b_1 - \Delta x_1 \, b_3  \\ 
\Delta x_1 \, b_2 - \Delta x_2 \, b_1 \end{pmatrix}
= \Delta\mathbf{x} \times \mathbf{b}
= -\mathbf{b} \times\Delta\mathbf{x} .
\end{align*}
I've taken the liberty of writing the matrices and vectors in bold,
it's just more familiar to me.
Actually, on the Wikipedia page, the equation is given in terms
of a sum over discrete masses rather than an integral over a mass density, but it's equivalent.
If we contract this matrix with two arbitrary vectors $\mathbf{a}$ and $\mathbf{b}$, we get your starting equation. I would prefer to write the left hand side as $\mathbf{a}\cdot\mathbf{J}\cdot\mathbf{b}$, or even as $\mathbf{a}^T\cdot\mathbf{J}\cdot\mathbf{b}$, not as $\mathbf{J}\,\mathbf{a}\cdot\mathbf{b}$, because your notation makes it look like $\mathbf{a}$ and $\mathbf{b}$ are being combined together in a scalar product, which is not the case. Your equation doesn't have the minus sign: the change in sign comes from one of the vector products on the right being $\mathbf{a}\cdot\big[ \Delta\mathbf{x} \big]$ and the other being $\big[ \Delta\mathbf{x} \big]\cdot\mathbf{b}$.
So, I believe your starting equation is obtained from mine by
\begin{align*}
\mathbf{a}\cdot\mathbf{J}\cdot\mathbf{b} &=
 -\int d\mathbf{x} \, \rho(\mathbf{x}) \, 
\mathbf{a}\cdot\big[ \Delta\mathbf{x} \big] \cdot \big[ \Delta\mathbf{x} \big]
\cdot\mathbf{b}
\\
&=
 -\int d\mathbf{x} \, \rho(\mathbf{x}) \, 
\left(\big[ \Delta\mathbf{x} \big]^T\cdot\mathbf{a}\right) 
\cdot 
\left(\big[ \Delta\mathbf{x} \big]\cdot\mathbf{b}\right)
\\
&=
 \int d\mathbf{x} \, \rho(\mathbf{x}) \, 
(\mathbf{a}\times\Delta\mathbf{x})
\cdot 
(\mathbf{b}\times\Delta\mathbf{x}) .
\end{align*}
I'm omitting the $T$ transpose sign on vectors, to avoid clutter; I don't believe that there is any ambiguity.
Now to your question. The main point is that $\mathbf{a}$ and $\mathbf{b}$ are arbitrary. Since they are arbitrary, your starting equation does completely specify $\mathbf{J}$. You can always choose one or both of $\mathbf{a}$ and $\mathbf{b}$ to be Cartesian basis vectors, to express any result in terms of components, if you wish. Alternatively, you can use the expression I gave above, and simply don't contract with the vector on the left. So I reckon
\begin{align*}
\mathbf{J}\cdot\mathbf{a} &= -\int d\mathbf{x} \, \rho(\mathbf{x}) \, 
\big[ \Delta\mathbf{x} \big] \cdot \big[ \Delta\mathbf{x} \big] \cdot \mathbf{a}
\\
&= -\int d\mathbf{x} \, \rho(\mathbf{x}) \, 
\big[ \Delta\mathbf{x} \big] \cdot (\Delta\mathbf{x}\times \mathbf{a}) 
\\
&= \int d\mathbf{x} \, \rho(\mathbf{x}) \, 
\big[ \Delta\mathbf{x} \big] \, (\mathbf{a}\times \Delta\mathbf{x})
\\
&= \int d\mathbf{x} \, \rho(\mathbf{x}) \, 
\Delta\mathbf{x} \times (\mathbf{a}\times \Delta\mathbf{x}) 
\\
&= \int d\mathbf{x} \, \rho(\mathbf{x}) \, 
\left(
|\Delta\mathbf{x}|^2 \mathbf{a}-
(\Delta\mathbf{x} \cdot\mathbf{a})\Delta\mathbf{x}) 
\right)
.
\end{align*}
I'm not completely sure about your second expression, 
because of the same notational concerns I raised above.
Clearly you don't mean 
$\mathbf{J} \cdot(\mathbf{a}\times\mathbf{a})$
because the quantity in parentheses vanishes identically.
So I guess you mean $\mathbf{a}\times(\mathbf{J}\cdot\mathbf{a})$,
or $(\mathbf{J}\cdot\mathbf{a})\times\mathbf{a}$.
If the first of these is true, then the answer is 
$$\mathbf{a}\times(\mathbf{J}\cdot\mathbf{a})
=
-\int d\mathbf{x} \, \rho(\mathbf{x}) \, 
(\Delta\mathbf{x} \cdot\mathbf{a})(\mathbf{a}\times\Delta\mathbf{x}) 
$$
while if it's the second, just drop the negative sign.
