Trace of linear operator on $\mathbb R^n$ as an integral ? Let $T : \mathbb R^n\to \mathbb R^n$ be a linear map. Is it true that $Trace (T)=\dfrac 1{\mu(B)}\int_{x \in B} \langle Tx, x\rangle d\mu(x)$, 
where $B$ is the closed unit ball in $\mathbb R^n$ and $\mu$ is the usual Lebesgue measure and $\langle.,.\rangle$ is the usual inner-product on $\mathbb R^n$? 
 A: The normalization of the formula is wrong, but let me propose a fix. Let
$$K_n=\int_{B\subset\Bbb{R^n}}x_1^2\,d\mu$$
be the integral we would get with $T=e_{11}$, the matrix with $1$ at position $(1,1)$ and zeros elsewhere. I think that we then have for all linear transformations $T:\Bbb{R}^n\to\Bbb{R}^n$ that
$$
tr(T)=\frac1{K_n}\int_{B}\langle Tx,x\rangle\,d\mu.
$$
I will think of $T$ as a matrix (use the natural basis). We can equally well use the symmetrized version of $T$, i.e. $(T+T^t)/2$ because the quadratic form $\langle Tx,x\rangle$ stays the same, and the symmetrized version shares the same trace.
Given that $T$ is symmetric the rest is easy. By linear algebra there exists an orthogonal matrix $P$ such that $P^tTP=D$ is a diagonal matrix. Furthermore, $P^t=P^{-1}$ and $\det P=1$. Also, the linear substitution $x\mapsto Px$ preserves the unit ball, because $P$ is length preserving. What this means is that it suffices to prove the formula for a diagonal matrix $D=diag(d_1,d_2,\ldots,d_n)$.
Obviously 
$$I_i=\int_{B\subset\Bbb{R^n}}x_i^2\,d\mu=K_n$$
for all $i=1,2,\ldots,n$ by the symmetries of the sphere. By linearity of the integral we then get that
$$
\begin{aligned}
\int_{B\subset\Bbb{R^n}}\langle Dx,x\rangle\,d\mu
&=\int_{B\subset\Bbb{R^n}}(d_1x_1^2+d_2x_2^2+\cdots+d_nx_n^2)\,d\mu\\
&=\sum_{i=1}^nd_i I_i\\
&=tr(D) K_n,
\end{aligned}
$$
and we are done.

In other words, instead of the measure of the $n$-dimensional ball you should factor out the integral of $x_1^2$ over that ball. I'm sure the exact value of $K_n$ is known. May be a friendly physicist calculated the moments of inertia of the $n$-dimensional homogeneous ball? Or a probability person has calculated the expected value of $x_1^2$ of a random point uniformly distributed over that ball?
A: This is the kind of nice formula one wishes were true, but it's not a stated.  Here's a simple counterexample:
Suppose
$n = 2, \tag 1$
and
$T = I, \tag 2$
the identity map.  Then 
$\text{Tr}(T) = \text{Tr}(I) = 2, \tag 3$
and $B$ is the unit disk, so
$\mu(B) = \pi, \tag 4$
whence
$\mu(B) \text{Tr}(T) = 2\pi; \tag 5$
for $\vec r = (x, y) \in B, \tag 6$
$\langle T\vec r, \vec r \rangle = x^2 + y^2 = r^2, \tag 7$
and
$\displaystyle \int_B r^2 d\mu = \int_B r^2\; r \; dr \; d\theta = \dfrac{2\pi \cdot 1^4}{4} = \dfrac{\pi}{2} \ne 2\pi,  \tag 8$
as we see the formula fails for $n = 2$, $T = I$.
A: This is not quite true but it is true up to a scale factor.  Let me first prove that it must be true up to a scale factor, and then figure out what the factor is.  Let us write $$A(T)=\dfrac 1{\mu(B)}\int_{ B} \langle Tx, x\rangle d\mu(x).$$  Note that $A$ is linear, and is also invariant under conjugation by orthogonal matrices.  Also, if $T$ is antisymmetric, then $$\langle Tx,x\rangle=\langle x,T^tx\rangle=\langle x,-Tx\rangle=-\langle Tx,x\rangle$$ for all $x$ so $A(T)=0$.  Now fix a rank $1$ orthogonal projection $T_0$.  Note that every symmetric matrix is a linear combination of conjugates of $T_0$ by orthogonal matrices, by the spectral theorem.  Moreover, every matrix is a linear combination a symmetric matrix and an antisymmetric matrix by writing $T=\frac{T+T^t}{2}+\frac{T-T^t}{2}$.  So, we can write an arbitrary matrix as a linear combination of antisymmetric matrices and conjugates of $T_0$ by orthogonal matrices.
Now, there exists a scalar $c$ such that $A(T_0)=c\operatorname{tr}(T_0)$ (in fact, $c=A(T_0)$ since $\operatorname{tr}(T_0)=1$).  Since $A$ and $\operatorname{tr}$ both are linear and are preserved by conjugation by orthogonal matrices and vanish on antisymmetric matrices, it follows that $A(T)=c\operatorname{tr}(T)$ for all $T$.
To determine the constant $c$, we can just take $T=I$ to be the identity.  We then have $$A(I)=\dfrac 1{\mu(B)}\int_{ B} \|x\|^2d\mu(x).$$  Letting $s$ denote the surface area of the unit $(n-1)$-sphere, we can compute the integral using polar coordinates: $$\int_{B} \|x\|^2d\mu(x)=\int_0^1 sr^2\cdot r^{n-1}\, dr=\frac{s}{n+2}.$$
On the other hand, we can also compute $\mu(B)$ from $s$ by polar coordinates to get $$\mu(B)=\int_0^1 s\cdot r^{n-1}\,dr=\frac{s}{n}.$$  Combining these results, we conclude that $A(I)=\frac{n}{n+2}.$  Since $\operatorname{tr}(I)=n$, this means $c=\frac{1}{n+2}$ and $$A(T)=\frac{1}{n+2}\operatorname{tr}(T)$$ for all $T$.
(To be ridiculously pedantic, several steps of this only work assuming $n>0$.  Of course, the final conclusion is still true for $n=0$ since in that case the only possible value of $T$ is $0$.)
