Consider a Borel measure $\mu$ on $\mathbb R^n$ ($n\geq2$) of finite second moment, i.e. such that $$ \int_{\mathbb R^n} \|x\|^2 \,\mathrm d\mu(x) < \infty. $$ It is said balanced in the direction $e_1$ (the first unit-vector) if for all Borel set $A\subset\mathbb R^{n-1}$ $$ \int_{\mathbb R\times A} x_1\,\mathrm d\mu(x)=0. $$ Intuitively, no matter the section $A$ considered, the infinite beam of section $A$, axis $e_1$ and weight distribution $\mu$ is balanced. Likewise, $\mu$ is said balanced in the direction $u$ (unit-vector) if $\phi_*\mu$ is balanced in the direction $e_1$, where $\phi$ is a rotation that maps $u$ to $e_1$. Finally, $\mu$ is said balanced if it is balanced in all directions, and isotropic if $\phi_*\mu=\mu$ for all linear isometry $\phi$ (including reflections).
Obviously if $\mu$ is isotropic, it is balanced. The question is: "If $\mu$ is balanced, is it necessarily isotropic?" It would seem so, but I cannot come up with a convincing argument.