 Solution

Put $\mathbf{r}=(x,y,z)$ with ${\mathbf{e}}_3$ along $z$, and interpret the symmetry as axisymmetry of the whole mass density. A half-turn about ${\mathbf{e}}_3$ sends $(x,y,z)$ to $(-x,-y,z)$ and leaves the integration domain and mass density invariant. Therefore $\int \rho xz{d}^{3}r=\int \rho yz{d}^{3}r=0$. A quarter-turn sends $(x,y,z)$ to $(-y,x,z)$; it givesAll off-diagonal entries of the inertia tensor consequently vanish. Writing $s^2=x^2+y^2$, the principal moments of inertia satisfyThus every orthonormal basis with third vector along the symmetry axis is a principal body frame. This is axisymmetry of the inertia tensor. In general there is no requirement that $I_3=I_1$.