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For a bounded volume $\Omega$ with piecewise smooth boundary and a continuously differentiable vector field on a neighborhood of its closure, the divergence theorem stateswhere $\mathbf{n}$ is the outward unit normal.

The boundary here consists of a paraboloid cap over the unit disk, a cylindrical wall of height one, and the bottom disk. Their joined surface enclosesThe cap meets the wall at $r=1,z=1$, and the wall meets the base at $r=1,z=0$. The sketches show each boundary piece and their closed union.

Differentiate the field components to find the divergence:The volume is invariant under reflecting $x$ or $y$, so odd-integrand cancellation by reflection removes both odd-power contributions. The outward flux equals the volume:The corners between the three pieces are permitted by the piecewise smooth boundary hypothesis.

For $i\ne j$, the inertia component is $I_{ij}=-\rho {\int }_V{x}_i{x}_{j}dV$. Reflecting just one of these coordinates preserves the cylinder and its uniform density but changes the integrand's sign. Odd-integrand cancellation by reflection therefore makes every off-diagonal entry zero. Rotations about the axis also give $I_{11}=I_{22}$.

Using cylindrical coordinates $x_1=r\cos \theta$, $x_2=r\sin \theta$, $x_3=z$, with $dV=rdrd\theta dz$, the required integrals areThus the inertia tensor isThe mass is $M=4\pi \rho$, so these entries are also $19M/12,19M/12,M/2$, respectively.