= Solution
For positive constant <mass density> $\rho_0$, reflection symmetry makes the horizontal coordinates of the <centre of mass> zero. The total mass and the vertical first moment are
$$
M=\rho_0\frac{\pi ah^2}{2},\qquad
\int z\rho_0\,dV=\rho_0\int_0^h z(\pi az)dz
=\frac{\rho_0\pi ah^3}{3}.
$$
Their ratio gives
$$
\boxed{(\bar x,\bar y,\bar z)=(0,0,2h/3).}
$$
This is the <centre> of the liquid <volume>, rather than of the curved wall. The density cancels because it is uniform; the larger disks near the top place the <centre> above half-height.
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