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Use cylindrical coordinates. The real square root restricts the projection to $0\le r\le 1$, and there $r-1\le 0\le \sqrt{1-r^2}$. Since $\nabla x={\mathbf{e}}_x$, the volume integral isFor an independent boundary calculation, split $\partial V$ into an upper hemisphere $S_{+}$ and a lower right circular conical surface $S_{-}$. They meet in the circle $r=1,z=0$ and there is no additional planar face.

On $S_{+}$, use the parametrizationThe outward unit normal vector is $\mathbf{R}$, and the surface area element is $\sin \alpha d\alpha d\theta$. The $y$ and $z$ components of ${\iint }_{S_{+}}x\mathbf{n}dS$ vanish by the angular integrals of $\sin \theta \cos \theta$ and $\cos \theta$. The $x$ component isOn $S_{-}$, the parametrization $(r\cos \theta ,r\sin \theta ,r-1)$ has outward vector area elementIts negative vertical component points out below the solid. Multiplying by $x=r\cos \theta$ and using the same angular symmetry in integration givesTherefore the independently evaluated surface integral isin agreement with the volume integral. The isolated apex of the cone does not contribute to the surface integral.

Write the centroid as $\overline{\mathbf{r}}=\overline{x}\mathbf{i}+\overline{y}\mathbf{j}$. The region has reflection symmetry in the $y$-axis: replacing $\theta$ by $\pi -\theta$ changes $x$ to $-x$ and preserves $y$ and $dA$. Thus symmetry in integration givesFor the other first moment of area, use $y=r\sin \theta$ in polar coordinates:Here the last integral has the value allowed in the paper. Dividing the first moment of area by the area found in i givesThe positive vertical coordinate agrees with the fact that the entire region lies above the $x$-axis.

The same integration by substitution gives the expected valueThe remaining integrand is an odd function. Its integral is absolutely convergent, since ${\mathrm{sech}}^{2}t\le 4e^{-2|t|}$ and $|t|e^{-2|t|}$ is integrable. Thus cancellation under symmetry in integration is legitimate, rather than merely a principal-value calculation. ConsequentlyEquivalently, the logistic distribution is symmetric about $\mu$, and its exponentially decaying tails ensure that its expected value exists.