What is the centroid of a hollow spherical cap? I have a unit hollow sphere which I cut along a diameter to generate two equivalent hollow hemispheres.  I place one of these hemispheres on an (x,y) plane, letting it rest on the circular planar face where the cut occurred. 
If the hemisphere was solid, we could write that its centroid in the above case would be given as $(0,0,\frac{3}{8})$.  Given that the hemisphere is hollow, can we now write its centroid as  $(0,0,\frac{1}{2})$?

In Cartesian coordinates, the hemisphere can be represented by a rotation about the z-axis of a circle, for which we can use, say, $y^2 + z^2 = 1$ to produce the infinitesimal "belt" of surface area $ dS = 2 \pi y  \ ds \ $ from
$$\frac{d}{dz}  [y^2 + z^2] \ = \ \frac{d}{dz}  [1] \ \Rightarrow \ \frac{dy}{dz}  \ = \ -\frac{z}{y}  $$
$$\Rightarrow \ ds \ = \ \sqrt{(\frac{dy}{dz})^2 + 1} \ \ dz \ = \ \sqrt{(-\frac{z}{y})^2 + 1} \ \ dz \ = \ \frac{\sqrt{y^2 + z^2}}{y} \  dz \ = \ \frac{1}{y} \ dz $$
$$\Rightarrow \ S \ = \ 2 \pi \int_a^b  y \ ds \ = \ 2 \pi \int_0^1  y \ \cdot  \frac{1}{y} \ dz \ = \ 2 \pi z \ |_0^1 \ = \ 2 \pi \ , $$
$$\Rightarrow \ M_{xy} \ = \ 2 \pi \int_a^b  z \cdot y \ ds \ = \ 2 \pi \int_0^1  z \ dz \ = \  \pi z^2 \ |_0^1 \ = \  \pi \ , $$
giving the height of the centroid as  $$\overline{z} = \frac{M_{xy}}{S} = \frac{1}{2} \ . $$ 
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But the actual reason I decided to chime in here at all is that there is a variant method we can employ by extending it to one dimension higher than it is often used.  Pappus' "centroid theorems" would give us the centroid of a semi-circular arc from 
$$2 \pi \overline{x} \ \cdot \ \pi R \ = \ 4 \pi R^2  \ \Rightarrow \ \overline{x} \ =  \frac{2}{\pi}R \ , $$
with $\pi R$ being the arclength of the curve and $4 \pi R^2$ the surface area of the sphere generated by revolution; by the same token, the centroid of a semi-circular region is given by
$$2 \pi \overline{x} \ \cdot \ \frac{1}{2}\pi R^2 \ = \ \frac{4}{3} \pi R^3  \ \Rightarrow \ \overline{x} \ = \ \frac{4}{3 \pi}R \ , $$
where $\frac{1}{2}\pi R^2$ is the area of the region and $\frac{4}{3} \pi R^3$ , the volume of the sphere of revolution.  (These are probably familiar enough results.)
Here, we are working with a hemispherical shell and a hemispherical volume.  It turns out that we can extend Pappus' theorems to deal with revolution generating the hypersurface area and hypervolume of a 4-sphere.  Thus, for the shell,
$$2 \pi \overline{x} \ \cdot \ 2 \pi R^2 \ = \ 2 \pi^2 R^3  \ \Rightarrow \ \overline{x} \ =  \frac{1}{2}R \ , $$
and for the solid hemisphere,
$$2 \pi \overline{x} \ \cdot \ \frac{2}{3}\pi R^3 \ = \ \frac{1}{2} \pi^2 R^4  \ \Rightarrow \ \overline{x} \ = \ \frac{3}{8}R \ , $$
from which your values for the unit sphere follow.  This works as it does because, although Pappus would not have said it that way, his relations are connected with moment integrals.
