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 $(0,0,\frac{1}{2})$?
$\endgroup$4 Answers
$\begingroup$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 $$\overline{z} = \frac{M_{xy}}{S} = \frac{1}{2} \ . $$
$$ $$
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.
$\endgroup$ $\begingroup$Use $z_0=\displaystyle {{\int z ds}\over {\int ds}}$ where $z=\cos\phi$ and $ds=\sin\phi d\theta d\phi$. That gives your 1/2.
$\endgroup$ $\begingroup$Oddly enough, there is a much faster way to verify that the centroid of a hollow spherical cap arranged, as the poster said, is at $(0,0,1/2)$. If the cap is infinitesimally thin-walled, then we need only look at its area $A$ and assume it has a mass $M$ evenly spread over that area. The area $A$ (~$M$) of the cap is $A = 2 \pi R h$, where $h$ is the height of the top of the cap above its base, in this case its base is the base of a hemisphere, so h equals $R$. But this thin shell can be divided by a horizontal plane into two regions, each having an $h$, or a thickness, of $R/2$: one region being a smaller cap resting on the upper base of a "spherical segment," whose lower base is the original base of the hollow hemisphere.
The area of the smaller cap is now $A_1 = 2 \pi R (R/2)$. By definition (see Wolfram, wikipedia..) the area of the lower "segment," excluding the area of its bases, is $A_2 = 2 \pi R (R/2)$. These are clearly identical areas which also correspond to identical masses. I.e., the geometric centroid, which is the same as the centroid of mass, is the point on the z-axis lying on the horizontal dividing plane, which is $R/2$ above the origin.
$\endgroup$ $\begingroup$A simpler argument appeals to the hat-box theorem ().
By symmetry the centroid of the hemispherical cap lies on the $z$-axis; all that remains is to find its $z$-coordinate. But by the hat-box theorem, that's the same as the $z$-coordinate of the centroid of the cylinder of radius 1 and height 1 resting on the $x,y$ plane, which by symmetry is 1/2.
(Maybe this is what horchler and user86742 had in mind?)
$\endgroup$
