[Pappus' Centroid Theorems](https://mathworld.wolfram.com/PappussCentroidTheorem.html) provide a slick way of computing the center of mass for plane curves and plane areas.


The first theorem states that the surface area $A$ of a surface of revolution generated by rotating a plane curve $\Gamma$ about an axis external to $\Gamma$ and on the same plane is equal to the product of the arc length $s$ of $\Gamma$ and the distance $d$ traveled by its geometric centroid.$$A=sd$$


The second theorem states that the volume $V$ of a solid of revolution generated by rotating a plane figure F about an external axis is equal to the product of the area $A$ of F and the distance $d$ traveled by its geometric centroid. $$V=ad$$


In both theorems, $d=2 \pi y\_{c}$, where $y\_{c}$ is the required centroid. When we were taught in class the techniques of evaluation of the centroids of three dimensional figures via integration, I remember that the evaluation of the centroid of the hollow hemisphere was particularly difficult for me. So I ask if there are Pappus-like theorems which one could apply for three dimensional bodies?



Are Pappus Theorems generalized?
There is a neat way of finding the centroid of surfaces/volumes of revolution. Suppose you have a planar figure $\Gamma$ and an axis $\ell$ in the plane that is disjoint from $\Gamma$. Now suppose that the centroid of $\Gamma$ projects on $\ell$ as $O$, and that it's distance from $\ell$ is $r$. Suppose we rotate $\Gamma$ around $\ell$ with an angle of $\alpha$, and call the resulting body $\Gamma(\alpha)$. 


To find the centroid of $\Gamma(\alpha)$, $G\_{\alpha}$, it is enough to measure it's distance from $O$, since $OG\_{\alpha}\perp \ell$ and $G\_{\alpha}$ is on the dihedral bisector of the angle of revolution. Let's denote $|OG\_{\alpha}|$ by $f(\alpha)$, so that by cutting $\Gamma(\alpha)$ into two $\Gamma (\alpha/2)$'s, and using Archimedes lemma that the centroid of the union of two objects is in the line joining the separate centroids, we get the functional equation $$f(\alpha)=f(\alpha/2)\cos (\alpha/4)$$ with initial condition $f(0)=r$, so that by continuity the solution is $f(\alpha)=2r\frac{\sin (\alpha/2)}{\alpha}$. This method takes care of all examples like the hollow hemisphere etc.


