I disagree with this statement. Consider $9$ points in the plane, called $x(i,j)$ where $(i,j)$ is in $\{ 0,1,2 \}$. There are $12$ triples $((i_1, j_1), (i_2, j_2), (i_3, j_3))$ such that $i_1 + i_2 + i_3 \equiv 0 \mod 3$ and $j_1 + j_2 + j_3 \equiv 0 \mod 3$. Let $L$ be the set of such triples.
Consider the following statement:
Suppose that, for all $((i_1, j_1), (i_2, j_2), (i_3, j_3)) \in L$, the points $x(i_1, j_1)$, $x(i_2, j_2)$ and $x(i_3, j_3)$ are colinear. Then either all of the $x(i,j)$ are colinear, or else two of the $x(i,j)$ are equal to each other.
It seems to me that this statement is an incidence theorem. It is true in $\mathbb{RP}^2$ and false in $\mathbb{CP}^2$, both of which obey Pappus theorem. I learned this example from Kiran Kedlaya.
To see this over $\mathbb{R}$, note that a counterexample to this claim is also a counterexample to the Sylvester-Gallai theorem. Over $\mathbb{C}$, the flexes of any cubic curve form a counter-example, as discussed on the above linked Wikipedia page. More precisely, I believe that the claim is true in $K\mathbb{P}^2$ if and only if $K$ does not contain a root of $x^2+x+1$.
More generally, I would consider an incidence theorem to be a first order statement about points and lines in $\mathbb{RP}^2$ where what we are allowed to say is that a given point does or does not lie on a given line. We can easily turn such a statement into an algebraic claim about $\mathbb{R}$.
By a result of Tarski, any true statement of this form follows from (1) the field axioms (2) the axioms that $\mathbb{R}$ has an ordering $\leq$ obeying the standard properties and (3) the "polynomial intermediate value theorem": for any polynomial $f \in \mathbb{R}[t]$, if
$f(a)<0$ and $f(b)>0$, then there exists $c \in (a,b)$ such that $f(c)=0$.
For example, to prove that $x^2+x+1$ has no roots in $\mathbb{R}$, just note that $x^2+x+1 = (x+1/2)^2+3/4 \geq 3/4$. Here we have used the field axioms (many times) and the basic properties of $\leq$.
Pappus theorem encodes the commutativity of multiplication, and I will believe you that the other field axioms can be deduced from it as well. However, it certainly doesn't include the properties related to inequalities. So, if I rig up an algebraic statement (like the above) whose proof requires the order properties of $\mathbb{R}$, you won't be able to prove it from Pappus theorem. I haven't actually worked this out, but presumably if you apply Tarski's algorithm to the above claim, it will give you a proof which, at some point, involves dividing by $x^2+x+1$ for some unknown quantity $x$.
I'll mention that the axioms of an oriented matroid are an attempt to systematize the properties of $\mathbb{RP}^2$ deducible from the order properties of $\mathbb{R}$. It might be true that every true incidence theorem in $\mathbb{RP}^2$ is deducible from Pappus theorem, plus the axiom that the set of points of the plane can be equipped with the structure of an oriented matroid of rank $3$, where the bases are the noncolinear triples.
Please forgive me if you are aware of this result (as it is linked from the Wikipedia page, albeit in another context), but there is a paper by Richard K. Guy called "The lighthouse theorem, Morley & Malfatti—a budget of paradoxes" in the American Mathematical Monthly. The eponymous theorem could be considered a generalization of Morley's theorem:
Lighthouse Theorem. Two sets of $n$ lines at equal angular distances, one set through each of the points $B$, $C$, intersect in $n^2$ points that are the vertices of $n$ regular $n$-gons.
Naturally, it is not clear how this would qualify as a generalization, but the connecting observation is the following:
The Morley Miracle. The nine edges of the equilateral triangles of the Lighthouse Theorem for $n=3$ are the Morley lines of a triangle.
Properly, the Lighthouse Theorem should be enlarged to include enough observations to make this connection. For example, the $n^2$ lines of the $n$ regular $n$-gons form $n$ families of $\binom{n}{2}$ parallel lines; if $n$ is odd, then the $n$-gons are homothetic. Moreover, there is an angle duplication result that establishes the presence of the trisectors.
From Guy's point of view, the particularly pleasant appearance of Morley's theorem is due to the fact that $\binom{n}{2} = n$ for $n=3$. For comparison, the case $n=2$ is even simpler and may be regarded as the statement that the altitudes of a triangle concur. (The $n$ $n$-gons are an orthocentric system.) The case $n=4$ gives some properties of Malfatti circles. For all of these interpretations, Guy wrestles with the "paradox" that you recover theorems about a triangle even though you don't start with any triangles.
Again, my apologies if you're aware of all of this. I imagine you may be, in which case I justify my answer as simply too long for a comment!
Best Answer
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.