tl; dr: The formulas work out for a cone of height $h$ and base radius $R$ in four-space. The volume is indeed
\begin{align*}
\tfrac{1}{3}\pi R^{3}h &= (\tfrac{1}{2}Rh)(\tfrac{2}{3}\pi R^{2}) \\
&= (\text{area of generating triangle})(\text{area of sphere through the triangle's $p$-centroid})
\end{align*}
for a suitable "$p$-centroid." This point is not the usual geometric centroid, however: Its location depends on $p$.
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Vec}[1]{\mathbf{#1}}\DeclareMathOperator{\Vol}{Vol}$Generalities: To start, here's a brief account of "volumes of rotation of species $p$." (This is doubtless the content of Sommerville's book, but Sommerville's notation is not how I visualize the geometry.) Let $p$ and $m$ be positive integers, and let $n = m + p + 1$. To describe what is meant by "revolving a region $D \subset \Reals^{m+1}$ about $p$-dimensional spheres," let $SO(p+1)$ denote the group of Euclidean rotations of $\Reals^{p+1}$, and decompose Cartesian $n$-space as
\begin{align*}
\Reals^{n} &= \Reals^{m} \times \Reals \times \Reals^{p} \\
&= \Reals^{m} \times \underbrace{\Reals^{p+1}}_{\text{$SO(p+1)$ acts}} \\
&= \underbrace{\Reals^{m+1}}_{\text{$D$ lives here}} \times \Reals^{p}.
\end{align*}
If $\Vec{x} \in \Reals^{m}$, $\Vec{y}_{0} \in \Reals^{p}$, $r$ is real, and $\Vec{y} = (r, \Vec{y}_{0}) \in \Reals^{p+1}$, the general element of $\Reals^{n}$ may be written
$$
(\Vec{x}, \Vec{y}) = (\Vec{x}, r, \Vec{y}_{0}).
$$
We're arranging that $D$ is contained in the half-space of $\Reals^{m+1}$ where $r \geq 0$ and $\Vec{y}_{0} = \Vec{0}$. A rotation $A \in SO(p+1)$ acts by
$$
(\Vec{x}, \Vec{y}) \mapsto (\Vec{x}, A\Vec{y});
$$
the orbit of the point $(\Vec{x}, r, \Vec{0}) \in D$ is a $p$-sphere of radius $r$, specifically $\{\Vec{x}\} \times S^{p}(r) \subset \{\Vec{x}\} \times \Reals^{p+1}$.
Under this action, a volume element $d\Vec{x}\, dr$ of $D$ at $(\Vec{x}, r, \Vec{0})$ sweeps an infinitesimal volume
$$
\Vol_{p} S^{p}(r)\, d\Vec{x} = (\Vol_{p} S^{p}) r^{p}\, d\Vec{x}.
$$
The volume swept by revolving $D$ is the integral over $D$,
$$
\Vol_{n}[SO(p+1)(D)] = \Vol_{p} S^{p} \int_{D} r^{p}\, d\Vec{x}.
$$
This framework reduces to that of calculus if $p = m = 1$: In Cartesian coordinates $(x, r, y)$, the region $D$ lies in the half-plane $r \geq 0$ and $y = 0$; the group $SO(p + 1) = SO(2)$ revolves the $(r, y)$-plane about $\Reals^{m}$, a.k.a. the $x$-axis.
If we define
$$
\bar{r}^{p} = \int_{D} r^{p}\, d\Vec{x}\bigg/\int_{D} d\Vec{x}
= \frac{\Vol_{n}[SO(p+1)(D)]}{\Vol_{p} S^{p} \Vol_{m+1}(D)},
$$
then $\bar{r}$ is (by fiat, see Note at the bottom) the distance from the $p$-centroid to $\Reals^{m}$. Pappus' theorem is a formal triviality:
\begin{align*}
\Vol_{n}[SO(p+1)(D)] &= (\Vol_{p} S^{p}) \bar{r}^{p} \Vol_{m+1}(D) \\
&= \Vol_{p} S^{p}(\bar{r}) \Vol_{m+1}(D),
\end{align*}
the $p$-dimensional volume of the sphere of radius $\bar{r}$ times the $(m + 1)$-dimensional volume of $D$.
The four-dimensional cone: Here $m = 1$ and $p = 2$. The region $D$ may be taken to be the triangle $0 \leq x \leq h$ and $0 \leq r \leq Rx/h$. The $2$-centroid (really, its distance $\bar{r}$ from the axis) satisfies
$$
\bar{r}^{2} = \frac{1}{\frac{1}{2}Rh} \int_{0}^{h} \int_{0}^{Rx/h} r^{2}\, dr\, dx
= \frac{2R^{2}}{3h^{4}} \int_{0}^{h} x^{3}\, dx
= \frac{R^{2}}{6}.
$$
The sphere of radius $\bar{r}$ has area $4\pi \bar{r}^{2} = \frac{2}{3}\pi R^{2}$, so as claimed above,
\begin{align*}
(\text{area of generating triangle})(\text{area of sphere through $p$-centroid})
&= (\tfrac{1}{2}Rh)(\tfrac{2}{3}\pi R^{2}) \\
&= \tfrac{1}{3}\pi R^{3}h \\
&= \text{volume of the (hyper-)cone}.
\end{align*}
Note: I don't see a natural mechanical interpretation of $\bar{r}$ Probabilistically, $\bar{r}^{p}$ is the mean of $r^{p}$, the $p$th power of the distance function from the "axis" (i.e., from $\Reals^{m}$). That is, $\bar{r}$ is the $p$-norm of the distance function $r$ over $D$, computed with respect to $(m + 1)$-dimensional Lebesgue measure.
Best Answer
$\newcommand{\dd}{\partial}$If $\bigl(x(t), z(t)\bigr)$, $a \leq t \leq b$, parametrizes a smooth plane curve $C$ in the half-plane $x > 0$, the surface $S$ swept out by revolving $C$ about the $z$-axis may be parametrized by $$ X(s, t) = \bigl(x(t) \cos s, x(t) \sin s, z(t)\bigr),\qquad a \leq t \leq b,\quad 0 \leq s \leq 2\pi. $$ The partial derivatives are \begin{align*} \frac{\dd X}{\dd s} &= \bigl(-x(t) \sin s, x(t) \cos s, 0\bigr), \\ \frac{\dd X}{\dd t} &= \bigl(x'(t) \cos s, x'(t) \sin s, z'(t)\bigr); \end{align*} Their cross product is $$ \frac{\dd X}{\dd s} \times \frac{\dd X}{\dd t} = -x(t) \bigl(z'(t) \cos s, z'(t) \sin s, x'(t)\bigr); $$ the area element is $$ \left\lVert\frac{\dd X}{\dd s} \times \frac{\dd X}{\dd t}\right\rVert ds\, dt = x(t) \sqrt{z'(t)^{2} + x'(t)^{2}}\, ds\, dt. \tag{1} $$ The surface area of $S$ is $$ \int_{a}^{b} \int_{0}^{2\pi} x(t) \sqrt{z'(t)^{2} + x'(t)^{2}}\, ds\, dt = 2\pi \int_{a}^{b} x(t)\sqrt{z'(t)^{2} + x'(t)^{2}}\, dt. $$ If $$ \ell = \int_{a}^{b} \sqrt{z'(t)^{2} + x'(t)^{2}}\, dt $$ denotes the arc length of $C$, the area of $S$ becomes $$ 2\pi \int_{a}^{b} x(t) \sqrt{z'(t)^{2} + x'(t)^{2}}\, dt = 2\pi\, \ell \left(\frac{1}{\ell} \int_{a}^{b} x(t) \sqrt{z'(t)^{2} + x'(t)^{2}}\, dt\right) = \ell\, (2\pi\, \bar{x}), $$ the length of $C$ times the circumference of the circle swept by the centroid of $C$.