I took a hint from this StackExchange answer. The volume of the sphere with a given surface area is fixed. The volume of the cylinder can vary based on variations in height and radius of the cylinder. We can calculate the maximum volume of a cylinder for a given surface area using ratio $\frac{h}{r} = 2$ as proved in the question I linked. If we use this relation, we have :
$$2r_2^2 = r_1(r_1+h)$$
$$\implies 2r_2^2 = 3r_1^2$$
$$\implies \frac{V_2}{V_1} = \frac{\frac{4}{3}\pi r_2^3}{\pi r_1^2h} = \frac{4}{3} \frac{r_2^3}{2r_1^3} = \sqrt{\frac{3}{2}} \gt 1$$
Thus, we can infer that the maximum volume of a cylinder with given surface area is also less than the volume of a sphere with that surface area. So, $V2 \gt V1$ can be inferred based on this argument.
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
The centroid of semi-circle should be $\dfrac{4r}{3\pi}$ from the origin.
So the volume is
$$V=2\pi \times \frac{4r}{3\pi}\times \frac{\pi r^2}{2}=\frac{4}{3}\pi r^3$$
The centroid of semi-circular arc should be $\dfrac{2r}{\pi}$ from the origin.
So the surface area is
$$S=2\pi \times \frac{2r}{\pi}\times \pi r=4\pi r^2$$