Instead of pentagon, consider the case we have a triangle with height $h$ and base $b$ with base lying on the $x$-axis, the centroid will be located at distance of the $\frac{h}{3}$ from the $x$-axis. If we rotate the triangle around the $x$-axis,
by Pappus's centroid theorem, the solid of revolution has volume
$$\left(2\pi \times \frac{h}{3}\right)\times\left(\frac12 hb\right) = \frac{\pi}{3}h^2b$$
We can obtain the pentagon at hand by removing two such triangles with height $2$ and base $3$ from a triangle width height $4$ and base $8$. This means the
volume of solid of revolution of the pentagon is
$$\frac{\pi}{3}\left( 4^2\cdot 8 - 2 \times 2^2\cdot 3\right) = \frac{104}{3}\pi$$
About what's going wrong in your approach, you have used the wrong formula for centroid!
The obvious extension of the formula for centroid of triangle
$$\begin{cases}
\bar{x} &= \frac13\sum\limits_{k=1}^3 x_i\\
\bar{y} &= \frac13\sum\limits_{k=1}^3 y_i
\end{cases}
\quad\longrightarrow\quad
\begin{cases}
\bar{x} &= \frac1n\sum\limits_{k=1}^n x_i\\
\bar{y} &= \frac1n\sum\limits_{k=1}^n y_i
\end{cases}
$$
gives you the center of mass of $n$ point masses instead of a solid polygon!
For a non-self-instersecting closed polygon with $n$ vertices
$(x_0,y_0), (x_1, y_1), \cdots, (x_{n-1}y_{n-1})$, the centroid $(C_x,C_y)$ of the polygon is given by another formula.
Let $(x_n,y_n) = (x_0,y_0)$ and $A_i = \frac12( x_i y_{i+1} - x_{i+1}y_i )$ for $i = 0,\ldots, n-1$.
$A_i$ is the signed area of the triangle with vertices $(0,0),(x_i,y_i),(x_{i+1},y_{i+1})$.
Their sum $A = \sum\limits_{i=0}^{n-1} A_i$ is the signed area of the polygon.
The centroid of the polygon can be computed as a weighted sum of centroids of these $n$ triangles:
$$\begin{align}
C_x &= \sum_{i=0}^{n-1} \frac13(x_i + x_{i+1}) \frac{A_i}{A} = \frac{1}{6A} \sum_{i=0}^{n-1} (x_i + x_{i+1})(x_i y_{i+1} - x_{i+1} y_i)\\
C_y &= \sum_{i=0}^{n-1} \frac13(y_i + y_{i+1}) \frac{A_i}{A} =\frac{1}{6A} \sum_{i=0}^{n-1} (y_i + y_{i+1})(x_i y_{i+1} - x_{i+1} y_i)\\
\end{align}
$$
