Volume made be a Revolving Pentagon I was looking to the volume created if I revolve the area of the pentagon having vertices
(1,0), (2,2), (0,4), (-2, 2), (-1,0) about the x-axis.
My work:
I used the Second theorem of Pappus, which states that the volume created is equal to the product
of circumference of the circle described by the centroid and the area. In short: $V = (2\pi \bar y)( A)$
I need to get the centroid of the pentagon.
$$\bar x = \frac{1+2+0+(-2)+(-1)}{5} = 0$$
$$\bar y = \frac{0+2+4+2+0}{5} = \frac{8}{5}$$
The area of the polygon laid out in Cartesian Coordinates is:
$$A = \frac{1}{2} \left[ \begin{matrix}
        x_1 & x_2 & x_3 & ... & x_n & x_1\\
        y_1 & y_2 & y_3 & ... & y_n & y_1\\
        \end{matrix} \right] $$
Turns out, the area described by the pentagon above is 10 square units.
We can now get the volume of solid of revolution of pentagon:
$$V = 2 \pi \bar y A$$
$$V = 2 \pi \left(\frac{8}{5}\right) (10)$$
We now get $32 \pi$ cubic units. 
But my book said the volume is $\frac{104}{3} \pi$
Where did I go wrong?
 A: 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:
$$\begin{align}
C_x &= \sum_{i=0}^{n-1} \frac13(x_i + x_{i+1}) \frac{A_i}{A} = \frac{1}{6A} \sum_{i=0}^{n-1} (x_i + x_{i+1})(x_i y_{i+1} - x_{i+1} y_i)\\
C_y &= \sum_{i=0}^{n-1} \frac13(y_i + y_{i+1}) \frac{A_i}{A} =\frac{1}{6A} \sum_{i=0}^{n-1} (y_i + y_{i+1})(x_i y_{i+1} - x_{i+1} y_i)\\
\end{align}
$$
A: It's
$$2\left(\frac{4\cdot\pi\cdot4^2}{3}-\frac{1\cdot\pi\cdot2^2}{3}-\frac{2\cdot\pi\cdot2^2}{3}\right)=\frac{104\pi}{3}$$
A: Partition the pentagon $P$ into three triangles by connecting $(0,4)$ with $(\pm2,0)$. In order to determine the $y$-coordinate $\eta$ of the centroid of $P$ we set up the "moment equation"
$$2\cdot ({3\over2}\cdot2)+{4\over3}\cdot 4+ 2\cdot({3\over2}\cdot2)=\eta\cdot 10\ .$$
On the LHS of this equation appear the $y$-moments $y_c\cdot{\rm area}$ of the three triangles, and on the RHS the $y$-moment of $P$. It follows that $\eta={26\over15}$. Pappus' theorem then gives the following volume of your donut $D$:
$${\rm vol}(D)=2\pi\eta\cdot 10={104\over3}\pi\ .$$
A: Consider the triangle $OAB$, which is made of half the pentagon, the pink and the blue triangles.

A: I think where you failed is assuming the "centroid" determined by averaging the vertex coordinates is the center of mass of a uniform area.  The center of mass is what enters the Theorem of Pappus.
For a triangle these points are the same but for higher polygons, nope.  As an example suppose you have a trapezoid.  Your centroid is halfway between the parallel bases but the real center of mass, like my center of mass when I gain weight where I don't want it, is shifted towards the "fatter" or longer parallel base.
Split your pentagon into triangular pieces by drawing two of the diagonals, then apply the Theorem of Pappus using the centroid and area of each triangle.  Then add the resulting volumes and you should get the right amount.
