Pappus's Centroid Theorem Pappus's Centroid Theorem may refer to one of two theorems. 
Theorem 1:
The surface area of a solid of revolution is the arc length of the generating curve multiplied by the distance traveled by the centroid of the curve. 
Theorem 2:
The volume of a solid of revolution is the area between the generating curve and the rotation axis multiplied by the distance traveled by the centroid of the curve. 
I've managed to prove 2nd theorem like this: 
Let the $x$-axis be the rotation axis and $y=f(x)$, $a \leq x \leq b$  be the generating curve. The $y$-coordinate of the centroid, $\bar y$, is then 
$$\frac{1}{2A}\int_{a}^{b}f(x)^2dx$$
where
$$A=\int_{a}^{b}f(x)dx$$
So, we have that 
$$2\pi\bar yA=2\pi A\left(\frac{1}{2A}\int_{a}^{b}f(x)^2dx\right)=\pi\int_a^bf(x)^2dx$$
My question is, how can we prove the second theorem?
 A: $\newcommand{\dd}{\partial}$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 $C$.
