Editing Pappus's centroid theorem (section)

===Proof 2===
Let <math>A</math> be the area of <math>F</math>, <math>W</math> the solid of revolution of <math>F</math>, and <math>V</math> the volume of <math>W</math>. Suppose <math>F</math> starts in the <math>xz</math>-plane and rotates around the <math>z</math>-axis. The distance of the centroid of <math>F</math> from the <math>z</math>-axis is its <math>x</math>-coordinate
<math display="block">R = \frac{\int_F x\,dA}{A},</math>
and the theorem states that
<math display="block">V = Ad = A \cdot 2\pi R = 2\pi\int_F x\,dA.</math>

To show this, let <math>F</math> be in the ''xz''-plane, [[Parametric equation|parametrized]] by <math>\mathbf{\Phi}(u,v) = (x(u,v),0,z(u,v))</math> for  <math>(u,v)\in F^*</math>, a parameter region. Since <math>\boldsymbol{\Phi}</math> is essentially a mapping from <math>\mathbb{R}^2</math> to <math>\mathbb{R}^2</math>, the area of <math>F</math> is given by the [[Integration by substitution#Substitution for multiple variables|change of variables]] formula:
<math display="block">A = \int_F dA = \iint_{F^*} \left|\frac{\partial(x,z)}{\partial(u,v)}\right|\,du\,dv = \iint_{F^*} \left|\frac{\partial x}{\partial u} \frac{\partial z}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial z}{\partial u}\right|\,du\,dv,</math>
where <math>\left|\tfrac{\partial(x,z)}{\partial(u,v)}\right|</math> is the [[determinant]] of the [[Jacobian matrix and determinant|Jacobian matrix]] of the change of variables. 

The solid <math>W</math> has the [[torus|toroidal]] parametrization <math>\boldsymbol{\Phi}(u,v,\theta) = (x(u,v)\cos\theta,x(u,v)\sin\theta,z(u,v))</math> for <math>(u,v,\theta)</math> in the parameter region <math>W^* = F^*\times [0,2\pi]</math>; and its volume is
<math display="block">V = \int_W dV = \iiint_{W^*} \left|\frac{\partial(x,y,z)}{\partial(u,v,\theta)}\right|\,du\,dv\,d\theta.</math>

Expanding,
<math display="block">
\begin{align}
\left|\frac{\partial(x,y,z)}{\partial(u,v,\theta)}\right| & = 
\left|\det\begin{bmatrix}
\frac{\partial x}{\partial u}\cos\theta & \frac{\partial x}{\partial v}\cos\theta & -x\sin\theta \\[6pt]
\frac{\partial x}{\partial u}\sin\theta & \frac{\partial x}{\partial v}\sin\theta &  x\cos\theta \\[6pt]
\frac{\partial z}{\partial u}           & \frac{\partial z}{\partial v}           &  0
\end{bmatrix}\right| \\[5pt]
& = \left|-\frac{\partial z}{\partial v}\frac{\partial x}{\partial u}\,x + \frac{\partial z}{\partial u}\frac{\partial x}{\partial v}\,x\right|
=\ \left|-x\,\frac{\partial(x,z)}{\partial(u,v)}\right| = x\left|\frac{\partial(x,z)}{\partial(u,v)}\right|.
\end{align}
</math>

The last equality holds because the axis of rotation must be external to <math>F</math>, meaning <math>x \geq 0</math>. Now,
<math display="block">
\begin{align}
V &= \iiint_{W^*} \left|\frac{\partial(x,y,z)}{\partial(u,v,\theta)}\right|\,du\,dv\,d\theta \\[1ex]
&= \int_0^{2\pi}\!\!\!\!\iint_{F^*} x(u,v)\left|\frac{\partial(x,z)}{\partial(u,v)}\right| du\,dv\,d\theta \\[6pt]
& = 2\pi\iint_{F^*} x(u,v)\left|\frac{\partial(x,z)}{\partial(u,v)}\right|\,du\,dv \\[1ex]
&= 2\pi\int_F x\,dA
\end{align}
</math>
by change of variables.