# derivation of Pappus’s centroid theorem

I.  Let $s$ denote the arc rotating about the $x$-axis (and its length) and $R$ be the $y$-coordinate of the centroid of the arc.  If the arc may be given by the equation

 $y\;=\;y(x)$

where  $a\leq x\leq b$, the area of the formed surface of revolution is

 $A\;=\;2\pi\!\int_{a}^{b}\!y(x)\sqrt{1\!+\![y^{\prime}(x)]^{2}}\,dx.$

This can be concisely written

 $\displaystyle A\;=\;2\pi\!\int_{s}\!y\,ds$ (1)

since differential-geometrically, the product $\sqrt{1\!+\![y^{\prime}(x)]^{2}}\,dx$ is the arc-element.  We rewrite (1) as

 $A\;=\;s\cdot 2\pi\cdot\frac{1}{s}\!\int_{s}\!y\,ds.$

Here, the last factor is the ordinate of the centroid of the rotating arc, whence we have the result

 $A\;=\;s\cdot 2\pi R$

which states the first Pappus’s centroid theorem.

II.  For deriving the second Pappus’s centroid theorem, we suppose that the region defined by

 $a\;\leq\,x\;\leq\;b,\quad 0\;\leq\;y_{1}(x)\;\leq\,y\;\leq\;y_{2}(x),$

having the area $A$ and the centroid with the ordinate $R$, rotates about the $x$-axis and forms the solid of revolution with the volume $V$.  The centroid of the area-element between the arcs  $y=y_{1}(x)$  and  $y=y_{2}(x)$  is $[y_{2}(x)\!+\!y_{1}(x)]/2$ when the abscissa is $x$; the area of this element with the width $dx$ is $[y_{2}(x)\!-\!y_{1}(x)]\,dx$.  Thus we get the equation

 $R\;=\;\frac{1}{A}\int_{a}^{b}\frac{y_{2}(x)\!+\!y_{1}(x)}{2}[y_{2}(x)\!-y_{1}(% x)]\,dx$

which may be written shortly

 $\displaystyle R\;=\;\frac{1}{2A}\int_{a}^{b}(y_{2}^{2}\!-\!y_{1}^{2})\,dx.$ (2)

The volume of the solid of revolution is

 $V\;=\;\pi\!\int_{a}^{b}(y_{2}^{2}\!-\!y_{1}^{2})\,dx\;=\;A\cdot 2\pi\cdot\frac% {1}{2A}\!\int_{a}^{b}(y_{2}^{2}\!-\!y_{1}^{2})\,dx.$

By (2), this attains the form

 $V\;=\;A\cdot 2\pi R.$
Title derivation of Pappus’s centroid theorem DerivationOfPappussCentroidTheorem 2013-03-22 19:36:11 2013-03-22 19:36:11 pahio (2872) pahio (2872) 7 pahio (2872) Derivation msc 53A05