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Homederivation of Pappus's centroid theorem

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# 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

$V\;=\;A\cdot 2\pi R.$ |

## Mathematics Subject Classification

53A05*no label found*

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