The Arc Length of the parabolic segment shown above is given by

(1) |

The Area contained between the curves

(2) | |||

(3) |

can be found by eliminating ,

(4) |

(5) |

(6) |

so the Area is

(7) |

Now,

(8) | |||

(9) |

So

(10) |

We now wish to find the maximum Area of an inscribed Triangle. This Triangle will have two of its Vertices at the intersections, and Area

(11) |

(12) |

The maximum Area will occur when

(13) |

(14) | |||

(15) |

so

(16) |

(17) |

(18) | |||

(19) |

so

(20) |

(21) |

which gives the result known to Archimedes in the third century BC that

(22) |

The Area of the parabolic segment of height opening upward along the *y*-Axis is

(23) |

(24) |

(25) |

**References**

Beyer, W. H. (Ed.) *CRC Standard Mathematical Tables, 28th ed.* Boca Raton, FL: CRC Press, p. 125, 1987.

© 1996-9

1999-05-26