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From: js2j@mhuxt.UUCP (sonntag)
Newsgroups: net.math
Subject: Re: volume of a tetrahedron
Message-ID: <887@mhuxt.UUCP>
Date: Fri, 24-May-85 14:45:22 EDT
Article-I.D.: mhuxt.887
Posted: Fri May 24 14:45:22 1985
Date-Received: Sat, 25-May-85 08:09:57 EDT
References: <758@gloria.UUCP>
Organization: AT&T Bell Laboratories, Murray Hill
Lines: 25

> Does anybody know of a _compact_ expression for the volume
> of a tetrahedron in terms of its edges--something comparable
> to Hero's and Brahmagupta's formulas in plane geometry?  I
> have an expression but it's rather long and hard to remember.
> Col. G. L. Sicherman

    I just computed it, using techniques from high-school level calculus,
and got: V = 1/(6*2^(1/2)) * L^3, where L is the length of a side.  Is that 
compact enough?
    Just in case anyone's interested, this is how I got that formula:
             h
	V = S (1/2)s^2 sin(60) dy
             0

    Where 'S' is the integral sign,
	   h  is the height of the tetrahedron.  h=(2/3)^(1/2)*L
	   s  is the length of the side of the triangle formed by the   
              intersection of the tetrahedron and y=constant.
	      s = L - y (L/h).
	   The (1/2)s^2 sin(60) term is the area of an equalateral triangle
	      with sides of length s.
-- 
Jeff Sonntag
ihnp4!mhuxt!js2j
    "You can be in my dream if I can be in yours." - Dylan