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Volume of tetrahedron

  1. Apr 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the tetrahedron with vertices at [itex] (0,0,0),(1,0,0),(0,1,0),(0,0,1) [/itex]


    3. The attempt at a solution

    I worked out the triple integral and found out that the volume is [itex] \frac{1}{6} [/itex]? Is this correct? I know there is probably a much quicker way working the volume by just using the volume formula for tetrahedron. However, I am not sure which value to substitute to the formula, so could you just tell me whether this answer is right or not?

    Thanks!
     
  2. jcsd
  3. Apr 22, 2010 #2
    The vertices you give do not make a tetrahedron. Try to draw the points in three dimensions. You'll see the volume is just half that of a cube with length 1. So the volume you're seeking should be 1/2.
     
  4. Apr 22, 2010 #3

    Dick

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    For a tetrahedron, like a cone, the area is (1/3)*(area of the base)*height. So (1/3)*(1/2)*1=1/6, yes.
     
  5. Apr 22, 2010 #4

    HallsofIvy

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    Mr. Miyagi is wrong. That is in fact a tetrahedron, it is 1/6 of a cube, not 1/2, and its volume is, indeed, 1/6.

    More generally, the volume of the tetrahedron is with vertices at (0, 0, 0), (a, 0, 0), (0, b, 0), and (0, 0, c) is (abc)/6.
     
  6. Apr 22, 2010 #5
    Ugh, sorry about that... Is it too late to claim temporary insanity? :uhh:

    Thanks for correcting it so quickly.
     
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