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Volume of a tetrahedron using triple integration

  1. Jul 25, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the tetrahedron in the first octant bounded by the coordinate planes and the plane passing through (1,0,0), (0,2,0) and (0,0,3)

    2. Relevant equations

    V=∫∫∫dV
    ...D


    3. The attempt at a solution

    I set up the problem as so:


    1 -2x+2...-3x+3
    ∫ ∫........∫......dz dy dx
    0 0........0

    (the dots are not significant, they are only for spacing)

    and integrated.....

    the result is -1, but this is wrong because:

    1. a volume cannot be negative
    2. the book answer is 1
     
  2. jcsd
  3. Jul 25, 2007 #2
    actually, i realize that the line should be

    -1½x+3 instead of -3x+3

    still, i get the answer as 2.....
     
  4. Jul 26, 2007 #3

    HallsofIvy

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    Science Advisor


    How did you get z= -3x+3 for the upper limit? Doesn't z change as y changes?

    Again, you should not be thinking of a line. The first integral should be from the xy-plane up to the plane forming the upper boundary.

    First, as I presume you have done, draw a picture. The base, in the xy-plane, is the triangle bounded by the x-axis, the y-axis, and the line from (1,0) to (0,2): its equation is y= 2- 2x. Obviously, we can cover the entire triangle by taking x running from 0 to 1 and, for each x, y running from 0 to 2-2x. The plane through (1, 0, 0), (0, 2, 0), and (0, 0, 3) has equation
    [tex]x+ \frac{y}{2}+ \frac{z}{3}= 1[/tex]
    or
    [tex]z= 1- 3x-\frac{3y}{2}[/tex]
    The volume is given by
    [tex]\int_{x=0}^1\int_{y= 0}^{2-2x}\int_{z=0}^{1-3x-\frac{3y}{2}} dzdydx[/tex]
    Integrate that and see what you get.
     
  5. Jul 27, 2007 #4
    honestly, this problem can be made very easy like this:
    volume= 1/3height * area of base. Height is z=3, area of base is the area of a right triangle with two sides being 1 and 2. Therefore, V=1.
     
  6. Jul 27, 2007 #5

    HallsofIvy

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    Yes. In fact, the volume of any tetrahedron with vertices at (a, 0, 0), (0, b, 0), and (0, 0, c) is (1/6)abc but I assumed from the original post that he wanted to do this without using specific formulas.
     
  7. Aug 24, 2007 #6
    thanks for your help everyone, indeed the issue was the Z limit.....and yea i needed to know how to do it without specific formulas (it was on a Calculus III test) but I did use the formulas to confirm my answer
     
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