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Homework Help: 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


    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


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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]
    [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


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    Science Advisor

    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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