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Volume by triple integral

  1. May 17, 2010 #1
    1. The problem statement, all variables and given/known data
    use triple integral to find the volume of tetrahedron enclosed by the coordinat planes "x=o , y=0 , z=0" and the plane 2x+y+z=0


    2. Relevant equations



    3. The attempt at a solution

    I will integrate the constant function f(x,y,z)=1 by the order : dzdydx

    the equation will be : z=-2x-y
    so the limits for the inner integral will be from 0 to -2x-y

    when z=0 ---> y=-2x
    so the limits for the middle integral will be from 0 to -2x

    THE PROBLEM HERE IS THAT
    when z=0,y=0 ---> x=0 .. !!!!
    so the limits for the outer integral will be from 0 to 0 .. !!
    and this means the triple integral will be 0 .. !!
    so there is no volume ??!
     
  2. jcsd
  3. May 17, 2010 #2
    I think there is a mistake in the plane's equation, right ?
    If I find the x&y&z intercepts, all will be (0,0,0)
    so there is no plane !
    Right?
     
  4. May 17, 2010 #3

    LCKurtz

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    Gold Member

    These type of homework problems typically ask for the volume in the first octant ...

    Your plane doesn't pass through the first octant because of the 0 on the right side of the equation. To get three positive intercepts you need a positive number on the right, then it will form a tetrahedron with the coordinate planes. Check the problem is copied correctly.
     
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