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Setting up a triple integral

  1. Dec 9, 2011 #1
    1. The problem statement, all variables and given/known data
    the function is xyz2

    V is bounded by y=1-x, z=0, and z=y.


    3. The attempt at a solution

    the limits are:

    x is from -1 to 1 ?
    y is from 0 to (1-x^2) ?
    z is from 0 to y ?

    the question asks for a picture ... how should that look? there are points on (1,0,0), (-1,0,0), (0,0,1) and (0,1,0) ?
     
  2. jcsd
  3. Dec 9, 2011 #2
    You need one more bound.
    This is likely x=c.
    c is come constant
     
  4. Dec 9, 2011 #3
    i dont get that, i mean like how that helps
     
  5. Dec 9, 2011 #4
    What you are likely to get is a tetrahedron, which requires 4 faces.
    The 4 faces are your 4 bounds, which are planes.
    But you have wrote 3 bounds, what about the 4th?
     
  6. Dec 9, 2011 #5

    HallsofIvy

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    Science Advisor

    Draw a graph. x is from -1 to 1 so draw two vertical lines at x=-1 and x= 1. y if from 0 to 1- x^2 so draw the the line y= 0 and the parabola y= 1- x^2. The region to be integrated is inside that parabola above y= 0. Finally, the plane z= y crosses the y-axis up to (x, 1, 1) and your three dimensional region comes up to that. As Quinzio said, you need another bound on z or that region is not bounded. As it is, z could go up from that plane to infinity of down to negative infinity. Is there another limit, perhaps 0, on the z-integration?
     
  7. Dec 9, 2011 #6
    Well I just gave what the problem gave. The upperimit for z is y?
     
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