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Changing Order of Integration on Triple Integral

  1. Jun 28, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\int^{1}_{-1}\int^{1}_{x^2}\int^{1-y}_{0} dz dy dx[/tex]

    2. Relevant equations

    See the attachment for graph. I am supposed to rewrite the order of integration to the following.

    a)dy dz dx
    b)dy dx dz
    c)dx dy dz

    ...and so on.

    3. The attempt at a solution

    First attempt is worthless as I don't understand how to do these. I understand doing them in 2d but with the third direction I am lost. My teacher, book, and solution manual are no help because I guess these are so easy you just get the answers not how they get the answers.

    The reason I know I don't get this is I cannot see how the limits change. For the z-axis it seems to me that z will always be from 0 to 1. I know that's wrong but I don't get why.

    Any tips to help me understand this?

    Attached Files:

  2. jcsd
  3. Jun 29, 2010 #2


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    Let's say that z is going to be our first integration variable. Then when x=0 and y=1, z only ranges from 0 to 0. For any given x and y, the smallest value z can be is going to be 0, but the largest value is going to be a function of x and y, depending on how high the top of your volume is. Basically by having z go first, you're saying for every fixed value of x and y, we're going to integrate over z only, and we only want to integrate over the choices of z so that we are still contained inside the volume (which is independent of the fact that on the z-axis z goes from 0 to 1)
  4. Jun 29, 2010 #3
    Hmm....I'm not sure if I understand. How does one know which values of x and y would be used to determine z?

    I see that z can be either 0 to 1 or 0 to 0 to 1-y, but have no idea which to use and when. Also the answer for one of them contains a small function that I don't even see in the set of limits.
  5. Jun 29, 2010 #4


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    The place where z goes from 0 to 1, is where y=0. So you're really going from 0 to 1-y. When it's going from 0 to 0, y=1 so you're still going from 0 to 1-y. 1-y is the correct answer because the way the top of your volume is defined is by z+y=1, so the largest z can be is z=1-y

    If you had a different shape to integrate over, you would need to find out how to describe z in terms of x and y for the top and bottom all over again
  6. Jun 29, 2010 #5
    I think this is where I am going to fail the class.:redface:

    Here's my next stupid question.

    In the original setup dy is from x^2 to 1. Why couldn't it be from x^2 to 1-z? I can see it is similar to your explanation above this post, but I don't see it yet.
  7. Jun 29, 2010 #6
    Ok....I really don't get this one.

    Finding dy dz dx

    The answer is [tex]\int^{1}_{-1}\int^{1-x^2}_{0}\int^{1-z}_{x^2} dy dz dx[/tex]

    I get all of them except the upper limit for dz. How is it 1-x^2? There isn't even a line with that equation??
  8. Jun 29, 2010 #7


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    No, but you are told, in your first integral, that, for all x, and y, z ranges from 0 to y. Further, you know that y ranges from y to [itex]1- x^2[/itex]. That is, z ranges from 0 to [/itex]y= 1- x^2[/itex].
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