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Triple Integral of minimum value

  1. Dec 5, 2011 #1
    1. The problem statement, all variables and given/known data
    The problem is to take the triple integral over B of min{x,y,z}dV where B = {x,y,z}: o<x<1 , 0<y<1, and 0<z<1. (These are all less than or equal to, I didn't know the notation).


    2. Relevant equations



    3. The attempt at a solution

    What I did was break it up into three cases; where x is the minimum, y is the minimum, and where z is the minimum. I got a value of 1/4 for all three of these, so that leads me to believe the answer for the triple integral is 1/4. Is this correct thinking
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 5, 2011 #2

    SammyS

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    Hello BioMan789. Welcome to PF.

    It would help greatly if you could show just what it was that you integrated. I suspect that you have miisunderstood the meaning of the function, min(x, y, z) .

    .
     
  4. Dec 5, 2011 #3
    In between my original post and your response I realized I was integrating incorrectly. I was integrating from 0 to x, 0 to y, 0 to z for each integral instead of 1 to x, 1 to y, 1 to z (see below for clarification). This is what I was integrating.

    ∫∫∫ x dydzdx (first integral is from x to 1, second is also from x to 1, final is from 0 to 1).
    ∫∫∫ y dxdzdy (first integral is from y to 1, second is also from y to 1, final is from 0 to 1).
    ∫∫∫ z dxdydz (first integral is from z to 1, second is also from z to 1, final is from 0 to 1).
     
  5. Dec 5, 2011 #4
    When I worked these integrals out I got 1/12
     
  6. Dec 5, 2011 #5

    SammyS

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    I'm pretty sure those integrals won't give the correct result.

    I've been trying to wrap my head around this for a while. Visualizing this has been a challenge.

    [itex]\displaystyle f(x,\,y,\,z)=\min(x,\,y,\,z)=\left\{\matrix{
    x\,,\ \ \text{ if }\ x\le y\ \ \text{ and }\ \ x\le z\\
    y\,,\ \ \text{ if }\ y\le x\ \ \text{ and }\ \ y\le z\\
    z\,,\ \ \text{ if }\ z\le x\ \ \text{ and }\ \ z\le y}\right.[/itex]

    My suggestion is to write f(x, y, z) as the sum of three functions, fx(x, y, z) + fy(x, y, z) + fz(x, y, z), where
    fx(x, y, z) = x when x is the minimum, otherwise it equals zero.

    fy(x, y, z) = y when y is the minimum, otherwise it equals zero.

    fz(x, y, z) = z when z is the minimum, otherwise it equals zero.​

    Then integrate all three functions over the entire cube. But since each is zero over 2/3 of the cube, simply integrate each over the region for which it's not zero. Considering symmetry, each of those integrals will give the same result.
    For example, integrate y first for x going from y to 1, then for z going from y to 1, then for y going from 0 to 1.​

    [STRIKE]Added in Edit:
    I struck part of the line above after looking at the figure in jackmell's post. That line should read:
    For example, integrate y first for x going from y to 1, then for z going from 0 to y, then for y going from 0 to 1.​
    [/STRIKE]
    Added in 2nd Edit:
    After further consideration, I withdraw the 1st Edit. and restore the post as it was first written. (This does not imply that jackmell's post was wrong -- only that I misinterpreted his figure.)
     
    Last edited: Dec 5, 2011
  7. Dec 5, 2011 #6
    I think it's gonna be six integrals over tetrahedrons. Here's a picture of three of them. Got the blue, red and yellow there and then three more behind them I'm not showing. So for example, we have:

    [tex]\mathop\iiint\limits_{\text{Red}} \text{min(x,y,z)}dV=\int_0^1\int_0^x\int_y^x y\;dzdydx[/tex]

    Ok, Bioman, you try and do the other five or at least do the yellow one or maybe the blue one would be easier to do first.
     

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  8. Dec 6, 2011 #7
    I got those figures by first segregating the cube along diagonal planes shown below. Also, I don't think my wonderful drawings would help when we go to 4D and beyond as in the general problem that was posted a few days ago:

    [tex]\int\cdots\int \text{min}(x_1,\cdots,x_n)dx_1\cdots dx_n[/tex]

    The link is:

    https://www.physicsforums.com/showthread.php?t=555766
     

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