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Why is it symmetrical?

  1. Mar 13, 2017 #1
    • Member warned that an effort toward solution must be shown
    1. The problem statement, all variables and given/known data
    Why is it symmetrical thus zero?

    2. Relevant equations
    The original equation is with main focus on the first part, cause therein lies the symmetry
    [tex] \int\int\int xy \quad dV + \int\int\int z^2 \quad dV [/tex]
    and
    [tex] 0 \leq z\leq 1 - x - y [/tex]
    See also the picture!

    3. The attempt at a solution
    I drew a drawing in the xy-plane and it was a line with equation [tex]y = 1-x[/tex].
    So that allowed me to set up the triple integral as:

    [tex]\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y} xy \quad dzdydx = \frac{1}{120}[/tex]
     

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    Last edited: Mar 13, 2017
  2. jcsd
  3. Mar 13, 2017 #2

    BvU

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    Could you please post your steps in detail for one of those, so we can help you ?
     
  4. Mar 13, 2017 #3

    jedishrfu

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    Can you show us the triple integral for the xy term and how you evaluated it?
     
  5. Mar 13, 2017 #4

    mfb

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    It would have been useful to write the problem statement here.
    If you integrate it over 1/4 of the set only, it is not zero. Integrate it over the full set (not what the image shows) and you should get zero.
     
  6. Mar 13, 2017 #5
    [tex] \int\int\int xy \quad dV + \int\int\int z^2 \quad dV [/tex]
    basically what is shown in the picture.
    I chose
    [tex]\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y} xy \quad dzdydx[/tex]
    from here on it is easy peasy, I dropped it in Mathmatica and it said [tex]\frac{1}{120}[/tex] so I started to cry.


    What exactly do you mean with 1/4?

    Yes, I plotted [tex]z = 1- x - y[/tex] in Wolframalpha and I saw that the shape was symmetrical, BUT the domain is for [tex]z\geq 0.[/tex] So I don't understand why it is okay. If you would say that the front in the picture is 1/4 and the back is 1/4 so a total of 1/2, then I understand but again the book wants you to calculate for z above 0, zo why multiply with 4 and not 2?

    I can figure that it is an odd function because [tex] f(-x) = -f(x)[/tex] but the calculation does not add up.

    NOTE: I used wolframalpa to be sure that I don't mess up the algebra, so I make a mistake however I don't understand where!
     
    Last edited: Mar 13, 2017
  7. Mar 13, 2017 #6

    LCKurtz

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    The integral you have calculated is correct, giving ##\frac 1 {120}##. But the ##xy## domain for your pyramid looks like this:
    diamond.jpg
    You have only done the integral of ##xy## over the shaded portion. If you do the remaining three regions you will find that you get the same ##\frac 1 {120}## for quadrant III and ##-\frac 1 {120}## in quadrants II and IV, so the total integral is ##0##. It is because of the signs of ##x## and ##y## in the various quadrants.
     
  8. Mar 13, 2017 #7

    Ray Vickson

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    You can see it is zero without doing any calculations.

    An integral ##\int\int\int_R f(x,y,x) \, dV## is really the limit of the sums ##\sum_{i} f(x_i,y_i, z_i) \Delta V## as the number of terms ##i## goes to ##\infty## and the individual box volumes ##\Delta V## go to 0. If we fix the box sides at length ##\delta > 0##, then ##\Delta V = \delta^3.## Now, for any box centered at ##(x_i,y_i,z_i)## in the first orthant, there are three other boxes centered at ##(-x_i, y_i, z_i), (x_i,-y_i,z_i), (-x_i -y_i,z_i).## The contribution of those four boxes to the sum is ##[x_i y_i + (-x_i) y_i + x_i (-y_i) + (-x_i) (-y_i)] \delta^3 = 0.##
     
    Last edited: Mar 14, 2017
  9. Mar 13, 2017 #8
    Yes! I understand, however I only drew the first line and not the other three. Why and how could I have distilled that from the equation?

    [tex] z = 1- x - y[/tex], with [tex]z = 0[/tex] in xy-plane gives [tex]y = 1- x[/tex]
     
  10. Mar 13, 2017 #9

    LCKurtz

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    The equation you were given was ##0 \le z \le 1 - |x| - |y|##, and that right side is ##1-x-y## only in the first ##xy## quadrant where both ##x## and ##y## are positive. The equation ##1 - |x| - |y| = 0## gives the diamond shape in the ##xy## plane.
     
  11. Mar 14, 2017 #10
    So that was my mistake! I was working on my problem for hours and couldn't find out what the problem was because I thought I did it right. The problem are the absolute values, I thought that meant it belonged in the first orthant only. However when I plotted the equation in wolframalpha.com with the absolute values I saw that it is a 4 sides pyramid.

    Now I am wondering what the absolute values in ##0 \le z \le 1 - |x| - |y|## mean.

    This is a different approach that asks a lot of insight. I had to think about it a long time, but I think I understand it.

    Thank you for helping!
     
  12. Mar 14, 2017 #11

    BvU

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    An almost essential key element is the picture LC showed in #6. From that such insight is almost trivial (certainly if the last line of the problem statement spells it out).
     
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