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Change the order of triple integration

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data

    Find all 5 other orders of intergration


    2. Relevant equations

    [tex]\int_{0}^{1}\int_{0}^{x^2}\int_{0}^{y}dzdydx[/tex]

    3. The attempt at a solution

    I am really confused as to how to represent it graphicaly so I don't have any visuals to help. Can you guys help me? Thank you SO much!
     
    Last edited by a moderator: Nov 30, 2011
  2. jcsd
  3. Nov 26, 2011 #2

    HallsofIvy

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    Use [ tex ] and [ /tex ] without the spaces:
    [tex]\int_{0}^{1}\int_{0}^{x^2}\int_{0}^{y}dzdydx[/tex]

    Draw an xy- coordinate system. x goes from 0 to 1 so pencil in vertical lines at x= 0 and x= 1. For every x, y goes from 0 to [itex]x^2[/itex] so draw the graph y= x^2. In the xy-plane the figure is bounded by [itex]y= x^2[/itex], y= 0, and x= 1. For every x and y, z goes from 0 to y. That's a plane at 45 degrees cutting through x-axis.


    To change the order of integration to, say, dxdydz, note that y reaches a max of 1 (at x= 1) and so z can have a max of 1. The "outer integral", with respect to z, has limits z= 0 and z= 1. For each z, the equation z= y becomes y= z. y ranges from 0 to z. Finally, since [itex]y= x^2[/itex] gives [itex]x= \sqrt{y}[/itex] (x is positive, remember), x ranges from 0 to [itex]\sqrt{y}[/itex].

    I realize now, this is an error. with y going from 0 to [itex]x^2[/itex], the region of integration is from below and to the right of the graph of [itex]y= x^2[/itex], the integration in x should be from [itex]\sqrt{y}[/itex] to 1.
     
    Last edited: Nov 30, 2011
  4. Nov 26, 2011 #3
    Thank you for getting me started.
    I drew 3 separate graphics to help me.

    I still am confused on how to find all 5. How do you go from one to the other? I am not certain i am keeping the right variables
     
    Last edited: Nov 26, 2011
  5. Nov 26, 2011 #4
    So basicaly, you are saying one of the answers would be

    [tex] \int_{0}^{1}\int_{0}^{z}\int_{0}^{sqrt(y)}dxdydz [/tex]

    Then how do i proceed to find, let's say, dz dx dy?

    I find it hard to visualize, once again. I find the graphic really helps, and I can't seem to get it right in 3D

    Thank you for you help

    A very confused Canadian student ;)
     
  6. Nov 26, 2011 #5
    Wow...i have been trying to figure it out since i first posted this, 2 hours ago. I really don't understand how to proceed. Any help would be much appreciated. Our teacher doesn't TEACH, he just scramble notes on the board.

    It can't be that complicated? With double intergrals i am doing fine....
     
  7. Nov 26, 2011 #6
    I no expert by any means, but I don't think this dxdydz works. Just to check, try having wolfram alpha or wolfram alpha's triple integrator evaluate it with f(x,y,z) equaling, say, 1 (or do it by hand, whichever), and compare that evaluation to your evaluation of f(x,y,z) with dzdydx bounds.

    If they don't match, try letting

    sqrt(y)≤x≤1
    z≤y≤1
    0≤z≤1

    for your dxdydz bounds.
     
  8. Nov 26, 2011 #7
    After a rather long and arduous amount of work I get these other four bounds:

    dxdzdy:
    0≤y≤1
    0≤z≤y
    sqrt(y)≤x≤1

    dydxdz:
    0≤z≤1
    sqrt(z)≤x≤1
    0≤y≤z

    dzdxdy:
    0≤y≤1
    sqrt(y)≤x≤1
    0≤z≤y

    dydzdx:
    0≤y≤z
    0≤z≤x^2
    0≤x≤1
     
  9. Nov 27, 2011 #8
    It really get complicated, but in 3d it's still manageable. It got me hard and arduous work too.
    We have 6 representation altogether. The last two are actually the trickier because you have to split the volume in 2 and manage 2 integrals then to sum them.

    What I get is:
    [tex]\int_{0}^{1} \int_{x^2}^{1} \int_{0}^{y} dz\ dy\ dx[/tex]

    [tex]\int_{0}^{1} \int_{0}^{\sqrt y} \int_{0}^{x^2} dz\ dx\ dy[/tex]

    [tex]\int_{0}^{1} \int_{y}^{1} \int_{0}^{\sqrt z} dx\ dz\ dy[/tex]

    [tex]\int_{0}^{1} \int_{0}^{z} \int_{0}^{\sqrt y} dx\ dy\ dz[/tex]

    [tex]\int_{0}^{1} \int_{\sqrt z}^{1} \int_{z}^{1} dy\ dx\ dz + \int_{0}^{1} \int_{0}^{\sqrt z} \int_{x^2}^{1} dy\ dx\ dz[/tex]

    [tex]\int_{0}^{1} \int_{x^2}^{1} \int_{\sqrt x}^{1} dy\ dz\ dx + \int_{0}^{1} \int_{0}^{x^2} \int_{x}^{1} dy\ dz\ dx[/tex]

    One must absolutely try to visualize the volume and to make intersections with the basic planes [tex]x=0, x=1, y=0, y=1, z=0, z=1[/tex] to get an idea of how it works.

    I hope it's ok, but it was the first of this kind, so I'm not really sure.
     
    Last edited: Nov 27, 2011
  10. Nov 27, 2011 #9
    I think you're right about needing to divide up the region in the last two. However, I also think that you're bounds on dzdxdy are not correct. If you integrate with 1 as your integrand, you find 1/5 as your answer. However, if you use the dzdydx bounds with the same integrand your answer is 1/10.
     
  11. Nov 27, 2011 #10
    Yep. I edited the 1st and the 2nd.
    In fact now there is more "simmetry" between the 1-2 3-4 5-6 couples.
    Thanks.
     
    Last edited: Nov 27, 2011
  12. Nov 27, 2011 #11
    Guys thank you so much for your help.
    I think that is why i blocked (amongst other reasons!!)
    I couldn't figure it out without adding two

    I am so thankful: you guys are better online than the actual teacher ;)
     
  13. Nov 27, 2011 #12
    The only problem is the given integral is
    [tex]\int_{0}^{1} \int_{0}^{x^2} \int_{0}^{y} dz\ dy\ dx[/tex]
    and you wrote
    [tex]\int_{0}^{1} \int_{x^2}^{1} \int_{0}^{y} dz\ dy\ dx[/tex]
     
  14. Nov 29, 2011 #13
    Arghm, another big mistake. I focussed on the method rather than on the actual problem data. Then the integrals should change accordingly.

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

    [tex]\int_{0}^{1} \int_{\sqrt{y}}^{1} \int_{0}^{x^2} dz\ dx\ dy[/tex]

    [tex]\int_{0}^{1} \int_{\sqrt{z}}^{1} \int_{0}^{x^2} dy\ dx\ dz[/tex]

    [tex]\int_{0}^{1} \int_{0}^{x^2} \int_{z}^{1} dy\ dz\ dx[/tex]

    [tex]\int_{0}^{1} \int_{z}^{1} \int_{\sqrt{y}}^{1} dx\ dy\ dz[/tex]

    [tex]\int_{0}^{1} \int_{0}^{y} \int_{\sqrt{z}}^{1} dx\ dz\ dy[/tex]
     
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