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Is there a simpler way to integrate this?

  1. Jul 14, 2012 #1
    Hi, I was wondering if there's a simpler way to integrate this? I got the answer by expanding one by one but that's such a long process! I got the answer 104/5 which is correct,though.


    ∫∫∫D (x + y + z)^4 dxdydz
    D = {(x,y,z) | -1≤ x ≤1, -1≤y≤1, -1≤z≤ }
     
  2. jcsd
  3. Jul 14, 2012 #2

    chiro

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    Hey aruwin.

    For your integral, you are dealing with a simple rectangular volume, so the integral in its form is best evaluated by doing each variable separately at a time.

    If you did it this way, then there's no point stressing about a better way to do it if you know and understand how to evaluate the integral.
     
  4. Jul 14, 2012 #3
    you could keep in your head that the integral if (x+a)^n = (1/n)(x+a)^(n+1) if you wanted but there's nothing wrong with expanding everything out
     
  5. Jul 14, 2012 #4
    How do I do each variable sperately at a time? You mean like ∫dx∫dy∫dz ?

    Ok,what about the function inside it? When say rectangle, it means
    (x+y+z)2(x+y+z)2,right? But still,how do I integrate it like this?
     
  6. Jul 14, 2012 #5

    chiro

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    You do it in the form of (x+a)^n with respect to dx and integrate out the x value. Then do the same for the y and the same for the z until you get a numerical quantity.

    Since x,y,z are all orthogonal variables, you can do this with no problem just like you would integrated say xydydx by keeping one variable 'constant' and then integrating it with respect to the other.
     
  7. Jul 14, 2012 #6
    OK,I am gonna ask you something first, is this correct?

    ∫∫ (x+y+z)5/5 dydz

    But what happens to the x inside the bracket?Shouldn't I do something about it?
     
  8. Jul 14, 2012 #7
    take [itex]\int \int \int (x+y+z)^4 dx dy dz[/itex] and stick some brackets in there [itex]\int \int ( \int (x+y+z)^4 dx ) dy dz[/itex]
    Substitute y+z = a
    perform integration
    substitute a = y + z
    remove brackets
    repeat

    :biggrin:
     
  9. Jul 14, 2012 #8
    Won't that change the range for dx?
     
  10. Jul 14, 2012 #9
    nope, all you've done is substitute a for y + z as to avoid confusion, you've not touched x or dx at all (you haven't really altered anything)
     
  11. Jul 14, 2012 #10

    chiro

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    It only changes the interval if the interval depends on those other variables.

    To understand this, you need to understand orthogonality: things are orthogonal if changing one variable doesn't change another. We can write this in terms of dA/dB = 0 and dB/dA = 0 where neither change in accordance with the other variable.

    If your limits were say from 0 to y^2 - z, then yes the limit would have to factor this in but because each variable is truly orthogonal, then this doesn't happen (in this case).
     
  12. Jul 14, 2012 #11
    So I did what you told me but I didn't get the answer :( Tell me what is wrong here.
    Check,please....

    Untitled-1.jpg
     
  13. Jul 14, 2012 #12

    chiro

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    You wrote in your final step 27 instead of 2x17
     
  14. Jul 14, 2012 #13
    Even so, I won't get the answer that is 104/5.
     
  15. Jul 14, 2012 #14

    chiro

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    You made a major sign error in your integral:instead of + (-2 + z)^6 you wrote - (-2 + z)^6.
     
  16. Jul 14, 2012 #15
    I will try again.But other than that,I'm doing this the right method,arent I?
     
  17. Jul 14, 2012 #16

    chiro

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    Yes. I checked the answer with your calculation using a calculator: it gave 20.8 which is 104/5.
     
  18. Jul 14, 2012 #17
    I calculated another 2 times but I didn't get the answer :( Are you sure my working there had only one mistake?
     
  19. Jul 14, 2012 #18

    chiro

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    If you fix up changing the - (-2 + z) to + (-2 + z) and fix up your mistake with the 2^7 and make it 2*(-1)^7 and plug everything in correctly you will get 104/5. Show us your working so we can see what you did.
     
  20. Jul 14, 2012 #19
    Oh,thank you!!!!I got it,I got it!!!:tongue:
    By the way, I have another question on multiple integration but must I create another thread or can I just continue in here?
     
  21. Jul 14, 2012 #20

    chiro

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    Just post it in here, it's on topic and a simple continuation of the original threads premise.
     
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