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Double integral question

  1. May 6, 2010 #1
    1. The problem statement, all variables and given/known data

    Looks like I'm back with another question already :frown: I need to change the order of integration for this double integral and then evaluate it, but I get to a point where I'm not sure what to do.

    2. Relevant equations

    [tex]\int^3_{0} \int^9_{y} \sqrt{x}cos(x) dx dy[/tex]

    3. The attempt at a solution

    With the changed order of integration it needs two integrals added together, this is what I came up with:

    [tex]\int^3_{0} \int^x_{0} \sqrt{x}cos(x) dy dx + \int^9_{3} \int^3_{0} \sqrt{x}cos(x) dy dx[/tex]

    And I planned to work them both out separately, but didn't get too far with the first one:

    [tex]= \int^3_{0} \left[y\sqrt{x}cos(x)\right]^{x}_{0} dx[/tex]

    [tex]= \int^3_{0} x\sqrt{x}cos(x) dx[/tex]

    I'm not sure if I've made a mistake getting here, but it looks like I need to integrate [tex]x\sqrt{x}cos(x)[/tex] and there doesn't seem to be an easy way to do that at all.
     
  2. jcsd
  3. May 6, 2010 #2

    tiny-tim

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    Hi Refraction! :smile:

    Your change of order looks fine.

    The only problem is how to integrate x1/2cosx or x3/2cosx … I don't know any way of doing that (other than using power series). :redface:
     
  4. May 6, 2010 #3
    That's what I was thinking as well, we've never done anything like that in this class before, and it's only supposed to be a small question so I'm not sure why it's like that.

    The only thing I can think of is it maybe meaning to change the order and just leave it like that, but it's worded a bit strangely then. Thanks anyway!
     
    Last edited: May 6, 2010
  5. May 6, 2010 #4
    Hi Refraction,

    How did you change the order?(there was a y ?)

    For the computation of the integral, the only way I can see is with power series too...
     
  6. May 6, 2010 #5
    The line was x = y in the original question, I just used it as y = x for when the order is reversed (so it's in the first half of the reversed order integral now).
     
  7. May 6, 2010 #6
    :confused: I still don't understand...
     
  8. May 6, 2010 #7
    Well the area bounded by the lines looks something like this:

    9qwcoi.png

    So with the reversed order of integration (dy dx) for the first double integral, R1, the inner integral is from y = 0 to y = x, and the outer integral is from x = 0 to x = 3.
     
  9. May 6, 2010 #8
    I got it. thank you very much!
     
  10. May 6, 2010 #9
    Ah, Grasshopper. The student has become the master!
     
  11. May 6, 2010 #10

    tiny-tim

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    :biggrin: Woohoo! :biggrin:
     
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