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Change of order of integration

  1. Nov 12, 2009 #1
    I need to change the order of integration of the integral

    [tex]\int dx[/tex][tex]\int f(x,y) dy[/tex]

    Where the x limits are 0 and 2[tex]\pi[/tex] and the y limits are 0 and sin(x)

    I've got that this should be

    [tex]\int dy[/tex][tex]\int f(x,y) dx[/tex]

    Where the y limits are -1 and 1 and the x limits are 0 and arcsin(y)

    But this doesn't give me the right answer, can anyone help please.
     
  2. jcsd
  3. Nov 12, 2009 #2
    draw a picture of the two-dimensional region of integration
     
  4. Nov 12, 2009 #3
    It's just the region between 0 and 2[tex]\pi[/tex] of the sine graph. The area between 0 and [tex]\pi[/tex] is the same as between [tex]\pi[/tex] and 2[tex]\pi[/tex] apart from this region is negative. Does that fact the region between [tex]\pi[/tex] and 2[tex]\pi[/tex] is negative affect the change of order of the integration?
     
  5. Nov 12, 2009 #4
    Well, when you change the order of integration, you need to find a way to describe the same region by selecting bounds for x first, then selecting bounds for y. Remember that Arcsin(x) only has the domain from -1 to 1 and a range from -pi/2 to pi/2. It's only that first interval. So, when you say that x goes from 0 to the arcsin(y), you're taking the area between the y axis and that first interval, which is actually on the outside of what you're trying to depict. I'd recommend splitting it according to symmetry.
     
  6. Nov 12, 2009 #5
    I'm still really confused I don't understand what the lower x limit should be even with a sketch of the arcsine graph. Are my y limits of -1 and 1 correct??
     
  7. Nov 12, 2009 #6
    What y-limits you choose depend on how you pick your x-limits. They won't actually have the variable 'x' in them, but the constants you choose will change.

    Try graphing the following functions on a single set of axes:
    x=arcsin(y)
    x=pi - arcsin(y)
    x=2pi + arcsin(y)

    What's special about this pattern of functions?
     
  8. Nov 12, 2009 #7
    These functions plotted together make up the sine function between -pi/2 and 5pi/2 but I'm sorry I still don't understand where to go with this. Feel like I'm being a bit of an idiot and missing something really simple!
     
  9. Nov 12, 2009 #8
    You're not being an idiot... I just wish I had a good plotting utility I could use to illustrate what I mean. I'll look around a bit and come back with some picture that might help.

    edit:
    ajt1lx.png

    The top graph is the correct bounds that you have to begin with. Then you want to change the order.
    The second graph is what you described when you took x from 0 to arcsin(y) and then y from -1 to 1.
    You need to come up with what should replace the '?' in the third graph that will get you what you want. Remember that the arcsin only goes up to the first peak there, then stops, so you can't use arcsin(y) to arcsin(y) (which would give you 0). You have to split it up into pieces.

    Also, will you get that portion of the area below the x-axis? Do you need to split it up into two integrals or can you do it all in one?
     
    Last edited: Nov 12, 2009
  10. Nov 14, 2009 #9
    Thanks for the graphs I think they've given me more of an idea of what I should be doing now. I tired to explain my working using the Latex Reference but I couldn't use it so I've attached what I've done so far. Any more comments would be great! Thanks!
     

    Attached Files:

  11. Nov 14, 2009 #10
    You broke up the function into working sections. The only thing is, in the original, the part of the shaded region that falls below the y-axis is treated as negative. The integrals you have will give you a positive value for the integral of that region.
     
  12. Nov 14, 2009 #11
    Even when I take the regions below the x-axis to be negative I'm still not getting my solution to be correct.
     
  13. Nov 14, 2009 #12
    You sure? It seems to be working for me, unless I'm making an arithmetic error. You don't need four integrals, by the way, but it should still work (you can go straight from one function to another without having the vertical x=pi/2 and x=3pi/2 in between). What are you checking it against?
     
  14. Nov 14, 2009 #13
    I'm checking it through Maple but it dosen't make sense that it's wrong. I can see exactly how the answer should be correct.
     
  15. Nov 14, 2009 #14
    Sorry! I'm being a massive idiot, I had a minus where I should have had a + it's completely correct! Thanks for all your help I really feel I understand this topic alot better now.
     
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