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On changing the order of integration

  1. May 13, 2013 #1
    1. The problem statement, all variables and given/known data

    This is a more general question, but I want to be sure I understand something here.

    Take any function f(x,y) and say you're doing a double integral, like this:

    [tex]\int^{\pi}_{\pi/2}\int^{sinx}_{\pi/2}f(x,y)dydx[/tex]

    So I want to reverse the order of integration. The x-coordinate interval is π/2 to π, and the y-interval is 0 to sinx.

    So that means that x=arcsin y. and the interval in those terms starts at 1. Am I correct so far?

    If I switch the order of integration I have to put the y interval on the outside. But since the y interval is in terms of x I have to change that. On this curve y is between 0 and 1. So I should make the outer integral from 1 to 0 since we were originally starting at π/2.

    The inner integral is therefore π/2 to arcsin y.

    And when you re-order the integral you should get

    [tex]\int^{1}_{0}\int^{sin^{-1}}_{\pi/2}f(x,y)dxdy[/tex]


    But I know this is wrong and I am trying to make sure I understand why that is.

    Any assistance is appreciated, as always.
     
  2. jcsd
  3. May 13, 2013 #2

    Mark44

    Staff: Mentor

    Did you mean this integral?
    $$\int^{\pi}_{\pi/2}\int^{sinx}_0f(x,y)dydx $$
    This is what you described, below.
    For one thing, sin-1 can't be one of the limits of integration. sin-1(y) would be OK, though.

    Assuming your first integral is as I corrected it, with the region of integration being bounded by the x-axis and y = sin(x), and between x = ##\pi/2## and ##\pi##, the integral in reversed order would be
    $$ \int_{y = 0}^1 \int_{x = \pi/2}^{sin^{-1}(y)}~f(x, y)~dx~dy$$
     
  4. May 13, 2013 #3
    Yes, the correction you made was right (my inexpertise with itex entries. Oh well).

    In any case, it seems i had my reasoning correct, even if my typing skills were off. So that's good to know.

    Thanks Mark 44.
     
  5. May 14, 2013 #4

    SammyS

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    I don't think all is right with the results.

    If the original integral is ##\displaystyle \ \int^{\pi}_{\pi/2}\int^{sinx}_0f(x,y)dydx \,,\ ## then the region of integration satisfies the inequalies:
    0 < y < sin(x) and π/2 < x < π .​
    I.e. It's the region above the x-axis and below y = sin(x) that is to the right of x = π/2 and to the left of x = π .

    The region of integration for the integral, ##\displaystyle \ \int_{y = 0}^1 \int_{x = \pi/2}^{sin^{-1}(y)}~f(x, y)~dx~dy \ ## is different because of the range of the arcsine function.
    -π/2 ≤ arcsin(u) ≤ π/2



    The upper limit of integration for the inner integral needs to be modified if the region of integration is to be to the right of x = π/2 .
     
  6. May 14, 2013 #5
    That wold mean that all I'd have to do tho, is subtract π from the arcsin function, no? So the range would be -π/2 to sin-1(y)?
     
    Last edited: May 14, 2013
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