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Double Integral question

  1. Oct 30, 2007 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral shown in the diagram

    2. Relevant equations



    3. The attempt at a solution

    The first step to evaluating the integral is shown in the diagram (labelled as 2). They said they changed the order of integration. I was wondering what they mean by changing the order of integration and how did they come up with that first step. Cheers!
     

    Attached Files:

  2. jcsd
  3. Oct 30, 2007 #2

    HallsofIvy

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    The first integral is
    [tex]\int_{y=0}^1\left[\int_{x= \sqrt{y}}^1 \sqrt{x^2+1}dx\right] dy[/tex]
    and the second is
    [tex]\int_{x=0}^1\left[\int_{y= 0}^{x^2}\sqrt{x^2+ 1}dy\right] dx[/tex]
    I've included "x= " and "y= " in the limits of the integrals and put in braces, to make it clearer that, in the first line, they are first integrating with respect to x and then with respect to y, while in the second line, they are first integrating with respect to y and then with respect to x. That's signaled in two ways: the first line has "dx dy" while the second line has "dy dx" and the "inner" integral in the first line has a function of y as a limit of integration while the "inner" integral in the second line has a function of x as a limit of integration.

    The way to see how to change the limits of integration when changing the order of integration is:

    First draw a picture of the region you are integrating over: the "outer" integral in the first line has limits of integration y= 0 and y= 1 (We know they are "y= " because the "outer" integral is with respect to y) so we draw horizontal straight lines at y= 0 and y= 1. The "inner" integral has limits of [itex]x= \sqrt{y}[/itex] and x= 1 so we draw the graphs of [itex]x= \sqrt{y}[/itex] (that's the right half of y= x2) and x= 1 (a vertical line). For each y, (looking at the inner integral again) x goes from [itex]\sqrt{y}[/itex] up to 1 so we are looking at the area to the right of the parabola.

    Now "reverse" the order. In order to cover the entire area it should be clear that x must go from 0 to 1. If the "outer" integral is with respect to x, then the limits must be x= 0 up to x= 1. For each x now, y must run from the lower line boundary, y= 0, up to the parabola boundary, y= x2. That gives the limits of integration for the "inner" integral.
     
    Last edited: Oct 30, 2007
  4. Oct 30, 2007 #3
    cheers mate. understand it loads better now!
     
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