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Reversing order of integration

  1. Aug 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Reverse the order of integration:

    [tex]\int_{0}^{1}\int_{y^2}^{2-y} f(x,y) \ dx\ dy[/tex]

    This was on a test I took today. I was having trouble with it and was wondering how I should have gone about it?
  2. jcsd
  3. Aug 3, 2011 #2


    Staff: Mentor

    Sketch a graph of the region over which integration is taking place. In the integral above, the inner integral runs from x = y2 to x = 2 - y. The outer integral runs from y = 0 to y = 1. When you reverse the order of integration for this problem you will end up with two iterated integrals.
  4. Aug 3, 2011 #3
    So x=2-y and x=[itex]y^{2}[/itex] so y=2-x y=[itex]\sqrt{x}[/itex]

    intersect at: [itex]2-x=\sqrt{x}[/itex] [itex]x=1[/itex] [itex]y=1[/itex]

    y at 0 is (0,2)

    I got confused because at y=1,x=1 each side can be explained differently so my first attempt at a solutions was:

    [tex]\int_{0}^{1}\int_{0}^{\sqrt{x}} f(x,y) \ dy\ dx + \int_{1}^{2}\int_{2-x}^{1} f(x,y) \ dy\ dx[/tex]

    I used some random logic in my head to get that and we've never seen anything like that in class before so I crossed it out and just wrote:

    [tex]\int_{0}^{2}\int_{2-x}^{\sqrt{x}} f(x,y) \ dy\ dx[/tex]
    Last edited: Aug 3, 2011
  5. Aug 3, 2011 #4


    Staff: Mentor

    This isn't correct, but it's closer than what you have below. Here's what you should get:
    [tex]\int_{0}^{1}\int_{0}^{\sqrt{x}} f(x,y) \ dy\ dx + \int_{1}^{2}\int_{0}^{2 - x} f(x,y) \ dy\ dx[/tex]
  6. Aug 3, 2011 #5
    Yea I made a mistake on my last post. I actually did write my second bound as [itex]{0}\leq{y}\leq{2-x}[/itex]

    We've never seen a problem like that before so I got so nervous I just crossed the entire answer out. The answer I decided to write down in the last couple seconds was actually:

    [tex]\int_{0}^{2}\int_{\sqrt{x}}^{2-x} f(x,y) \ dy\ dx[/tex]

    I kind of just tried to push the two integrals from my past answer together so I wouldn't have two iterated integrals.
    Last edited: Aug 3, 2011
  7. Aug 3, 2011 #6


    Staff: Mentor

    You understand why a single integral won't work, right?
  8. Aug 3, 2011 #7
    Yea, I have a bad habit of writing something that I know can't work when I get nervous.
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