1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Mark44

    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

    Mark44

    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

    Mark44

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Reversing order of integration
Loading...