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Change the order of Integrtation

  1. May 15, 2007 #1
    1. The problem statement, all variables and given/known data
    Change the order of integration and perform the integration.

    [tex]\int_0^2\int_{2x}^{4x-x^2} dydx[/tex]
    2. Relevant equations



    3. The attempt at a solution
    I've tried changing it to this but I end up with the wrong answer..

    [tex]0<=y<=2, \sqrt{4-y}+2 <=x<=\frac{1}{2}y[/tex]
     
  2. jcsd
  3. May 15, 2007 #2

    Hootenanny

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    Sketch the region, what does it look like to you? Have you met polar coordinates yet?
     
  4. May 15, 2007 #3
    I've sketched it, it's looks a little cut of a parabola.. I can't see how I can describe it using polar coordinates :/
     
  5. May 15, 2007 #4

    Hootenanny

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    Nevermind, it seems that polar coordinates are not nesscary, but you should reconsider your limits. Lets look at how the region is bounded. We have;

    [tex]0\leq x \leq 2[/tex]

    [tex]0\leq y \leq 4[/tex]

    [tex]x\leq \frac{1}{2}y \equiv y \geq 2x[/tex]

    [tex]y \leq 4x-x^2[/tex]

    Would you agree?
     
    Last edited: May 15, 2007
  6. May 15, 2007 #5
    Yes, I do agree. What I did was solve the last equation and so I got
    [tex]0<=y<=2, \sqrt{4-y}+2 <=x<=\frac{1}{2}y[/tex] ..

    Still can't see what I'm doing wrong. I'm kinda slow
     
  7. May 15, 2007 #6

    Hootenanny

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    You've chosen the wrong solution to your equation, [itex]y=4x-x^2[/itex] has two solutions;

    [tex]x = 2\pm\sqrt{4-y}[/tex]

    Now, your original equation [itex]y=4x-x^2[/itex] goes through the point (x,y)=(0,0), therefore, your new solution must also go through the same point if it is to describe the same region. Which of your solutions goes through the origin?
     
  8. May 15, 2007 #7
    Ah, darn I should have seen that! Thank you for being so patient!
     
  9. May 15, 2007 #8

    Hootenanny

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    Nah, its no problem, I didn't spot it at first until I started sketching the region :smile:
     
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