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Homework Help: Understanding the graph for reversing an integral

  1. Jul 28, 2008 #1
    1. The problem statement, all variables and given/known data
    I'm trying to reverse the order of integration of
    [tex]\int_0^{81}\int_{y/9}^{\sqrt{y}} dx dy[/tex]
    first integral is from 0 to 81
    second integral is from y/9 to [tex]\sqrt{y}[/tex]

    The problem is, I identified the inequalities for the regions of integration, but I'm having problems understanding how the graph is formed.
    2. Relevant equations


    3. The attempt at a solution
    I understand the inequalities are y/9 < x < [tex]\sqrt{y}[/tex]
    and 0 < y <81.

    But I don't understand the graph.
    x-axis is 0 to 9
    y-axis is 0 to 81
    y = x^2
    y = 9x

    And since according to the homework problem I'm doing, I can't progress the problem until I understand why the graph is so.

    Also, can someone help me in formating the integrals in latex?
     
    Last edited by a moderator: Jul 28, 2008
  2. jcsd
  3. Jul 28, 2008 #2

    HallsofIvy

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    Draw a picture. In particular, graph x= 1/y and [itex]x= \sqrt{y}[/itex] which are the same as the hyperbola y= 1/x and the parabola y= x2. Now one problem you have here is that those two graph cross at (1,1). The area over which you are integrating is the area from the curve y= x2 on the left to the curve y= 1/x on the right up until y= 1, then from the curve y= 1/x on the left up to the curve y= x2 on the right.

    Reversing the order of integration, you will need to divide this into four integrals. To cover the lower area, below y= 1, because of the change at (1,1), you will need to integrate from 0 up to y= x2 with x from 0 to 1, then from 0 up to y= 1/x, with x from 0 to 1/81. To cover the upper area, above y= 1, you need to integrate from y= 1/x up to y= 81, with x from 1/81 to 1, then from y= x2 up to y= 81, with x from 1 to up to 9:
    [tex]\int_{x=0}^1\int_{y= 0}^{x^2} dy dx+ \int_{x=1}^\infty \int{y= 0}^{1/x} dy dx+ \int_{x= 1/81}^1 \int_{y=1/x}^81 dy dx+ \int_{x=1}^9\int_{y= x^2}^{81} dy dx[/tex]
     
  4. Jul 29, 2008 #3
    I'm suck on that step. How do you determine which formulas to graph? I'm looking at the equations and inequalities, but I can't arrive to that step.

    I can do reversing of integration, but this is the pothole in the bridge.
     
  5. Jul 29, 2008 #4

    Defennder

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    Homework Helper

    The graphs you should sketch are those determined by the limits of the integrands. The limits for x are x=y/9 and x=sqrt{y}. I think HallsOfIvy meant x=y/9 instead of 1/y.
     
  6. Jul 29, 2008 #5
    Oh yes, I get it now. Thanks! :)
     
  7. Jul 29, 2008 #6

    HallsofIvy

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    Right! my eyes are going!
     
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