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!

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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    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

    User Avatar
    Staff Emeritus
    Science Advisor

    Right! my eyes are going!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Understanding the graph for reversing an integral
  1. Graph of integral (Replies: 3)

  2. Reversing integration (Replies: 4)

  3. Reversing Integration (Replies: 3)

Loading...