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Changing the order of integration

  1. Feb 23, 2009 #1
    1. The problem statement, all variables and given/known data

    change the order of integration in the follwing double integral:

    intgral [0,a] integral [0,sqrt(2ay-y^2)] f(x,y) dx dy




    if x= sqrt(2ay-y^2),, do i solve for y or something,,ive used the computer to graph it but that didnt help me.
     
  2. jcsd
  3. Feb 23, 2009 #2

    HallsofIvy

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    Then presumably you see, from the graph, that the region of integration is a semi-circle with center at (0,a), radius a and extending to the right of the y-axis.

    If you had not graphed it you could still get that by noting that [itex]x= \sqrt{2ay- y^2}[/itex] can be expanded, by squaring both sides, to [itex]x^2= 2ay- y^2[/itex] or [itex]x^2+ y^2- 2a= 0[/itex]. Completing the square, [itex]x^2+ (y- a)^2= a^2[/itex] which is a circle of radius a, center at (x, 0). Of course the square root is never negative so that is just the semi-circle with x non-negative.

    However you determine region, to reverse the order of integration, so that you are integrating with respect to y first, and then with respect to x, the limits of integration on x must be numbers so they should be the smallest and largest possible values of x in that semi-circle. To get the limits of integration for y, draw a vertical linel, representing a particular value of x. The limits of integration for y are the lowest and highest value of y on that line- the values at the endpoints (and, of course, they depend on x).
     
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