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Evaluating a basic double integral

  1. May 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Integrate the following function f over the given region Ω:

    f(x,y) = xy; Ω bounded by y = 0, x = 2a, and x^2 = 4a



    3. The attempt at a solution

    The given answer is (a^4)/3

    The answer I got was (a^4)...ignore the answer I put in the attachment at the end, slight computational error.

    http://img831.imageshack.us/img831/1600/dsc0017ye.jpg [Broken]

    http://img225.imageshack.us/img225/7238/dsc0016xd.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 1, 2012 #2

    HallsofIvy

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    So the boundaries are given by y= 0, x= 2a, and [itex]x^2= 4ay[/itex]? (You forgot the "y" in the last equation.) Then x, over all, goes from 0 on the leftt ([itex]x^2= 4ay[/itex] and y= 0 intersect where [itex]x^2= 0[/itex]) to x= 2a on the right. Then for each x y goes from y= 0 to [itex]y= x^2/4a[/itex]. So your integral would be
    [tex]\int_{x=0}^{2a}\int_{y=0}^{x^2/4} xy dydx[/tex]

    You may have misled yourself by miswriting [itex]x^2= 4ay[/itex] as [itex]x^2= 4a[/itex].
     
  4. May 1, 2012 #3
    Ah. I mistakenly typed it like that, however I still calculated it with [itex]x^2 = 4ay[/itex] on paper, my mistake was with the bound of the second integral.

    I don't understand how that upper bound of y is [itex] y = (x^2)/4a[/itex]

    I thought if the upper bound of x was 2a, then we substitute that into y-equation and hence get the upper bound of y. That's how I ended up getting "a" as the upper bound, which I can now see is incorrect, but I don't understand why.
     
  5. May 1, 2012 #4

    HallsofIvy

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    No, what you are trying to do is wrong. In order that the result of a double integral be a number, the limits of integration on the "outer" integral must be constants but the "inner" integral may have the "outer" variable in its limits of integration. In fact, an integral like
    [tex]\int_a^b \int_c^d f(x,y) dy[/tex]
    is over the rectangle with boundaries x= a, x= b, y= c y= d. To have an integration over any region other than a rectangle, limits of integration on the "inner" integral must be functions, not constants.

    Yes,, x goes from 0 to 2a and, for each x, y goes from y=0 up to the upper boundary, [itex]y=x^2/4a[/itex].
     
  6. May 1, 2012 #5
    Thank you!
     
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