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Double Integrals: Changing the order of integration

  1. Apr 29, 2012 #1
    Hi,

    I am able to manipulate and use double integrals, but I am having a bit of mental block when trying to visual how they actually work.

    First, would you agree that a double integral is simply summing a function over a region by taking lots of tiny squares (or rectangles?) of sides dx and dy => dA = dxdy.

    So let's take http://gyazo.com/39f756b268c4e1cf4d8e09b7213e58ed

    So we want to integral the function over the triangular region. This is where the trouble comes in. As I said I can do it, but I cannot really visualize exactly how the limits work

    So the internal integral is integrating with respect to x. To me the area we need to compute is

    http://gyazo.com/ed588350139dd383b783542ce8e3ab11

    So the limits go from 0 to a NOT x=y to a. All the limits tell you is where to begin integrating from. y is simply the height of the function at each strip (dx - shown in red), not an actual limit

    Similarly, if we had y as the internal integral

    http://gyazo.com/eafcfb2b70f96c9c6b517167ceea29fb

    The limits would be (y = x to y=a) but y=x tells you how wide your strip (dy - shown in blue is). If you look where the y strips start from , they go from y = 0 to y = a,

    Can someone please explain to me where I have gone wrong in my reasoning/provide a better visualization/interpretation

    Thanks
    Thomas
     
  2. jcsd
  3. Apr 29, 2012 #2

    sharks

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    Gold Member

    You should always describe your region of integration. That's what i do. Maybe someone else can provide you with some clever tricks to go about this?

    In the meantime, you need to get the basics right. Since the internal integral is with respect to dx, meaning x varies. That automatically means that in an (x,y) integral, the other variable, y, will be fixed. Similarly, if dy is internal, then y varies, while x stays fixed. Based on those conditions, you can now derive the limits either horizontally or vertically (depending on which axis is fixed).
     
  4. Apr 29, 2012 #3

    HallsofIvy

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    Staff Emeritus
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    It would be better to write the first one as
    [tex]\int_{y= 0}^a\int_{x= y}^a f(x,y)dxdy[/tex]
    so that you can see that y can go from 0 to a and, for each y x goes from y to a. Mark horizontal lines at 0 and a to give the y- limits. Then draw x= y and x= a. You should see that, as you say, you are integrating over a triangle with vertices at (0, 0), (a, 0), and (a, a).
    Now, to reverse the order of integration, what are the smallest and largest values of x in that triangle? That should be easy. Those give the limits on the "outer", "dx", integral. Now, for each x, draw (or imagine) a horizontal line at that x crossing the triangle. You should see that the lower limit is the line y= 0 and the upper limit is y= x. Those give the limits on the "inner", "dy", integral.
     
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