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Drawing a Region of Integration

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data

    http://img803.imageshack.us/img803/6397/skjermbilde20111130kl50.png [Broken]

    3. The attempt at a solutionI seriously can't even figure out how this region is drawn. I think in the second integral I need y to range from -x to 3. Then the answer could conceivably be a). But, as is, I can't do this.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 30, 2011 #2

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    Hi TranscendArcu! :smile:


    Perhaps you can start with a drawing in which you mark the end points of the boundaries?
    And if you have that, a couple of points in between?

    If you switch the integrals around, the outer integral(s) will be with respect to y.
    y can not be dependent on x then, since x does not exist anymore when you integrate y.
    The outer integral should typically cover the largest possible range, while the inner integral sets a condition on what to integrate.
     
  4. Nov 30, 2011 #3
    Okay. Here's what I have: in the first quadrant, a triangle with vertices (0,0), (3,3), (0,3). This is the region I get when I map out the region described by the first integral. In the second integral I have x≤y≤3, which is exactly the same constraint on y I had before. If I let -3≤x≤0, in the first quadrant I have a square with vertices (0,0), (0,3), (-3,3), (-3,0). In the third quadrant I have another triangle with vertices (-3,-3), (0,0), (-3,0).

    !!!

    Is it c?
     
  5. Nov 30, 2011 #4

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    Hmm, before saying yeah or nay, let's explore your analysis.

    If -3≤x≤0, then you're not in the first quadrant...
    Furthermore, if x=-3, can you recheck the bounds?


    So you think it's c?
    What is your analysis then of the region that c covers?
     
  6. Nov 30, 2011 #5
    I think I mistyped something: If I let -3≤x≤0, in the second quadrant I have a square with vertices (0,0), (0,3), (-3,3), (-3,0).

    But I still think it's c, since my picture did actually have the square in Q2.
     
  7. Nov 30, 2011 #6

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    All right!

    So your problem gives you 2 triangles and a square.
    And c gives you a square in Q2.

    Are these the same figures?
     
  8. Nov 30, 2011 #7
    I think so. If I graph the bounds in c I seem to get the same region.
     
  9. Nov 30, 2011 #8

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    And what about a, b, and d?
     
  10. Nov 30, 2011 #9
    a is a triangle in the first and second quadrants. b I found more tricky, but I think it's the region in (a) and also its reflection over the x-axis. Who knows what d is? As you said in #2, it doesn't make sense to have variables in the outermost integral's bounds.
     
  11. Nov 30, 2011 #10

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    Sounds like you've got things down!

    So?
     
  12. Nov 30, 2011 #11

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    You still seem unsure...?
    I was kind of hoping for a statement like: "I'm sure it is c!".
     
  13. Nov 30, 2011 #12
    Pick c! I was pretty sure it was c after I really mapped out the region being described. Just by inspection, c seemed to be the appropriate answer once graphed correctly. Still, it took me a minute or two to realize what region b described.
     
  14. Nov 30, 2011 #13

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