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Drawing domain of R^2 integral

  1. May 11, 2010 #1
    [tex]\int _1 ^e \int _{1/e} ^{1/y} f dx dy[/tex]
    If asked how to sketch the bounds for that:

    "x starts at x=1/e and ends at x=1/y
    1/e is just a constant so that's a straight line
    and x=1/y is the exact same line as y=1/x"

    That's a decent enough engineer's explanation right?
    (no one in university should need the y-bounds explained)
     
  2. jcsd
  3. May 11, 2010 #2
    well, y=1/x is not a straight line,it's a rectangular hyperbola present in 1st and 4th quadrant,but we have to consider only 1st quadrant as limits of x and y are +ive.
     
  4. May 11, 2010 #3
    The domain in the first quadrant is bounded by the curves y = 1, y = e and x = 1/y.
    The points of intersection are (1/e, 1), (1/e, e) and (1, 1).
     
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