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Difficulty finding integral answer to set function

  1. Dec 13, 2011 #1
    Note: as i was typing this post, i think i figured out the answer but i want to confirm.
    let A denote the set of points that are interior to, or on the boundary of, a square with opposite verties at the points (0,0) and (1,1). let Q(A) = ∫∫dydx
    if C, a subset of A, is the set {(x,y): 0 < x/2 ≤ y ≤ 3x/2 < 1}, compute Q(A)

    I kept getting 1/2 as the answer, but i think my limits of integration were wrong. is the following correct?


    this setup gives the answer as 2/9... unfortunately i do not have the correct answer which is why i needed to verify.
  2. jcsd
  3. Dec 14, 2011 #2


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    The question is not completely clear. For x > 2/3 is anything included?
    This would involve: x/2 < y < 1.
  4. Dec 15, 2011 #3
    Unfortunately, i typed the question verbatim and not sure how to respond to you.

    note that the 2/3 upper limit on x was defined by me. I figured that y <1 (from y ≤ 3x/2 < 1). so therefor x cannot be more than 2/3, since y must also be less than 3x/2. for x> 2/3, y>1. e.g. when x=1, y=3/2.

    does that make sense?
  5. Dec 15, 2011 #4


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    No. Your condition x/2≤y≤3x/2 for x > 2/3 restricts y to x/2≤y≤1, but does not remove it altogether.
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