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Double Integrals, Rectangular Region

  1. Jan 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Using ∫∫kdA = k(b-a)(d-c), where f is a constant function f(x,y) = k and R = [a,b]x[c,d], show that 0 ≤ ∫∫sin∏xcos∏ydA ≤ 1/32, where R = [0,1/4]x[1/4,1/2].


    2. Relevant equations

    ∫∫kdA = k(b-a)(d-c)
    0 ≤ ∫∫sin∏xcos∏ydA ≤ 1/32

    3. The attempt at a solution

    I tried to integrate ∫∫sin∏xcos∏ydA, but that didn't utilize the ∫∫kdA = k(b-a)(d-c) part and I didn't use the fact that ∫∫sin∏xcos∏ydA is between 0 and 1/32. I don't know how to incorporate all of these aspects to show that 0 ≤ ∫∫sin∏xcos∏ydA ≤ 1/32.
     
  2. jcsd
  3. Jan 13, 2013 #2

    haruspex

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    If you were to show that 0 ≤ sin∏xcos∏y ≤ k everywhere in R, what would you need k to be in order to prove the desired answer?
     
  4. Jan 13, 2013 #3
    k=1/32.
    Sorry, I still don't know how to proceed.
     
  5. Jan 13, 2013 #4

    haruspex

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    No, you're not taking into account the area of R. Try that again.
     
  6. Jan 13, 2013 #5
    Ohh is it (1/32)(1/4-0)(1/2-1/4)?
     
  7. Jan 13, 2013 #6
    Wait hold on is it ∫∫kdA = 1/32, for R = [0,1/4]x[1/4,1/2]? So k(1/4-0)(1/2-1/4)=1/32.
     
  8. Jan 13, 2013 #7

    haruspex

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    Yes. So what is k? Can you see why sin∏xcos∏y <= that everywhere in R?
     
  9. Jan 14, 2013 #8
    k=0.5 I graphed sin∏xcos∏y on Mathematica and saw that for x in [0,1/4] and y in [1/4,1/2], z is between 0 and 0.5. And solving for k=0.5 agrees with the graph.

    I'm also wondering if you can know that z is from [0,0.5] without graphing it on a computer or something?
     
  10. Jan 14, 2013 #9

    Dick

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    sin(pi*x) is less than sqrt(2)/2 on [0,1/4]. cos(pi*x) is less than sqrt(2)/2 on [1/4,1/2]. So?
     
  11. Jan 15, 2013 #10
    So since the min of sin(pi*x) is 0 and max is sqrt(2)/2, and it's the same for cos(pi*x), then the min of their product is 0 and max is 0.5. Is that correct?
     
  12. Jan 15, 2013 #11

    Dick

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    Right, on those particular intervals.
     
  13. Jan 15, 2013 #12
    Alright then. Thank you!
     
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