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Homework Help: Area under de sine curve

  1. Oct 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the area of: {(x, y) R^2 : 0 ≤ x ≤ π, 0 ≤ y ≤ 1/2, 0 ≤ y ≤ sin x}


    3. The attempt at a solution

    y = sinx on the interval 0 to π but with y < 1/2.

    if Sinx = 1/2, then x = pi/6, or x = 5pi/6

    Because of symmetry I integrate one part and the multiply by two.

    [0,pi/6) In: (sinx) dx = (-cosx) = -cos(π/6) - -cos(0) = cos(0) - cos(π/6) = 1 - (1/2)√3.

    2(1 - (1/2)√3.)

    I stop there, but then I saw that to the answer that I had, I must add the rectangular area from pi/6 to 1/2(pi) with height 1/2, or (1/3)pi(1/2) = pi/6

    So the answer was: 2(1 - (1/2)√3 + pi/6) = 2 - √3 + (1/3)pi).

    I don't understand that last part. Which rectangular area? I don't see it.
     
    Last edited: Oct 23, 2011
  2. jcsd
  3. Oct 23, 2011 #2

    ehild

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    Homework Helper

    Have you plotted the function and the boundaries?

    ehild
     
  4. Oct 23, 2011 #3
    lol, ok. Now I saw the rectangular area...

    Thank you!
     
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