1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper
    Gold Member

    Have you plotted the function and the boundaries?

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

    Thank you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Area under de sine curve
Loading...