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Area problem

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Well, the problem is to find the area of:

    {(x,y), -(pi/2) <= x <= Pi/2, 1/2< y <=cosx}

    3. The attempt at a solution

    Well, I know that for an angle of 60 degrees, the cosine is 1/2. So I guess that the limits of my integral will be from 0 to Pi/3.

    But I'm getting confused with the domain of x: for 270, and 90 degrees... So on the graph of cosine, and considering the condition of 1/2< y <=cosx = 1, it seems to me that the only point of intersection between the condition and the cosine graph is just a small piece of the cosine graph. So after I evaluate the inegral I got sqrt(3)/2.

    I think my answer is wrong. Any suggestions please?
     
  2. jcsd
  3. Feb 19, 2012 #2

    Dick

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    If you drew a graph of y=cos(x) and y=1/2 for x between -pi/2 and pi/2 you should have found that the crossed twice. And what integral did you work out to get sqrt(3)/2?
     
  4. Feb 19, 2012 #3
    Well, the integral of cos(x) from 0 to Pi/3. Did the graph really cross it twice? But 270 is on the negative side and the condition says that y is in between 1/2< y <= cosx = 1
     
  5. Feb 19, 2012 #4

    Dick

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    cos(-pi/3) is also equal to 1/2. And if you want to find the area between two curves you integrate the difference between the two curves. You want to find the area between y=cos(x) and y=1/2.
     
  6. Feb 19, 2012 #5
    Ok, I understand now. I would by symmetry multiply the integral by 2, and get sqrt(3).

    The problem now is that I cannot visualize where is (-pi/3). So I don't know if there is another segment which I should consider. Is 360 - 60 = 300? I think I'm wrong.

    ...
     
  7. Feb 19, 2012 #6

    Dick

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    -pi/3 is -60 degrees. And I'm trying saying you should integrate cos(x)-1/2. Not just cos(x). It's the difference between the two curves you should be integrating.
     
  8. Feb 19, 2012 #7
    Hello! Ok.

    Well, I got 2 - 3Pi/4. My limits where -Pi/2, Pi/2. And the integral was (cos(x) - 1/2), right?
     
  9. Feb 19, 2012 #8

    Dick

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    You want to integrate between the points where the two curves cross. cos(x) is only greater than 1/2 between -pi/3 and pi/3.
     
  10. Feb 19, 2012 #9
    Hello there!

    Well, I tried the integration with the new limits. My answer is 1 - Pi/3

    Is this fine?
     
  11. Feb 19, 2012 #10

    Dick

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    No, the -pi/3 part is ok. The 1 isn't. Can you show your steps?
     
  12. Feb 19, 2012 #11
    Sure.

    Integral of (Cosx - 1/2) dx = Sinx - (1/2)x,

    f(b) - f(a) = [Sin(Pi/3) - Pi/6] - [Sin(-Pi/3) + Pi/6] = 1/2 - Pi/6 + 1/2 - Pi/6 = 1 - Pi/3.

    Sin(-Pi/3) = -1/2. Or?
     
  13. Feb 19, 2012 #12
    Ohhh, no. My bad. I messed it up.

    Yeah, the answer is - Pi/3.

    But what does a negative area in this case means exactly?
     
  14. Feb 20, 2012 #13

    Dick

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    You are still messing it up. sin(pi/3) isn't 1/2. And the answer isn't negative. -pi/3 is only part of the answer.
     
  15. Feb 20, 2012 #14
    Yes, indeed, I'm still messing it up...


    f(b) - f(a) = [Sin(Pi/3) - Pi/6] - [Sin(-Pi/3) - (-Pi/6)] = sqrt(3)/2 - Pi/6 + sqrt(3)/2 - Pi/6 = sqrt(3) - Pi/3.

    That should be fine!
     
  16. Feb 20, 2012 #15

    Dick

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    That looks better.
     
  17. Feb 20, 2012 #16
    Thank you!
     
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