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Volume and symmetry

  1. Sep 1, 2013 #1
    1. The problem statement, all variables and given/known data

    Your supposed to set up the integral and do it from the calculator.
    y = , y = (cosx)^2, -pi/2 <= x <= pi/2
    a. x axis
    b. y axis
    2. Relevant equations


    I = ∏∫(cosx)^4 dx = 3.70110 and for part b
    I = ∏∫((1)-(cosx)^2)^2 dx = 3.70110

    3. The attempt at a solution

    Did I set this up right? It would be the same volume in both cases right?
     
  2. jcsd
  3. Sep 1, 2013 #2

    vela

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    It's not clear to me what exactly you've been asked to do. Could you state explicitly what you're trying to do?
     
  4. Sep 1, 2013 #3
    Yes, given y =0 , y = (cosx)^2, -pi/2 <= x <= pi/2
    set up the integral to revolve it around a.) The x axis and b.) the y axis
    It is just a volume of revolution problem the book just says do it on the calculator. That's it I just want to know if my integral are right.Thanks
     
  5. Sep 1, 2013 #4

    vela

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    Your answer for part (a) looks fine. I don't understand what you're doing for part (b).
     
  6. Sep 1, 2013 #5
    Part b wants you to revolve it around the line y = 1.
    Does that help?
    So I did 1 - cosx^2
     
  7. Sep 2, 2013 #6

    vela

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    Yes, it helps when you tell us what the actual problem is. :wink: In the original post, you said part (b) was about revolving the area about the y-axis, and then you repeated that in post 3. And somehow we were supposed to read your mind and realize the problem was actually about revolving around the horizontal line y=1?

    In any case, your integral for part (b) isn't correct. The volume you calculated is if you took the area bounded below by ##\cos^2 x## and above by y=1 and rotated it about y=1. You want the area bounded above by ##\cos^2 x## and below by y=0 and rotate that area about y=1.
     
  8. Sep 2, 2013 #7
    I'm sorry. I don't understand your above and below. I think just the second part. I want the area bounded above by ## cos(x)^2## and below y =0 ? I don't understand this. Do you mean that I need the area below y = 1 and above ## cos(x)^2##? So could I just multiply my original answer by two? Sorry for the confusion from the earlier post.
     
  9. Sep 2, 2013 #8
    I don't understand why the integral would not be the same. Can anyone clarify? I just looked at this thing I can't figure out what you are saying about the areas.
     
  10. Sep 2, 2013 #9

    vela

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    You were given an area bounded by y=0 and y=cos2 x, right? You're supposed to revolve this area around the line y=1. You're looking at a completely different area if you look at the region between y=1 and y=cos2 x.
     
  11. Sep 2, 2013 #10
    Yes, you are correct. The first part (a) you rotate about (cosx)^2 around the x axis. Next you rotate about y = 1. Look at the graph. I wish I had a way to draw it. When you do around y = 1 you have (1-cos(x)^2).
    So you have ∏∫(1-(cos(x)^2)^2 dx = ∏∫(sin(x))^4 dx
    When you do this between (-∏/2) and (∏/2) you get 3.7011
    Which is equivalent to the answer you get when you integrate cos(x)^2 around the x axis. So that integral is ∏∫(cos(x))^4 dx between (-∏/2) and (∏/2) you get 3.7011.
    I don't see how I'm wrong here.
     
  12. Sep 2, 2013 #11

    vela

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    Depending on how the problem was worded, you could be right, but I think you're not. Could you please post the problem here exactly as it was written?
     
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