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Homework Help: Parametrize a surface

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Parametrize a surface obtained by revolving the curve y = lnx, as x goes from 1 to 10 about the y axis



    3. The attempt at a solution

    y=lnx
    x=lnxcos(theta)
    z=lnxsin(theta)

    vector r(x,theta) = lnx cos(theta) i + lnx j + lnxsin(theta) k

    0[tex]\leq[/tex] (theta) [tex]\leq[/tex] 2pi
    1[tex]\leq[/tex] x [tex]\leq[/tex] 10

    I'm not really sure how to complete, or if this is correct-any ideas would be helpful, thanks
     
  2. jcsd
  3. Dec 8, 2008 #2

    tiny-tim

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    Hi hils0005! :smile:
    Nooo … how can x be lnx times something?

    Draw a diagram (with pretty circles on :wink:), and try again! :smile:
     
  4. Dec 8, 2008 #3
    Obviously I 'm confused by Parametrizations
    the given function is in the form y= f(x,z)
    x=cos(theta)
    y=lnx
    z=sin(theta)

    theta from 0 to 2pi
    x from 1 to 10
     
  5. Dec 8, 2008 #4

    tiny-tim

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    hmm …

    i] it's a surface, so it'll need two parameters

    ii] how can x=cos(theta) and z=sin(theta)? that means x2 + z2 = 1.

    Choose another parameter, and try again! :smile:
     
  6. Dec 8, 2008 #5

    HallsofIvy

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    If y= ln(x) is revolved around the y-axis, then y= ln(r) where r= [itex]\sqrt{x^2+ z^2}[/itex].

    You are going to have "tilt" your head and think about "cylindrical" coordinates where y rather than z is one of the coordinates.
     
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