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Area of Surface of Revolution of Plane Curve - Use 1/2 Interval?

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the area of the surface generated by revolving the curve

    2. Relevant equations
    x = 5(cos3t), y = 5(sin3t), 0 ≤ t ≤ [itex]\pi[/itex], about the y axis.

    3. The attempt at a solution
    x' = -15(cos2t)(sin t)
    y' = 15(sin2t)(cos t)

    (I think this forms the top part of an astroid)

    [x'(t)]2 = 225(cos4t)(sin2t)
    [y'(t)]2 = 225(sin4t)(cos2t)

    S = 2[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) 5(cos3t) sqrt[225(cos4t)(sin2t) + 225(sin4t)(cos2t)] dt

    S = 2[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) 5(cos3t) sqrt[225(cos2t)(sin2t)*(cos2t + sin2t)] dt

    S = 2[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) 5(cos3t) sqrt[225(cos2t)(sin2t)(1)] dt

    S = 2[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) 5(cos3t)*15(cos t)(sin t) dt

    S = 150[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) (cos4t)(sin t) dt

    S = -150[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) (cos4t)(-sin t) dt

    S = -150[itex]\pi[/itex] [(cos5t) / 5] from 0 to [itex]\pi[/itex]/2

    S = -150[itex]\pi[/itex] (0 - 1/5)

    S = 30[itex]\pi[/itex] units2

    Please excuse my formatting. My basic question is, was I correct to take the integral only from 0 to [itex]\pi[/itex]/2, even though the interval given for the plane curve is from 0 to [itex]\pi[/itex]?

    My secondary question is, do you see anything else I messed up?

    Thank you much.
     
    Last edited: Apr 14, 2012
  2. jcsd
  3. Apr 25, 2012 #2
    Follow-up: The instructor admitted that this question was not valid, because the curve does not lie entirely on one side of the axis of revolution.
     
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