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Another parametric question

  1. Apr 6, 2005 #1
    If [tex]f(\theta)[/tex] is given by:[tex]f(\theta) = 6cos^3(\theta)[/tex] and [tex]g(\theta)[/tex] is given by:[tex]g(\theta) = 6sin^3(\theta)[/tex]
    Find the total length of the astroid described by [tex]f(\theta)[/tex] and [tex]g(\theta)[/tex].
    (The astroid is the curve swept out by ([tex]f(\theta)[/tex],[tex]g(\theta)[/tex]) as [tex]\theta[/tex] ranges from 0 to 2pi )

    [tex]f/d(\theta) = -18*cos(x)^2*sin(x)[/tex]
    [tex]g/d(\theta) = 18*sin(x)^2*cos(x)[/tex]

    this is asking for arclength right?
    my integral:


    anyone what's wrong with my integral? cause i keep getting the wrong answer.
    Last edited: Apr 6, 2005
  2. jcsd
  3. Apr 6, 2005 #2
    I suppose it's all right with your integral. What a result do You receive? In which way You integrated it?
  4. Apr 6, 2005 #3


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    Do not forget that


    Integral from 0 to pi/2 and multiple the result by 4.

  5. Apr 6, 2005 #4
    i used my calculator to integrate my function. I also used another math program on my computer to verify it. i integrated from 0 to pi/2 and got 9.558*4 = 38.232 but it's incorrect and i dont understand why. Also tried to integrate from 0 to 2pi and got 50.253217, but it wont take that either.
  6. Apr 6, 2005 #5
    From 0 to pi/2 i and "Mathematica" got 9. Thus the total length is 9*4=36. Agree?
  7. Apr 6, 2005 #6
    awesome thanks
  8. Apr 6, 2005 #7


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    Simplify your integrand. It becomes


    Integral from 0 to pi/2, multiply by 4. The result should be 36.

    You could have made the mistake with your programs that you did not set to radians and the program calculated with degrees.

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