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Parametric/Polar Equations

  1. Apr 19, 2006 #1
    a) find the area bounded by the polar curves r=3sinx, r=1+sinx

    first i find the points of intersection, x=pi/6, and x=5pi/6
    so to find the area i set up the integral
    A= 1/2 integral from pi/6 to 5pi/6 [(3sinx)^2 - (1+sinx)^2]dx

    using trig identities, it simplifies to

    1/2 the integral from pi/6 to 5pi/6 [3 -4cos2x - 2sinx] dx

    integrate and i get 1/2[3x - 2sin2x + 2cosx] from pi/6 to 5pi/6

    which the sqrt 3's cancel out so i get (1/2)[ 5pi/2 - pi/2 ] = pi

    i think this is correct but my answer is not on any of the answer choices
    can someone please check what i did wrong?

    b) find the length of the curve from t=0 to t=2pi given the curve has parametric equation
    x= t- sint
    y= 1 - cost

    since the arc length ds is integral of sqrt[ (dx/dt)^2 + (dy/dt)^2]

    dx/dt = 1 - cost
    dy/dx = sint

    thus ds = integral from 0 to 2 pi sqrt[(1-cost)^2 + (sint)^2]

    which simplifies to integral from 0 to 2pi sqrt[ 2-2cost]
    i couldnt integrate sqrt[ 2-2cost], or did i do something wrong in the steps prior to integration?

  2. jcsd
  3. Apr 19, 2006 #2


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    Homework Helper

    a) I don't see anything wrong with your working or result. I get the same.

    b) You're correct up to where you last got to. The integral is doable.
    Use the substutution: v = tan(t/2) and it will come out.
  4. Apr 19, 2006 #3
    are you suggesting trig substitutions?
    i thought you can only do that when the integral is in the form A^2 + V^2
  5. Apr 19, 2006 #4


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    Homework Helper

    You can use trig substitutions at most anytime.

    v = tan(t/2)
    dv = (1/2)sec²(t/2) dt
    2 dv = (1+v²) dt
    dt = 2/(1+v²) dv


    cos t = (1- v²)/(1+v²)
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