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Area between two polar curves

  1. Jan 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the area between the two curves:

    [tex]r=2sin(\theta), r=2(1-sin(\theta))[/tex]

    2. Relevant equations

    [tex]
    A=\frac{1}{2} \int_{\beta}^{\alpha} r^2 d\theta
    [/tex]

    3. The attempt at a solution

    I've got the points of intersection at [tex](1,\frac{1}{6}\pi) and (1,\frac{5}{6}\pi)[/tex] and worked out the answer to be [tex]\frac{8}{3}-4\sqrt{3}[/tex] using the angles in the above polar co-ordinates as the limits, however my textbook says that the answer is [tex]\frac{7}{3}-4\sqrt{3}[/tex] Is anyone able to confirm which is the correct answer.

    Thanks in advance
    Charismaztex
     
  2. jcsd
  3. Jan 9, 2010 #2
    your book is right

    2sin(t) is a circle
    next is a kind of cycloidal
    this is the formula i used in matlab
    int((2*sin(t))^2,0,pi/6)+int((2-2*sin(t))^2,pi/6,pi/2)
    gives area of one lobe
    multiply by 2 for both the lobes
     
    Last edited: Jan 9, 2010
  4. Jan 9, 2010 #3
    Ahhh, got the answer thanks!
     
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