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Polar curve area question

  1. Mar 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the area of the region that lies inside the curve r^2=8cos(2θ) and outside r=2


    2. Relevant equations
    area of polar curves = .5∫R^2(outside)-r^2(inside) dθ


    3. The attempt at a solution
    r^2=8cos(2θ) and r=2, so....

    4=8cos(2θ)
    .5=cos(2θ) since .5 is positive, we need the angles in the first and fourth quadrant
    2θ=∏/3 and -∏/3
    θ=∏/6 and -∏/6

    .5∫8cos(2θ)-4 from -pi/6 to pi/6
    =difference of .5[4sin(2θ)-4θ] evaluated at pi/6 and -pi/6. but when i plug the values in, i cant seem to get the right answer
     
  2. jcsd
  3. Mar 29, 2013 #2

    SammyS

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    Consider what happens when cos(2θ) < 0 .

    Ignore this post. I misread the problem. DUH !
     
    Last edited: Mar 30, 2013
  4. Mar 29, 2013 #3
    not sure what your trying to get at. :<
     
  5. Mar 29, 2013 #4
    What answer should you be getting?
     
  6. Mar 29, 2013 #5
    2.739
     
  7. Mar 29, 2013 #6
    im so dumb
     
    Last edited: Mar 29, 2013
  8. Mar 29, 2013 #7

    LCKurtz

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    Are you getting exactly half the desired answer? Are you aware there is a symmetric area on the left side of the y axis?
     
  9. Mar 29, 2013 #8
    yes, i was getting exactly half the right answer. forgot about the symmetry, thanks!
     
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