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Area of polar curve

  1. Apr 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the area enclosed by the graph r=2+sin(4θ)


    2. Relevant equations
    Area = .5∫r^2



    3. The attempt at a solution
    Area = .5∫(2+sin(4θ))^2
    =.5(4.5θ-1/16sin(8θ)-cos(4θ))

    I can do the integration and all, but I am having trouble finding the limits of integration

    I set sin(4θ)=0 and solve, and get 0 and pi/4. but when i use these limits, i get the wrong answer.

    correct answer is 4.5pi
     
  2. jcsd
  3. Apr 1, 2013 #2

    CompuChip

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    Have you tried plotting the curve (either using a calculator or computer, or by hand)?
    That should give you a clue (hint: the correct answer is related to the period of r(theta)).
     
  4. Apr 2, 2013 #3

    haruspex

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    It is? i don't see that. Since r > 0 for all theta, the curve cannot join up until it has gone all the way around 2π.
     
  5. Apr 2, 2013 #4

    CompuChip

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    Exactly. So why the pi / 4?
     
  6. Apr 3, 2013 #5

    haruspex

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    I agree the pi/4 is wrong, but the period of r would be pi/2, which is also wrong.
     
  7. Apr 3, 2013 #6

    CompuChip

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    Ah, you are right, the problem was in my wording. Thanks!
     
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