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Finding Area in Polar Coordinates

  1. Nov 9, 2006 #1
    The problem is to find the area of r = 4cos3θ.

    I know the formula for finding the area in polar coordinates is ∫ (from α to β) ½r²dθ.

    I substituted into this formula the given equation and got:
    A = ½ ∫ (from 0 to 2π) (4cos3θ)²dθ
    = ½ ∫ (from 0 to 2π) (16cos²9θ)dθ
    = 8 [(9/2)θ + (9/4)sin18θ) |(2π - 0)]
    = 8 (9π + 0)
    = 72π

    This answer seems really high and the answer in the books gives an answer of 4π... can someone help show me where exactly I am messing up? Thanks :)
     
  2. jcsd
  3. Nov 9, 2006 #2

    Office_Shredder

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    Something nice to remember is that cos and sin squared have an average value of 1/2 over a period of 2pi
     
  4. Nov 9, 2006 #3

    marcusl

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    Do not square 3*theta when you square the cosine function! Also you have gotten the integral (line 3 above) wrong, although it won't change your answer. Please look up the integral of cos^2 again. Finally, are you sure you are supposed to integrate from 0 to 2pi?
     
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