Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor
    Gold Member

    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook