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Area under Polar Curves: Where did I go wrong?

  1. Sep 4, 2009 #1
    1.Where did I go wrong in finding the area enclosed inside r = 3 cos θ?



    2. Relevant equations

    I used the formula 1/2 ∫ ((f(θ)) squared dθ from alpha to beta




    3. The attempt at a solution

    I looked for the area of the semicircle from 0 to pi and then multiplied the whole thing by 2, since the equation is a circle, so it looks like this:

    =>2*1/2 ∫ from 0 to pi (3 cos θ)squared dθ
    => ∫ from 0 to pi (9 cos squared θ) dθ
    => ∫ from 0 to pi (9 (1/2) (1+cos 2θ) dθ [Using half angle identities]
    => 9/2 ∫ from 0 to pi (1+cos 2θ) dθ
    Integrating I get :
    => 9/2 (θ + 1/2 sin 2θ ] from 0 to pi)
    => 9/2 [(pi + 1/2 sin 2pi) - (0 + 1/2 sin 2o)]
    => 9/2 [pi + 0 - 0-0]

    So I get the final answer of 9/2pi. Thing is, the answer in the book is 9/4pi and I can't figure out they go there. Help, anyone?
     
    Last edited: Sep 5, 2009
  2. jcsd
  3. Sep 4, 2009 #2

    Dick

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    If you trace the curve carefully from theta=0 to pi, you'll realize it's not a semicircle. It's the whole circle. r is negative for theta between pi/2 and pi.
     
  4. Sep 5, 2009 #3
    Sorry, I phrased that wrong. I took the area of the half circle first, then multiplied it by two.

    Anyway, I am quite confused by the limits. How do you get the limits? >_<
     
    Last edited: Sep 5, 2009
  5. Sep 5, 2009 #4

    Dick

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    You get the limits by sketching the curve. It's starts at (3,0) (in xy coordinates) at theta=0, r=3 and returns to (3,0) at theta=pi, r=(-3). Calling that either a semicircle or a half circle are BOTH wrong. Don't multiply by 2!
     
  6. Sep 5, 2009 #5
    I see! Is there a general way for finding the limits? It seems that I always get messed up in that part...
     
  7. Sep 5, 2009 #6

    Dick

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    Just what I've already suggested. You'll have to learn to look at enough points on the curve to draw a good sketch.
     
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