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Area under a polar curve

  1. Nov 9, 2012 #1
    r2 = 4cos(2θ)

    First I graph it. Then I set up the integral.

    _____π
    (1 / 2)∫ 4cos(2θ) dθ
    _____0

    ________π
    = [sin(2θ)]
    ________0

    I thought the limits ought to be π and 0, but that comes out to zero. I pick other limits and they come out to 0. My graph matches the one in the back of the book. I know I've integrated correctly. I think the problem is the limits I'm picking somehow. The correct answer is 4.

    [Edit] I also picked (π / 4) and -(π / 4) as limits, but that certainly didn't get me 4. [Edit]
     
  2. jcsd
  3. Nov 9, 2012 #2
    Forget it I see what I'm doing. I'm only finding the area of one loop. My bad. Feel free to delete this post if you're a mod.
     
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