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Math Help - Area of polar curves

  1. #1
    Member
    Joined
    Apr 2009
    Posts
    85

    Area of polar curves

    Trying to the find this area
    The equation is r= 3+2sin\theta


    Heres what i did:
    A= \frac{1}{2}\int_0^\pi \! (3+2sin\theta)^2 \, d\theta
    A= \frac{1}{2}\int_0^\pi \! (9+6sin\theta+4sin^2\theta) \, d\theta
    A= \frac{1}{2}\int_0^\pi \! (9+6sin\theta+4*\frac{1}{2}(1-cos2\theta) \, d\theta
    A= \frac{1}{2}\int_0^\pi \! (11+6sin\theta-2cos2\theta) \, d\theta
    A= \frac{1}{2}[11\theta-6cos\theta-sin2\theta] \bigg|_0^\pi
    A= \frac{1}{2}[11\pi+12]
    A= \frac{11\pi}{2}+6

    Where did i go wrong?
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  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    715
    I think the upper limit is  \frac{\pi}{2}

    and the lower limit is  - \frac{\pi}{2}

    so  6 should disappear since  \cos(\theta) \bigg|_{-\pi/2}^{\pi/2} = 0

    Other is OK ( Happy)
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