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Thread: polar coordinates

  1. #1
    Junior Member
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    polar coordinates

    any one here good with finding area of surface of revolution or arc length in polar coordinate system?
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  2. #2
    Junior Member
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    I finally got one of the problems I was working on. Now stuck on find the length of the curve over the given interval.

    r=8(1+cos theta) [0, 2pi]
    r=8+8 cos theta
    r`=-8 sin theta

    Arc length= 2 \int_0^\pi\sqrt{(8+8~cos\theta)^2+(-8~sin\theta)^2}~d\theta=2\int_0^\pi\sqrt{64+128~co  s\theta+64~cos^2\theta+64~sin^2\theta}~d\theta

    I am stuck at this point.
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  3. #3
    Senior Member tukeywilliams's Avatar
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     2 \int_{0}^{\pi} \sqrt{64 + 128 \cos \theta + 64 \cos^{2} \theta + 64 \sin^{2} \theta} \ d \theta =  2 \int_{0}^{\pi} \sqrt{128(1 + \cos \theta)} \ d \theta  = 2 \sqrt{128} \int_{0}^{\pi} \sqrt{1 + \cos \theta} \ d \theta = 4 \sqrt{128} \sqrt{\cos x +1} \tan \left(\frac{x}{2} \right) .
    Last edited by tukeywilliams; Jul 25th 2007 at 04:38 PM.
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