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Math Help - Area in Polar Coordinates 2

  1. #1
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    Area in Polar Coordinates 2

    Ok so the equation is

    r= 6 +4 sin theta

    Its is a limacon so the area will be symmetrical from 0 to pi/2

    Plugging this into the area formula I get

    Integral (6+4 sin theta)^2
    Integral (36 + 48 sin theta + 16 sin^2 theta)
    Integral (36 + 48 sin theta + 16 [(1-cos 2 theta)/2]
    Integral (36 + 48 sin theta + 8 (1 - cos 2 theta)
    Integral (36 + 48 sin theta + 8 - 8 cos 2 theta)
    Integral (44 + 48 sin theta - 8 cos 2 theta)

    which gives me

    44 theta - 48 cos theta - 8 sin 2 theta + C

    I worked it out and I am still unable to arrive at the answer. Can anyone tell me what I'm doing wrong?
    Last edited by shogunhd; December 9th 2007 at 08:17 PM.
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  2. #2
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    Hello, shogunhd!

    Find the area enclosed by: . r \:= \:6 +4\sin\theta

    It is a limacon, symmetric to the "y-axis".
    We can integrate from -\frac{\pi}{2} to \frac{\pi}{2} . . . and double.

    We have: . A \;=\;2 \times \frac{1}{2}\int^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}} (6 + 4\sin\theta)^2\,d\theta

    The function is: . 36 + 48\sin\theta + 16\sin^2\!\theta \;=\;36 + 48\sin\theta + 16\left(\frac{1-\cos2\theta}{2}\right)

    . . = \;36 + 48\sin\theta + 8 - 8\cos2\theta \;=\;44 + 48\sin\theta - 8\cos2\theta


    The integral is: . A \;\;=\;\;\int^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}}\left(44 + 48\sin\theta - 8\cos2\theta\right)\,d\theta \;\;=\;\;44\theta - 48\cos\theta - 4\sin2\theta\,\bigg|^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}}

    . . = \;\;\bigg[44\left(\frac{\pi}{2}\right) - 48\cos\left(\frac{\pi}{2}\right) - 4\sin(\pi)\bigg] - \bigg[44\left(\text{-}\frac{\pi}{2}\right) - 48\cos\left(\text{-}\frac{\pi}{2}\right) - 4\sin(-\pi)\bigg]

    . . = \;\;\bigg[22\pi - 0 - 0\bigg] - \bigg[-22\pi - 0 - 0\bigg] \;\;=\;\;\boxed{44\pi}

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  3. #3
    Ife
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    Why is this doubled though? If the integral is between \pi/2 and -\pi/2 isn't that for the entire region?
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