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Greens' Theorem (finding area) -

  1. Apr 18, 2005 #1
    Ok, so I am obviously doing something wrong... but I don't know what it is. Here is the problem:

    Use Green's theorem to compute the area inside the curve.
    [tex] \vec r(t)=(2\sin(t)\cos(t),\sin(t))\,\,\,\,\,\,0\leq t\leq \pi [/tex]

    Greens Theorem
    [tex] \frac{1}{2} \oint_c x\,dy - y\,dx [/tex]

    [tex] x = 2\sin(t)\cos(t) [/tex]
    [tex] y = \sin(t) [/tex]

    I'm guessing this is where I'm making my error, since I'm making an assumption about the notation.
    [tex] dy = \frac{dy}{dt} = \cos(t) [/tex]

    [tex] dx = \frac{dx}{dt} = 4(\cos(t))^2 - 2 [/tex]

    [tex] A = \frac{1}{2}\oint_0^\pi [(2\sin(t)\cos(t))(\cos(t)]\,dt-[(\sin(t))(4(\cos(t))^2 - 2)]\,dt [/tex]

    Letting my trusty TI-89 eat up the integral and I get:

    [tex]2\pi-\frac{4}{3} [/tex]

    Well the online homework doesn't like this answer, so it must be wrong. I'm just unsure where I'm making the mistake. The notation is a little off the wall for me, so I'm not 100% sure with what I'm doing. Any help is appreciated. Thank you.
  2. jcsd
  3. Apr 18, 2005 #2
    [tex] x = 2sin(t)cos(t) = sin(2t), dx = 2cos(2t) [/tex]

    [tex] y = sin(t), dy = cos(t) [/tex]

    [tex] \frac{1}{2} \oint_c x\,dy - y\,dx [/tex]

    [tex] \frac{1}{2} \oint_c(sin(2t)cos(t) - 2sin(t)cos(2t))dt [/tex]

    Maple says that equals 4/3
  4. Apr 18, 2005 #3
    I think you miscalculated 'dx'.

    [tex] x = 2sin(t)cos(t) [/tex]

    [tex] dx = 2(cos(t)cos(t) - sin(t)sin(t)) [/tex]

    [tex] dx = 2(cos^2(t)-sin^2(t) = 2cos(2t) \mbox{ by a double angle identity.}[/tex]
  5. Apr 18, 2005 #4
    Thanks man, the help was very much appreciated
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