Greens' Theorem (finding area) -

[FrogPad]

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.
\vec r(t)=(2\sin(t)\cos(t),\sin(t))\,\,\,\,\,\,0\leq t\leq \pi


Greens Theorem
\frac{1}{2} \oint_c x\,dy - y\,dx

x = 2\sin(t)\cos(t)
y = \sin(t)

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

dx = \frac{dx}{dt} = 4(\cos(t))^2 - 2

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

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

2\pi-\frac{4}{3}

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.

[whozum]

x = 2sin(t)cos(t) = sin(2t), dx = 2cos(2t)

y = sin(t), dy = cos(t)

\frac{1}{2} \oint_c x\,dy - y\,dx

\frac{1}{2} \oint_c(sin(2t)cos(t) - 2sin(t)cos(2t))dt

Maple says that equals 4/3

[whozum]

I think you miscalculated 'dx'.

x = 2sin(t)cos(t)

dx = 2(cos(t)cos(t) - sin(t)sin(t))

dx = 2(cos^2(t)-sin^2(t) = 2cos(2t) \mbox{ by a double angle identity.}

[FrogPad]

Thanks man, the help was very much appreciated