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