- #1

FrogPad

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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.