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I want to use Green's theorem for computing the area of the region bounded by the $x$-axis and the arch of the cycloid:

$$ x = t- \sin (t),\;\;\; y = 1 - \cos (t),\;\; 0 \leq t \leq 2\pi $$

So basically, I know the radius of this cycloid is 1. And to use Green's theorem, I will need to find $Q$ and $P$.

$$\int_C P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$$

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pretty sure actually. – Ichbin Dec 8 '12 at 17:24
up vote 2 down vote accepted

You can take $P = y$ and $Q = 0$. Then, $$ \oint\limits_C-ydx = \iint\limits_D dA = \text{Area}(D). $$ Along the $x$-axis, you have $y = 0$, so you only need to compute the integral over the arch of the cycloid. Note that your parametrization of the arch is a clockwise parametrization, so in the following calculation, the answer will be the minus of the area: $$\int_0^{2\pi} (\cos(t) - 1)(1 - \cos(t)) dt = - \int_0^{2\pi} 1 - 2\cos(t) + \cos^2(t) dt = -3\pi. $$

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Hmm.. 3pi is right. But the answer, according to the key is positive 3pi though. – Ichbin Dec 8 '12 at 17:24
Why is Q= 0 and P = y – Ichbin Dec 8 '12 at 20:18
I wrote that you must take into the consideration the orientation of the curve. Why $Q = 0$ and $P = y$? You can use any $(P,Q)$ such that $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$. Common choices are $Q = 0, P = -y$ or $Q = x, P = 0$ or $P = -\frac{y}{2}, Q = \frac{x}{2}$. In your problem, since some part of the curve is the $x$-axis, if we take the first option, the integral over any segment of the $x$-axis will be zero, and so this choice simplifies the calculation. – levap Dec 8 '12 at 22:38

Since you are asked to find the area which corresponds to $ \iint_D dA,$ so you need to have the following condition

$$ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}=1. $$

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