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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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  • $\begingroup$ pretty sure actually. $\endgroup$ – 40Plot Dec 8 '12 at 17:24
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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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  • $\begingroup$ Hmm.. 3pi is right. But the answer, according to the key is positive 3pi though. $\endgroup$ – 40Plot Dec 8 '12 at 17:24
  • $\begingroup$ Why is Q= 0 and P = y $\endgroup$ – 40Plot Dec 8 '12 at 20:18
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    $\begingroup$ 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. $\endgroup$ – levap Dec 8 '12 at 22:38
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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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  • $\begingroup$ I like this answer because it clears my confusion of how the curl came into the equation. Everyone assumes that everyone knows already. The other mystery is that it lets you know the intention of the problem. Line integrals are for finding work done.It just so happens area and work can be the same thing. So in this case you are using the theorem to calculate an area under a curve and this corresponds to work calculated by a line integral. Ideal if you have a swirling vector field inside a closed area. $\endgroup$ – Sedumjoy Sep 3 '17 at 13:42

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