Green's theorem - area of a cycloid

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SUMMARY

This discussion focuses on applying Green's Theorem to calculate the area under one arch of a cycloid defined by the parameterization p(t) = < t-2sin(t),2-2cos(t)> for 0≤t≤2π. The correct area is determined to be 8π, achieved by selecting the appropriate force field, either F=<0,x> or F = <-y,0>. The key takeaway is that while both force fields can be used, the accuracy of the integral calculation is crucial, as demonstrated by the miscalculation of the integral of t*sin(t) over the interval [0, 2π].

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Feodalherren
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Homework Statement



Use Green’s Theorem to find the area of the region between the x – axis and one arch of the cycloid
parameterized by p(t) = < t-2sin(t),2-2cos(t)> for 0≤t≤2∏
p

Homework Equations


The Attempt at a Solution


My problem here is that I get different answers depending on if I use
F=<0,x> or F = <-y,0>

The 8∏ answer that I have to the right is the correct one. How do I know which force field to pick?

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Feodalherren said:

Homework Statement



Use Green’s Theorem to find the area of the region between the x – axis and one arch of the cycloid
parameterized by p(t) = < t-2sin(t),2-2cos(t)> for 0≤t≤2∏
p

Homework Equations





The Attempt at a Solution


My problem here is that I get different answers depending on if I use
F=<0,x> or F = <-y,0>

The 8∏ answer that I have to the right is the correct one. How do I know which force field to pick?

You can pick either one. The only problem is that you did the first integral wrong. The integral of t*sin(t) between 0 and 2pi isn't zero.
 

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