anonymous
  • anonymous
Need help: use Green's Theorem to find the area under one arch of the cycloid: x= a*(z-sin(z)), y = a*(1-cos(z))...z is theta
OCW Scholar - Multivariable Calculus
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\int\limits_{2\pi}^{0}ydx-xdy+\int\limits_{0}^{2\pi}0dx =\int\limits_{2\pi}^{0}\lefta*a( 1-\cos(\theta) (1+\cos(\theta)d \theta\-a*a(\Theta -si(\theta)(-\sin(\theta)dtheta)\right)-\]
anonymous
  • anonymous
\[xdy-ydx=a*2(zsin z-2+2\cos z)dt (now integrate \it with \in \limits 0 \to 2\pi ) then answer will be -3\pi z2\]
anonymous
  • anonymous
let C1 x=a(z-sinz) y=a(1-cosz) C2 x=2pi-z y=0 so we have closed curve C =C1+C2 then area \[A=\int\limits_{0}^{2\pi}(1-cosz)(1-cosz)dz-\int\limits_{0}^{2\pi}0d(-z)=3\pi\]

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anonymous
  • anonymous
but in the above result the radius of the circle through which the cycloid is made is not present. in the integration the 'a' present in the equation should come. and the final answer should be 3pi a^2

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