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LeoMessi Group Title

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

  • 3 years ago
  • 3 years ago

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  1. ducthinh Group Title
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    \[\int\limits_{2\pi}^{0}ydx-xdy+\int\limits_{0}^{2\pi}0dx =\int\limits_{2\pi}^{0}\lefta*a( 1-\cos(\theta) (1+\cos(\theta)d \theta\-a*a(\Theta -si(\theta)(-\sin(\theta)dtheta)\right)-\]

    • 3 years ago
  2. eletronic_rajesh Group Title
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    \[xdy-ydx=a*2(zsin z-2+2\cos z)dt (now integrate \it with \in \limits 0 \to 2\pi ) then answer will be -3\pi z2\]

    • 3 years ago
  3. ducthinh Group Title
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    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 \[A=\int\limits_{0}^{2\pi}(1-cosz)(1-cosz)dz-\int\limits_{0}^{2\pi}0d(-z)=3\pi\]

    • 3 years ago
  4. eletronic_rajesh Group Title
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    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

    • 3 years ago
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