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LeoMessi
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Need help: use Green's Theorem to find the area under one arch of the cycloid: x= a*(zsin(z)), y = a*(1cos(z))...z is theta
 3 years ago
 3 years ago
LeoMessi Group Title
Need help: use Green's Theorem to find the area under one arch of the cycloid: x= a*(zsin(z)), y = a*(1cos(z))...z is theta
 3 years ago
 3 years ago

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ducthinh Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{2\pi}^{0}ydxxdy+\int\limits_{0}^{2\pi}0dx =\int\limits_{2\pi}^{0}\lefta*a( 1\cos(\theta) (1+\cos(\theta)d \theta\a*a(\Theta si(\theta)(\sin(\theta)dtheta)\right)\]
 3 years ago

eletronic_rajesh Group TitleBest ResponseYou've already chosen the best response.0
\[xdyydx=a*2(zsin z2+2\cos z)dt (now integrate \it with \in \limits 0 \to 2\pi ) then answer will be 3\pi z2\]
 3 years ago

ducthinh Group TitleBest ResponseYou've already chosen the best response.0
let C1 x=a(zsinz) y=a(1cosz) C2 x=2piz y=0 so we have closed curve C =C1+C2 then area \[A=\int\limits_{0}^{2\pi}(1cosz)(1cosz)dz\int\limits_{0}^{2\pi}0d(z)=3\pi\]
 3 years ago

eletronic_rajesh Group TitleBest ResponseYou've already chosen the best response.0
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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