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anonymous
 5 years ago
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
anonymous
 5 years ago
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

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

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

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but 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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