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LeoMessi
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
\[\int\limits_{2\pi}^{0}ydx-xdy+\int\limits_{0}^{2\pi}0dx =\int\limits_{2\pi}^{0}\lefta*a( 1-\cos(\theta) (1+\cos(\theta)d \theta\-a*a(\Theta -si(\theta)(-\sin(\theta)dtheta)\right)-\]
\[xdy-ydx=a*2(zsin z-2+2\cos z)dt (now integrate \it with \in \limits 0 \to 2\pi ) then answer will be -3\pi z2\]
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 \[A=\int\limits_{0}^{2\pi}(1-cosz)(1-cosz)dz-\int\limits_{0}^{2\pi}0d(-z)=3\pi\]
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