# Find the area bounded by one arc of the cycloid x = a(0-sin(0)), y = a(1-cos(0)),where a > 0, and 0 <= 0 <= 2pi, and the axis(use Green's theorem).

Find the area bounded by one arc of the cycloid $x=a\left(0-\mathrm{sin}\left(0\right)\right),y=a\left(1-\mathrm{cos}\left(0\right)\right)$,where a > 0, and $0\le 0\le 2\pi$, and the axis(use Green's theorem).
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Step 1
Note that the atea of the region D bounded by C del D is $A\left(D\right)=\frac{1}{2}{\int }_{C}xdy-ydx$.
The given cycloid and the x-axis intersect each other at the points (0,0) and (2 a $\pi$,0).
Note that, $\frac{1}{2}{\int }_{C}xdy-ydx=\frac{1}{2}{\int }_{{y}_{1}^{+}}xdy-ydx+\frac{1}{2}{\int }_{{y}_{2}^{+}}xdy-ydx$.
The parametrization of the segment .
Step 2
The derivative of the above function is computed as follows,
$\frac{dx}{dt}=1$
$\frac{dy}{dt}=0$
Now,
$\frac{1}{2}{\int }_{{y}_{1}^{+}}xdy-ydx=\frac{1}{2}{\int }_{0}^{2a\pi }\left(x\frac{dy}{dt}-y\frac{dx}{dt}\right)dt$
$=\frac{1}{2}{\int }_{0}^{2a\pi }\left(1\left(0\right)-\left(0\right)\left(1\right)\right)dt$
=0
Step 3
The parametrization of the segment .
The derivative of the above function is computed as follows,
$\frac{dx}{d0}=a\left(1-\mathrm{cos}0\right)$
$\frac{dy}{d0}=a\mathrm{sin}0$
Now,