We want to calculate the area of the cycloid: $x = a(t-\sin t)$, $y=a(1-\cos t)$, with $t\in [0,2π]$.
I know that the solution has to do with the Green's Theorem and calculating the surface integral of $-y\,dx$.
My question is: can we calculate the area by the surface integral of $x\,dy$? Will we get the same result?
When calculating the area of a parametric curve you should express the area as follows,
$$A=\int y(x)~dx=\int y(t)\dot x(t)~dt$$
In your case we have $\dot x=a(1-\cos t)=y$, therfore
$$ \begin{align} A &=a^2\int_0^{2\pi} (1-\cos t)^2 ~dt\\ &=a^2\left( \frac{3 t}{2} - 2 \sin t + \frac14 \sin 2t\right)\biggr|_0^{2\pi}\\ &=3\pi a^2\approx 9.4248a^2 \end{align} $$
