The area bounded by half of a cycloid $\alpha(t)=(R(t-\sin(t), R(1-\cos(t))$ where $R>0$ and $0 \leq t \leq \pi$ and the x-axis is:
I've tried to use Green's theorem to solve this, by "closing" the original curve, creating two curves $\mathcal{C}_2$ and $\mathcal{C}_3$ (the red one and the blue one respectively). They're all clockwise oriented.
The parametric equations for both curves are:
$\begin{cases}\alpha_2(t)=(\pi R, t) \\ \alpha _2:[0,2R] \to \mathbb{R}^2 \end{cases}$
and
$\begin{cases}\alpha_3(t)=(t, 0) \\ \alpha _3:[0,\pi R] \to \mathbb{R}^2 \end{cases}$
And the vector field $F:\mathbb{R}^2 \to \mathbb{R}^2$ $F(x,y)=(0,x)$ such that the curl of $F$ is (0,0,1).
So by Green's theorem, using $\mathcal{C}'=\underbrace{\mathcal{C}_1}_{\text{Original curve}} \cup \mathcal{C}_2 \cup \mathcal{C}_3$:
$\oint\limits_{\mathcal{C}'} F dl = \iint\limits_{D} dx dy = A(D)$
Where D is the surface bounded by the curve $\mathcal{C}'$.
$A(D)=\int\limits_{\mathcal{C}_1}F dl-\int\limits_{\mathcal{C}_2}F dl-\int\limits_{\mathcal{C}_3}F dl$ (negative signs to change the orientation of the curves).
$\int\limits_{\mathcal{C}_1}F dl=\int\limits_{0} ^{\pi} R^2(t-\sin t)\sin t dt = \frac{\pi R^2}{2}$
$\int\limits_{\mathcal{C}_2}F dl=\int\limits_{0} ^{2R} \pi R dt = 2 \pi R^2$
$\int\limits_{\mathcal{C}_3}F dl=\int\limits_{0} ^{\pi R} 0 dt = 0 $
So, I end up with
$$ A(D)=\frac{-3 \pi R^2}{2} $$
Why am I getting a negative area? Is this correct?