If $$Q_x\ - P_y=1$$ use Green's Theorem to find the area between the $x$-axis and $1$ arch of the cycloid $$x=t\ -sin(t)\ ,\ y=1-cos(t)$$What i've done so far is set $$P(x,x)\ =\ -0.5y, \ Q(x,y)\ =\ 0.5x$$ $$A=\int\int_R(1)\ dA\ ... $$And then I applied Greens Theorem from there by integrating over $0$ to $2\pi$. The answer I'm getting is $0$.
Thanks
You are supposed to get the area not by computing $\iint_D 1 \ dA$ but by the associated line integral $\oint_{\partial{D}} \langle P, Q \rangle \cdot d\mathbf{r}$ instead. If you choose $P(x,y) = - \frac{y}{2}$ and $Q(x,y) = \frac{x}{2}$ then Green's Theorem gives you that $\oint_{\partial{D}} \langle P, Q \rangle \cdot d\mathbf{r}$ is indeed $\iint_D 1 \ dA$, your area. If you compute this line integral over your curve by brute force (parameterize everything in sight) it comes out to be doable by integration by parts. Carry on...