Finding the area under the curve $y=3-3\cos(t),x=3t-3\sin(t)$

Question

I need to find the area under the curve $\color{blue}{y=3-3\cos(t),x=3t-3\sin(t)}$ and between $\color{blue}{x=2\pi,x=0\text{, above axis}}$ using $\color{blue}{\text{Green's theorem}}$.

My attempt

I thought to extract $\color{blue}t$ from $y=3-3\cos(t)$ and then to place $\color{blue}t$ here: $x=3t-3\sin(t)$

$$y-3=-3\cos t$$

$$\frac{y-3}{-3}=\cos t$$

$$\Longrightarrow$$

$$t=\arccos\left(\frac{y-3}{-3}\right)$$

$$x=3\left[\arccos\left(\frac{y-3}{-3}\right)\right]-3\sin\left(\arccos\left(\frac{y-3}{-3}\right)\right)$$

but now it looks weird.

Is my approach correct?

Edit:

this is cycloid https://en.wikipedia.org/wiki/Cycloid

Answer 1

No, there is no point trying to render this into Cartesian form: it is better to stick to parametric form

Hint...if you can find the right $t$ values for the limits, you only need $$\int y\frac{dx}{dt}dt$$