Hello mates, there's this wild calculus problem in my text that I can't figure out.

I have to find the area under this parametric curve with

x = 6 ( T - Sin(T) )

y = 6 ( 1 - Cos(T) )

and $\displaystyle 0 \le T \le 2\pi$

I tried solving both for T but I don't think that was a good first step.

Thanks for any help, I really appreciate it.

I GOT HERE and said UGH: I solved y =6 ( 1 - Cos(T) ) for T and pluged T into x = 6 ( T - Sin(T) )

$\displaystyle \theta = arccos( 1 - \frac{y}{6})$

$\displaystyle x = 6 ( arccos( 1 - \frac{y}{6}) - sin(arccos( 1 - \frac{y}{6}))) $

Yeah, Kinda lost. Any help would be great, thanks a lot for looking guys.