# Thread: Area under a curve with parametric equations? help plz.

1. ## Area under a curve with parametric equations? help plz.

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 $0 \le T \le 2\pi$

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) )
$\theta = arccos( 1 - \frac{y}{6})$
$x = 6 ( arccos( 1 - \frac{y}{6}) - sin(arccos( 1 - \frac{y}{6})))$
Yeah, Kinda lost. Any help would be great, thanks a lot for looking guys.

2. Hello, Intsecxtanx!

Find the area under this parametric curve with:

. . $\begin{array}{ccc}x &=& 6 (\theta - \sin\theta) \\ y &=& 6 (1 - \cos\theta)\end{array}\quad 0\le \theta \le 2\pi$

Recall the ancient formula: . $A \;=\;\int^b_a y\,dx$

We have: . $\begin{array}{ccc}y &=& 6(1-\cos\theta) \\ dx &=& 6(1 - \cos\theta)\,dx \end{array}$

Substitute: . $A \;=\;\int^{2\pi}_0 \underbrace{6(1-\cos\theta)}_y\cdot \underbrace{6(1-\cos\theta)\,d\theta}_{dx} \;=\;36\int^{2\pi}_0(1-\cos\theta)^2d\theta$

Got it?

3. ugh it makes perfect sense now, it's so obvious and i was staring at it for so long, thanks math god!!!!