Hello, Intsecxtanx!
Find the area under this parametric curve with:
. .
Recall the ancient formula: .
We have: .
Substitute: .
Got it?
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
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) )
Yeah, Kinda lost. Any help would be great, thanks a lot for looking guys.