The vector field is equal to $F=⟨6y,−6x⟩$, what is the integral over the circle $x^2+y^2=4$.
I have tried $c'(t)=<-2sin(t), 2cos(t)>$, since the points for a unit circle would be $<cos(t), sin(t)>$ and $F(c(t))=<12cos(t),-12sin(t)>$. This comes out to the integral $\int^{2\pi}_{0}-48sin(t)cos(t)dt$..which is giving me $0$, which is wrong.
On
Looks like you made a simple transposition error. If you set $c(t)=\langle 2\cos t,2\sin t\rangle$, then $F(c(t)) = \langle 12\sin t,-12\cos t\rangle$ and $F(c(t))\cdot c'(t)=-24$.
Besides that serious error, you’ve also got the wrong orientation for $c$. You can fix that by negating the result, switching the order of the limits of integration, or using $-t$ instead of $t$ in your parameterization.
HINT What is the flux of $F$ through the circle? And how is that related to the line integral (you could get "stoked" by this part of the problem).
Doing the integral by brute force is possible but much harder.