Flux through the boundary of a rectangle

Question

given the rectangle described by:

$0\leq x \leq 6$, $0 \leq y \leq 2$

And a fluid following the vector field:

$F = \langle x^3+2, y \cos(6x) \rangle$

We get the 2D curl:

$curl(F) = -6y\sin(6x)-0=-6y\sin(6x)$

By green's theorem, the flux through the boundary is thus:

$\int^6_0\int^2_0 -6y\sin(6x) dydx = -4\sin^2(18)$ (by wolfram, manually I got $-2\cos(36)$)

Neither of these answers is correct and I am not sure where the problem is.

Answer 1

\begin{align} \int^6_0\int^2_0 -6y\sin(6x) dydx &= \int^6_0\sin(6x)dx\int^2_0 -6y dy \\ &= \dfrac{-\cos6x}{6}\Big|^6_0\dfrac{-6y^2}{2}\Big|^2_0 \\ &= \dfrac{1-\cos36}{6}\dfrac{-24}{2} \\ &= \color{blue}{-4\sin^218} \end{align}