I'll go ahead and assume you meant to say that S is the surface given by
[tex]S = \left\{(x,y,z) \mid z = \cos(y)\text{ with } 0\le y\le \pi\text{ and }0\le x\le4\right\}[/tex]
This immediately gives us a parameterization for the surface,
[tex]\vec r(x, y) = \left\langle x, y, \cos(y)\right \rangle[/tex]
The upward-pointing normal vector to this surface is then
[tex]\vec n = \dfrac{\partial\vec r}{\partial x} \times \dfrac{\partial\vec r}{\partial y} = \left\langle0,\sin(y),1\right\rangle[/tex]
Then the flux of [tex]\vec F(x,y,z) = \left\langle e^{-z}, 4z, 6xy\right\rangle[/tex] across S is
[tex]\displaystyle \iint_S \vec F(x,y,z)\cdot\mathrm d\vec s = \int_0^4\int_0^\pi \vec F(x,y,\cos(y))\cdot\vec n\,\mathrm dy\,\mathrm dx \\\\ = \int_0^4\int_0^\pi \left\langle e^{-\cos(y)},4\cos(y),6xy\right\rangle \cdot \left\langle0,\sin(y),1\right\rangle \,\mathrm dy\,\mathrm dx \\\\ = \int_0^4\int_0^\pi (4\sin(y)\cos(y)+6xy)\,\mathrm dy\,\mathrm dx \\\\ = 2 \int_0^4\int_0^\pi (\sin(2y) + 3xy)\,\mathrm dy\,\mathrm dx = \boxed{24\pi^2}[/tex]
