Calculate the flux of the vector field $$\mathbf{v} = (4x,-2z,2y)$$ over the surface $$\mathbf{r} = (u, vcos(u), vsin(u)), \:\:\: \: 0\leq u \leq \pi, \:\:\: \: 0 \leq v \leq 1, \:\: \:\: \mathbf{n} \cdot \mathbf{e_x} > 0 $$
I calculate the normal vector as $$\frac{\partial \mathbf{r}}{\partial v} \times \frac{\partial \mathbf{r}}{\partial u} = (v, sin(u), -cos(u)) $$ which has a non-negative x-component which may be a bit troublesome since we want it to be positive but maybe it works. Anyway, how do I express my $x, y, z$ in terms of $u$ and $v$? I'm having trouble setting up the surface integral...
In general, you can always write $$ \iint_S \mathbf{v} \cdot \mathbf{dS} = \iint_{D(u,v)} \mathbf{v}(u,v)\cdot \frac{\partial \mathbf{r}}{\partial u}\times \frac{\partial \mathbf{r}}{\partial v} \; dudv $$ Substituting with the expressions you computed yields $$ \iint_S \mathbf{v} \cdot \mathbf{dS} = \iint_{D(u,v)} 4uv-2v\sin^2u-2v\cos^2u\; dudv=\iint_{D(u,v)} 4uv-2v\; dudv $$ where $D(u,v)=\{(u,v)\;|\; 0\le u \le \pi, 0\le v\le 1 \}$.
Can you take it from there?
Note. The orientation of your normal vector $\frac{\partial \mathbf{r}}{\partial u}\times \frac{\partial \mathbf{r}}{\partial v}$ is fine, it will always be non negative as $v\in [0,1]$.