A solid $\Gamma$ in $\mathbb{R}^{3}$ is bounded by $$0 \leq x \leq 1, \hspace{.5cm} 0 \leq y \leq 1, \hspace{.5cm} 0 \leq z \leq g(x, y),$$ where $z = g(x, y)$ is an unknown differentiable surface.
Suppose the flux of the vector field $\mathbf{F} (x, y, z) = (x, -2y, z + 3)$ across the side of $\Gamma$ which is on the plane $y = 1$ is $-3,$ and across the side of $\Gamma$ which is on the plane $x = 1$ is $1.$ Then:
(1) What is the flux of $\mathbf{F}$ across the side of $\Gamma$ which is on the plane $x = 0?$
(2) What is the flux of $\mathbf{F}$ across the side of $\Gamma$ which is on the plane $y = 0?$
Okay, now my work: since the flux of a vector field is defined as $$ \iint_{\Gamma} \mathbf{F} \cdot \mathbf{n} \hspace{.1cm}d\Gamma,$$ where the unit normal vector of $\Gamma,$ denoted by $\mathbf{n},$ defines the orientation on the surface, and $\Gamma$ is a closed surface, then the side of $\Gamma$ which is on the plane $x = 0$ has orientation opposite to the orientation on the side of $\Gamma$ which is on the plane $x = 1.$ Hence the answer to (1) is: $Flux = -1.$
By a similar argument the answer to (2) would be: $Flux = 3.$
Is the above reasoning correct?