In showing my thought process in solving the following problem using the divergence theorem, I hope someone could point out where the selection of the sign comes into play. While the question below does define the orientation of the normal vector, it seems, at least in my working, that this piece of information is not used...
Question: Find $ \iint_{S}^{} \textbf{F} \cdot d \textbf{S}$, where $\textbf{F}(x,y,z) = x^2 \textbf{i} + xy \textbf{j} + x^3y^3 \textbf{k}$ and $S$ is the surface of the rectangular region bounded by the three coordinate planes and the planes $x=1$, $y=2$ and $z=-3$. The orientation of $S$ is given by the outward pointing normal vector.
$ \iint_{S}^{} \textbf{F} \cdot d \textbf{S} = \iiint_{}^{} div(\textbf{F})dV $
$div(\textbf{F}) = \frac{∂x^2}{∂x}+\frac{∂(xy)}{∂y}+\frac{∂(x^3y^3)}{∂z} = 3x$
We set the bounds of the triple integral to $x=0$ to $x=1$, $y=0$ to $y=2$ and $z=0$ to $z=-3$:
$ \iint_{S}^{} \textbf{F} \cdot d \textbf{S} = \int_{0}^{-3}\int_{0}^{2}\int_{0}^{1} (3x) dx dydz = -9$
Is the sign correct? Any help is appreciated
Let $R$ be the box defined in the question. Then Gauss' theorem says that $$\int_{\partial R}{\bf F}\cdot dS=\int_R {\rm div}({\bf F})\>dV\ .$$ You have computed ${\rm div}({\bf F})(x,y,z)=3x$. On the other hand $$R=[0,1]\times[0,2]\times[-3,0]\ .$$ This implies that $$\int_R {\rm div}({\bf F})\>dV=\int_0^1\int_0^2\int_{-3}^0 3x \>dz\>dy\>dx=6\int_0^1 3x\>dx=9\ .$$ You wrote $\int_0^{-3}\ldots dz$, which causes the wrong "orientation" of the volume element $dV$.