In what case does the area vector $\vec \Delta A$ equal $\|A\|\vec i$?

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I'm trying to find the flux out of a cube (with side length $2$) centered at the origin. The vector field is $\vec F = -x\vec i$, so the only relevant faces of the cube are the two perpendicular to the $x$-axis (the other faces are parallel to the vector field and so don't affect the flux).

Let's say the orientation of the cube is outward, so for the face at $(1,0,0)$ the orientation vector $\vec n$ is equal to $\vec i$. For the face at $(-1,0,0)$, $\vec n = -\vec i$

And I know the area vector for a face to be the area of the face (a scalar) multiplied by the orientation vector (a vector): $\vec A = \vec n A $. The two area vectors we have are $\vec i A$ and $-\vec i A$.

When we want to create a flux integral, we have to consider an infinitesimal portion of the surface in question, so we consider $\Delta \vec A$. This should be equal to $\vec i \Delta A$, right? But for some reason my textbook says that $\Delta \vec A = \vec i \|\vec A\|$, i.e. the magnitude of the infinitesimal area vector is equal to the magnitude of the ENTIRE area of the surface...can anyone explain why this is?

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Since in this case the entire face has the same normal vector $\vec i$, it seems that the book is indicating with $\Delta \vec A$ not an infinitesimal portion but indeed just $\vec i ||\vec A||$.