A little puzzle with the way to compute the formula for Surface Integral of a Vectors Field

Question

I have just found a website that explains how to construct the formula of Surface Integrals of Vectors Fields in an intuitive way (link here). However, there's a very confusing step, that is:

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$$Volume = (\left\lvert \vec F\right\rvert cosθ)dS = (\left\lvert \vec F\right\rvert\left\lvert \vec n\right\rvert cosθ)dS = (\vec F \cdot \vec n)dS$$

$$Flux = \iint_S (\vec F \cdot \vec n)dS = \iint_S F(x,y,z) \cdot \langle -g_x (x,y), -g_y (x,y), 1\rangle dA $$

I don't understand why there is $\left\lvert n\right\rvert$ suddenly appearing in the formula above ?

To me, it should be as follows:

$$cosθ = \frac{\left( \vec F \cdot \vec n\right)}{\left\lvert \vec F\right\rvert\left\lvert \vec n\right\rvert}$$

$$dS = \left\lvert \vec n\right\rvert dydx$$

$$Volume = (\left\lvert \vec F\right\rvert cosθ)dS = (\left\lvert \vec F\right\rvert\frac{\left( \vec F\cdot\vec n\right)}{\left\lvert \vec F\right\rvert\left\lvert \vec n\right\rvert})\left\lvert \vec n\right\rvert dydx = \left( \vec F\cdot\vec n\right) dydx$$

$$Flux = \iint_S (\vec F\cdot\vec n) dydx = \iint_S F(x,y,z)\cdot\langle -g_x (x,y), -g_y (x,y), 1\rangle dydx $$

Is my way of thinking correct ?

Answer 1

We have $|n|=1$, hence

$|F| \cos \theta = |F|\cdot|n|\cos \theta=F \cdot n$.

Answer 2

When you write $dS = |\vec n|\,dx\,dy$, the $\vec n$ here is not the unit normal, whereas in the flux integral $\vec n$ is supposed to be the unit normal.

What's going on is this: Letting $\vec N = \langle -g_x,-g_y,1\rangle$ be the fundamental cross product, we do in fact have $dS = |\vec N|\,dx\,dy$. Then $$\vec F\cdot\vec n\, dS = \vec F\cdot \underbrace{\left(\frac{\vec N}{|\vec N|}\right)}_{\vec n} \underbrace{|\vec N|\,dx\,dy}_{dS} = \vec F\cdot\vec N\,dx\,dy.$$