Q:

Calculating the mass flux through the curve $AB$

Flux through a flat curve

We want to calculate the mass flux through the curve $AB$
$$\Delta m= \delta \cdot \Delta s \cdot \Delta t \cdot \overrightarrow{v} \cdot \hat{n}$$
$$\frac{dM}{dt}=\int_A^B \delta \cdot \overrightarrow{v} \cdot \hat{n} ds$$
$$\overrightarrow{F}=\delta \cdot \overrightarrow{v}$$
$$\text{ Flux }=\int_A^B \overrightarrow{F} \cdot \hat{n}ds$$
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I haven't understood what we are doing here... Could you explain it to me?
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EDIT
Then there is the following in my notes:

$$\frac{d \overrightarrow{R}}{ds}=\frac{dx}{ds} \hat{\imath}+\frac{dy}{ds} \hat{\jmath}$$
$$\hat{n}=\hat{T} \times \hat{k}$$
$$\hat{n}=(\frac{dx}{ds} \hat{\imath}+\frac{dy}{ds} \hat{\jmath}) \times \hat{k}$$
$$\hat{n}=-\frac{dx}{ds} \hat{\jmath}+\frac{dy}{ds} \hat{\imath}$$
$$\overrightarrow{F}=M \hat{\imath}+N \hat{\jmath}$$
$$\overrightarrow{F} \hat{n}=M \frac{dy}{ds}-N \frac{dx}{ds}$$
$$\int_A^B\overrightarrow{F} \hat{n} ds=\int_A^B(M \frac{dy}{ds}-N \frac{dx}{ds})ds$$
So $$\int_A^B \overrightarrow{F} \hat{n}ds=\int_A^B(Mdy-Ndx)$$
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Why is $\hat{n}=\hat{T} \times \hat{k}$ ?
$\hat{n}$ is the normal vector to the curve, so also to the tangent vector $\hat{T}$, right?
And what's with $\hat{k}$?
And also isn't $\hat{T}$ the unit tangent vector? So shouldn't it be: $\displaystyle{\hat{T}=\frac{\frac{d \overrightarrow{R}}{ds}}{|\frac{d \overrightarrow{R}}{ds}|}}$?

A:

It's the mass flux of a fluid reduced to 2 dimensions.
The fluid has a velocity $\overrightarrow v$ which flows at an angle with the curve.
Note that $\overrightarrow v$ is a vector field that may have different directions and magnitudes at different locations and times.
If we look at a particular small amount of fluid close to the curve, we can say the following:
In an amount of time $\Delta t$ the fluid flows a distance: 
$$\overrightarrow{\Delta l} = \overrightarrow v \Delta t$$
The amount that would flow through a small perpendicular width $\Delta s$ is: 
$$|\overrightarrow v \Delta t| \Delta s$$
To find the amount that actually flows through the curve, we need to take the dot product with the normal $\hat n$ of the curve.
The amount that flows through the curve is:
$$\Delta V = (\overrightarrow v \Delta t \cdot \hat n) \Delta s$$
We find the corresponding mass by multiplying with the density $\delta$ (usually $\rho$):
$$\Delta m = \delta (\overrightarrow v \Delta t \cdot \hat n) \Delta s$$
The total mass $\Delta M$ that flows through the curve in time $\Delta t$ is:
$$\Delta M = \sum_i \delta (\overrightarrow v \Delta t \cdot \hat n) \Delta s_i $$
where $\Delta s_i$ is some partition.
Or per unit time that is:
$$\frac{\Delta M}{\Delta t} = \sum_i \delta (\overrightarrow v \cdot \hat n) \Delta s_i $$
Applying the definition of the Riemann integral gives us:
$$\frac{dM}{dt} = \int_A^B \delta (\overrightarrow v \cdot \hat n) ds $$
EDIT:
The vector $\hat k$ is the so called binormal vector (usually $\hat B$). It is perpendicular to the plane of the curve. In other words $\hat T, \hat n, \hat k$ form an orthonormal basis.
That is also why each of them is the cross product of the other two.
And yes, $\hat T$ must be normalized. However, if $\overrightarrow R(s)$ has "unit speed" it is implicitly normalized. The term "unit speed" means that $\frac{d\overrightarrow R(s)}{ds} = 1$. In other words, the parameter $s$ is the same as the path length.

