### Flux in 2D

1. What is flux in 2D?

 Recall that: The flux of any two dimensional vector field $\mathbf v =v_1(x,y)\,\mathbf i +v_2(x,y)\,\mathbf j$ across a plane curve $C$ is defined by $$\text{Flux}=\int_C\mathbf v \cdot \mathbf n\, dS$$ where $\mathbf n$ is a unit normal vector to $C$.

A particular context that helps to understand the definition of flux is considering the velocity field of a fluid. Let $C$ be a plane curve and let $\mathbf v$ be a velocity vector  in the plane. Now imagine that $C$ is a membrane across which the fluid flows, but does not impede the flow of the fluid. In other words, $C$ is an idealised membrane invisible to the fluid. In this context, the flux of $\mathbf v$ across $C$ is the quantity of fluid flowing through $C$ per unit time, or the rate of flow.

2. Flux across line segments

The following applet shows a representation of a fluid flowing through a line segment. Activate the boxes to show the field and flow. Observe what happens to the flux when you change the angle, length or position of the segment.

Example 1. Consider a fluid flowing with a velocity vector $\mathbf v =-y\, \mathbf i + x\, \mathbf j$. Calculate the flux of $\mathbf v$ across the line $C=\{(x,y)|\,x=2, 2\leq y\leq 6\}$.

First, we parametrise the line as
$$\mathbf r (t)=2\,\mathbf i +(4t+2)\,\mathbf j \qquad (0\leq t\leq 1).$$
Then we have that $\mathbf r ' (t)=4\,\mathbf j.$ A unit tangent vector to $C$ is then given by
$$\mathbf T=\frac{\mathbf r ' (t)}{|\mathbf r ' (t)|}=\frac{4\,\mathbf j}{4}=\mathbf j$$
The unit vector normal is then
$$\mathbf n =\mathbf T \times \mathbf k = \mathbf j \times \mathbf k = \mathbf i.$$
With this parametrisation,
$$\mathbf v = -(4t+2)\,\mathbf i+ 2 \,\mathbf j \qquad \implies \qquad \mathbf v \cdot \mathbf n= -4t-2.$$
Note also that within the integral the infinitesimal element of arc can be expressed in terms of the parametrisation as $dS=|\mathbf r ' (t)|dt=4\,dt$ in this case. The flux across $C$ is then
$$\int_0^1(-16t-8)dt=-8(t^2+t)\Big|_0^1=-16.$$

 Recall that: If $\mathbf r (t)=x(t)\,\mathbf i +y(t)\,\mathbf j$ is a parametrisation of $C$ for $a\leq t\leq b$, then $$\text{Flux}=\int_a^b\left(v_1(t)\dot{y}-v_2(t)\dot{x}\right)dt.$$

You can check the previous result in the following applet, which shows a representation of the fluid flowing with a velocity vector $\mathbf v =-y\, \mathbf i + x\, \mathbf j$. Play with the applet changing the position of the endpoints of the line segment. Observe what happens to the flux.

Activity 1: Use the applet to answer the following questions.

1. What is the flux across the line segment oriented from $(0,0)$ to $(2,3)$?

2. What is the flux if the line segment is oriented from $(2,3)$ to $(0,0)$?

3. What is the flux if the line segment is oriented from $(-2,-2)$ to $(2,2)$ or from $(-3,3)$ to $(3,-3)$?

The general case:

Calculate the flux across the curve defined as the line segment oriented from $(a,b)$ to $(c,d)$, with $a,b,c,d$ real constants. Check the values obtained in the previous activity with the general solution.

Example 2. Consider a fluid flowing with a velocity vector $\mathbf v =x\, \mathbf i -y \, \mathbf j$. Calculate the flux of $\mathbf v$ across the line segment from $(-1,1)$ to $(-1,3)$ followed by another line segment from $(-1,3)$ to $(-2,5)$.

To calculate the flux, we shall calculate the flux across each of the two curves which make the complete curve. That is, we shall calculate
$$\text{Flux}=\int_{C_1}\mathbf F \cdot \mathbf n_1 dS+\int_{C_2}\mathbf F \cdot \mathbf n_2 dS.$$

For $C_1$:

First, we parametrise the line as
$$\mathbf r (t)=-\mathbf i +(2t+1)\,\mathbf j \qquad (0\leq t\leq 1).$$
Then we have that $\mathbf r ' (t)=2\,\mathbf j.$ A unit tangent vector to $C_1$ is then given by
$$\mathbf T=\frac{\mathbf r ' (t)}{|\mathbf r ' (t)|}=\frac{2\,\mathbf j}{2}=\mathbf j$$
The unit vector normal is then
$$\mathbf n_1 =\mathbf T \times \mathbf k = \mathbf j \times \mathbf k = \mathbf i.$$
With this parametrisation,
$$\mathbf v = -\mathbf i-(2t+1) \,\mathbf j \qquad \implies \qquad \mathbf v \cdot \mathbf n_1= -2.$$
Note also that $dS=|\mathbf r ' (t)|dt=2\,dt$ in this case. The flux across $C$ is then
$$\int_0^1-2dt=-2t\Big|_0^1=-2.$$

For $C_2$:

We parametrise the line as
$$\mathbf r (t)=-(t+1)\,\mathbf i +(2t+3)\,\mathbf j \qquad (0\leq t\leq 1).$$
Then we have that $\mathbf r ' (t)=-\mathbf i+2\,\mathbf j.$ A unit tangent vector to $C_2$ is then given by
$$\mathbf T=\frac{\mathbf r ' (t)}{|\mathbf r ' (t)|}=\frac{-\mathbf i+2\,\mathbf j}{\sqrt{5}}.$$
The unit vector normal is then
$$\mathbf n_2 =\mathbf T \times \mathbf k = \frac{2\mathbf i+\mathbf j}{\sqrt{5}}.$$
With this parametrisation,
$$\mathbf v = -(t+1)\,\mathbf i- (2t+3) \,\mathbf j \qquad \implies \qquad \mathbf v \cdot \mathbf n_2= \frac{-4t-5}{\sqrt{5}}.$$
Note also that $dS=|\mathbf r ' (t)|dt=\sqrt{5}\,dt$ in this case. The flux across $C$ is then
$$\int_0^1(-4t-5)dt=\left(-2t^2-5\right)\Big|_0^1=-7.$$
Therefore the flux is $-2-7=-9$.

You can check the previous result in the following applet, which shows a representation of the fluid flowing with a velocity vector $\mathbf v =x\, \mathbf i - y\, \mathbf j$. Play with the applet changing the position of the endpoints of the line segmen. Observe what happens to the flux.

Activity 2:  Within the applet.

1. Move around the three points that define the line segments. Observe what happens to the flux.

2. If we fix the endpoints of the path and we just move the point in the middle, you can observe that the flux remains constant. Why?

3. Finally, drag the three points to create the path from $(3,0)$ to $(3,6)$ and then to $(0,6)$. Observe that the flux is zero. Why? What happens if you move the endpoints of the path along the axis.

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Key Concepts

A key concept expressed in terms of line integrals is flux. Flux measures the rate that a field crosses a given line. The idea of flux is especially important for Green’s theorem, and in higher dimensions for Stokes’ theorem and the divergence theorem.

Key Equations

Calculating a scalar line integral:

$$\int_C f(x,y)dS=\int_a^bf\left(x(t),y(t)\right)\sqrt{(x'(t))^2+(y'(t))^2}.$$

Calculating a vector line integral:

$$\int_C\mathbf F (\mathbf r)\cdot \mathbf d \mathbf r =\int_C\left(F_1dx+F_2dy\right)=\int_a^b \mathbf F (\mathbf r (t) ) \cdot \mathbf {r'}(t) dt.$$

Calculating flux:

$$\int_C \mathbf F \cdot \mathbf n dS=\int_C \left(F_1dy-F_2dx\right).$$