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Analiza matematyczna funkcji wielu zmiennych

Kurs: Analiza matematyczna funkcji wielu zmiennych > Rozdział 4

Lekcja 4: Całki krzywoliniowe na polach wektorowych (artykuły)

Flux in two dimensions

How line integrals can measure flow rate through a curve. Learning this is a good foundation for Green's divergence theorem.

Kontekst

Do czego zmierzamy

Given a region enclosed by a curve start color #bc2612, C, end color #bc2612, and a fluid flow determined by a vector field start color #0c7f99, F, end color #0c7f99, left parenthesis, x, comma, y, right parenthesis, the rate at which fluid is exiting that region (assuming it has density 1) can be measured with the following line integral:
$\begin{aligned} \underbrace{ -\dfrac{d(\text{fluid mass in region})}{dt} }_{\text{Rate at which mass leaves region}} = \oint_{\redE{C}} \blueE{\textbf{F}} \cdot \greenE{\hat{\textbf{n}}}\;\redE{ds} \end{aligned}$
Here, start color #0d923f, start bold text, n, end bold text, with, hat, on top, left parenthesis, x, comma, y, right parenthesis, end color #0d923f is some function that returns the outward unit normal vector at every point on the curve start color #bc2612, C, end color #bc2612.
This integral integral, start subscript, start color #bc2612, C, end color #bc2612, end subscript, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, start color #bc2612, d, s, end color #bc2612 is called a flux integral, or sometimes a "two-dimensional flux integral", since there is another similar notion in three dimensions.
In any two-dimensional context where something can be considered flowing, such as a fluid, two-dimensional flux is a measure of the flow rate through a curve. In general, the curve isn't necessarily a closed loop.

Changing fluid mass in a region

Suppose you have some two-dimensional vector field, start color #0c7f99, start bold text, F, end bold text, left parenthesis, x, comma, y, right parenthesis, end color #0c7f99

And consider some closed curve C wandering through this field.

As we like to do with vector fields, let's say this represents some kind of fluid flow. But let's limit ourselves to thinking about what happens over a very small amount of time, just a quick instance really. For example, you can imagine each fluid particle moving from the tail of one of the vectors drawn above to its tip.

Key question: How can we measure the instantaneous rate of change of the mass of fluid inside the region enclosed by start color #bc2612, C, end color #bc2612?

Specifically, let's say our fluid has a constant density throughout the plane, perhaps 1 kilogram per square meter. If we let the fluid flow for a small amount of time, delta, t, what is the total mass of fluid which leaves/enters the region? The answer, of course, will be some function of the vector field start bold text, F, end bold text and the curve C.

If this question reminds you of the intuition for divergence, it's for a very good reason. In fact, in another article, we will use the formula we are now deriving to give the formal definition of divergence, and to show a special relationship divergence has with line integrals in two dimensions.

One bit of length at a time

One way to answer this question is to break up the curve into many tiny segments and consider how much fluid leaves or enters each segment. If we zoom in on a small enough segment, we can basically treat it as a straight line, and the fluid particles going through it are pretty much all moving at the same speed and in the same direction.

As we let the fluid flow for the small change in time delta, t, the fluid passing through this segment will form a parallelogram.

(I happened to draw a case where the fluid is going out of the region, but you could just as easily imagine the fluid coming into the region if the velocity vectors at that point were turned the other way.)

Since we are assuming a uniform density of 1, start text, k, g, end text, slash, start text, m, end text, squared for the fluid, the mass of fluid leaving the region equals the area of this parallelogram. Let's break down that area.

The base of the parallelogram is the length of our tiny segment. Let's call that delta, s.
The displacement vector of a fluid particle which starts at some point on the tiny segment will be start bold text, v, end bold text, delta, t, where start bold text, v, end bold text is the velocity vector of fluid at that point, and delta, t is the amount of time the fluid flowed.
The height of the parallelogram will be the component of the start bold text, v, end bold text, delta, t displacement vector which is perpendicular to the segment. You can extract this by taking the dot product between start bold text, v, end bold text, delta, t and a unit normal vector to the curve start color #bc2612, C, end color #bc2612. Let's name that unit normal vector start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f.

Concept check: What is the total mass exiting the tiny segment of length delta, s over the short time delta, t?

(Choice A)
left parenthesis, start bold text, v, end bold text, with, vector, on top, dot, start bold text, n, end bold text, with, hat, on top, right parenthesis, left parenthesis, delta, t, right parenthesis, left parenthesis, delta, s, right parenthesis
(Choice B)
left parenthesis, start bold text, v, end bold text, with, vector, on top, delta, t, right parenthesis, left parenthesis, delta, s, right parenthesis

[Odpowiedź]

Dividing out by delta, t, we have the amount of mass passing through this tiny line segment per unit time:

Exiting mass per unit time equals, left parenthesis, start bold text, v, end bold text, with, vector, on top, dot, start bold text, n, end bold text, with, hat, on top, right parenthesis, left parenthesis, delta, s, right parenthesis

Bringing it together with an integral

Now think about all of the tiny segments which make up the curve start color #bc2612, C, end color #bc2612, each with a tiny amount of mass exiting or entering through it per unit time. If you want to add up all those tiny changes in mass along the curve, what better tool is there than a line integral?

Specifical, the line integral will look like this:

$\begin{aligned} \oint_{\redE{C}} \blueE{\textbf{F}} \cdot \greenE{\hat{\textbf{n}}}\;\redE{ds} \end{aligned}$

where

The vector field start color #0c7f99, start bold text, F, end bold text, left parenthesis, x, comma, y, right parenthesis, end color #0c7f99 gives the fluid velocity at each point along the curve.
The term start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f should be considered a function, start color #0d923f, start bold text, n, end bold text, with, hat, on top, left parenthesis, x, comma, y, right parenthesis, end color #0d923f, which takes in a point on start color #bc2612, C, end color #bc2612 and outputs the unit normal vector to start color #bc2612, C, end color #bc2612 at that point.
Using the symbol \oint instead of integral is just to emphasize that the line integral is around a closed loop.
start color #bc2612, d, s, end color #bc2612 indicates a tiny change in arc length along the curve. Conceptually it is no different from the delta, s term in the previous section, but now we are considering it as an infinitesimal quantity used for integration.
As you walk along the curve start color #bc2612, C, end color #bc2612, the value start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f measures how much the fluid is leaving/entering the region enclosed by start color #bc2612, C, end color #bc2612 at each point. It is positive when the fluid is leaving, and negative when the fluid flows in, so the integral as a whole will return the total mass leaving the region enclosed by start color #bc2612, C, end color #bc2612 per unit time.

More elaborately, you might write:

$\begin{aligned} \underbrace{ -\dfrac{d(\text{fluid mass in region})}{dt} }_{\text{Rate at which mass leaves region}} = \oint_{\redE{C}} \underbrace{ \blueE{\textbf{F}} \cdot \greenE{\hat{\textbf{n}}}\;\redE{ds} }_{\substack{ \text{Mass leaving each }\\ \text{tiny piece of size $\redE{ds}$} }} \end{aligned}$

Definition: This flow rate through a curve is called flux. More specifically, I should say the component of the flow rate which is perpendicular to the curve is called flux.

Many things other than fluids in physics are considered to "flow", such as heat, and (loosely speaking) the electromagnetic field, and the word flux is used very broadly in reference to any one of these circumstances. It's most common for the word "flux" to refer to flow rate through a surface in three-dimensions. I will talk about this in the context of surface integrals later on. As such, you might call this flow rate through a curve "two-dimensional flux".

This integral \oint, start subscript, start color #bc2612, C, end color #bc2612, end subscript, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, start color #bc2612, d, s, end color #bc2612 is sometimes called a "flux integral", and as with all new operations, the best way to get a feel for it is to work through an example.

Example: Flux through a circle

Consider the vector field

$\begin{aligned} \textbf{F}(x, y) = \left[ \begin{array}{c} x^2 \\ y \end{array} \right] \end{aligned}$

And draw a circle of radius 3 centered at the origin.

From this picture, you might be able to guess that the fluid flowing along this vector field tends to go out of the circle. But how much? To apply the integral from the last section, we first need to do two things:

Parameterize the circle.
Find a function for start bold text, n, end bold text, with, hat, on top on the circle.

We parameterize the circle using our friendly neighborhood cosine-sine parameterization:

$\begin{aligned} \textbf{r}(t) = \left[ \begin{array}{c} 3\cos(t) \\ 3\sin(t) \end{array} \right] \quad \leftarrow \text{Draws a circle with radius $3$} \end{aligned}$

For the parameterization to traverse the circle once and only once, let t range from 0 to 2, pi.

Which of the following functions gives the unit normal vector?

(Choice A)
$\hat{\textbf{n}}(x, y) = \begin{aligned} \left[ \begin{array}{c} x \\ y \end{array} \right] \end{aligned}$
(Choice B)
$\hat{\textbf{n}}(x, y) = \begin{aligned} \left[ \begin{array}{c} -y \\ x \end{array} \right] \end{aligned}$
(Choice C)
$\hat{\textbf{n}}(x, y) = \begin{aligned} \left[ \begin{array}{c} x/3 \\ y/3 \end{array} \right] \end{aligned}$
(Choice D)
$\hat{\textbf{n}}(x, y) = \begin{aligned} \left[ \begin{array}{c} -y/3 \\ x/3 \end{array} \right] \end{aligned}$

[Odpowiedź]

We can now apply this data to the line integral we found in the last section. (If you are uncomfortable with the computation of the line integral, consider reviewing the article on line integrals in a scalar field).

$\begin{aligned} \oint_C \blueE{\textbf{F}} \cdot \greenE{\hat{\textbf{n}}} \; \redE{ds} &= \int_0^{2\pi} \blueE{\underbrace{ \textbf{F}(\textbf{r}(t)) \cdot }_{\substack{ \text{Velocity} \\ \text{at a point} }}} \greenE{\overbrace{ \hat{\textbf{n}}(\textbf{r}(t)) }^{\substack{ \text{Normal vector} \\ \text{at a point} }}} \; \redE{\underbrace{ ||\textbf{r}'(t)|| dt }_{ds}} \\ &= \int_0^{2\pi} \left( \blueE{\left[ \begin{array}{c} (3\cos(t))^2 \\ (3\sin(t)) \end{array} \right]} \cdot \greenE{\left[ \begin{array}{c} (3\cos(t))/3 \\ (3\sin(t))/3 \end{array} \right]} \right) \;\; \redE{\left| \left| \left[ \begin{array}{c} \frac{d}{dt} (3\cos(t)) \\ \frac{d}{dt} (3\sin(t)) \end{array} \right] \right| \right| dt} \\ &= \int_0^{2\pi} \left( \blueE{\left[ \begin{array}{c} 9\cos^2(t) \\ 3\sin(t) \end{array} \right]} \cdot \greenE{\left[ \begin{array}{c} \cos(t) \\ \sin(t) \end{array} \right]} \right) \;\; \redE{\left| \left| \left[ \begin{array}{c} -3\sin(t) \\ 3\cos(t) \end{array} \right] \right| \right| dt} \\ &= \int_0^{2\pi} \tealE{\left( 9\cos^3(t) + 3\sin^2(t) \right)} \redE{\sqrt{3^2 \sin^2(t) + 3^2 \cos^2(t)} dt} \\ &= \int_0^{2\pi} \tealE{\left( 9\cos^3(t) + 3\sin^2(t) \right)} \redE{3 \;\, \underbrace{ \sqrt{\sin^2(t) + \cos^2(t)} }_{=1} \; dt} \\ &= \int_0^{2\pi} \tealE{\left( 9\cos^3(t) + 3\sin^2(t) \right)} \redE{3 dt} \\ &= 9\int_0^{2\pi} \left(3\cos^3(t) + \sin^2(t) \right) dt \\ \end{aligned}$

Once you have an integral in this form, you can just plug it into a calculator (or something like Wolfram Alpha) to get the numerical result:

$\begin{aligned} 9\int_0^{2\pi} \left(3\cos^3(t) + \sin^2(t) \right) dt = \boxed{9\pi \approx 28{,}274} \end{aligned}$

Computing a unit normal vector

You might be wondering how to compute the unit normal vector start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, left parenthesis, x, comma, y, right parenthesis for an arbitrary curve that is given to you. It felt like kind of a special case for the circle, didn't it? Well, luckily for you, constructing such a unit normal vector just so happens to be the topic of the next article.

I actually kind of lied to you when I said the unit normal vector is given with a function start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, left parenthesis, x, comma, y, right parenthesis with inputs on the x, y-plane. In practice, it is typically given as a parametric function start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, left parenthesis, t, right parenthesis that "lines up" with the parameterization of the curve start color #bc2612, C, end color #bc2612, so to speak. In principle, you should still think of it as acting on points in the x, y-plane, it's just that in practice you only end up defining it for points on the curve start color #bc2612, C, end color #bc2612 itself, since that's all you need. Don't worry, you'll see what I mean in the next article.

Podsumowanie

In any context where something can be considered flowing, such as a fluid, two-dimensional flux is a measure of the flow rate through a curve. The flux over the boundary of a region can be used to measure whether whatever is flowing tends to go into or out of that region.

The flux through a curve start color #bc2612, C, end color #bc2612 can be measured with the line integral
$\begin{aligned} \int_\redE{C} \blueE{\textbf{F}} \cdot \greenE{\hat{\textbf{n}}} \, \redE{ds} \end{aligned}$
where
- start color #0c7f99, start bold text, F, end bold text, left parenthesis, x, comma, y, right parenthesis, end color #0c7f99 defines the vector field which indicates the flow rate.
- start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f is some function which return the outward unit normal vector at every point on the curve start color #bc2612, C, end color #bc2612.

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