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## Analiza matematyczna funkcji wielu zmiennych

### Kurs: Analiza matematyczna funkcji wielu zmiennych>Rozdział 4

Lekcja 13: Strumień pola w 3D (artykuły)

# Flux in 3D example

After learning about what flux in three dimensions is, here you have the chance to practice with an example.

## The steps

In the last article, I talked about how the flux of a flowing fluid through a surface is a measure of how much fluid passes through that surface per unit of time. If that fluid flow is represented with a vector field $F\left(x,y,z\right)$, and if $S$ represents the surface itself, the flux is computed with the following surface integral:
$\begin{array}{r}{\iint }_{S}\mathbf{\text{F}}\cdot \stackrel{^}{\mathbf{\text{n}}}\phantom{\rule{0.167em}{0ex}}d\mathrm{\Sigma }\end{array}$
The vector-valued function $\stackrel{^}{\mathbf{\text{n}}}\left(x,y,z\right)$ gives the unit normal vector to $S$. For closed surfaces, you typically choose an outward facing unit normal vector.
In practice, there is quite a lot that goes into solving this integral.
• Krok 1: Rewrite the integral in terms of a parameterization of $S$, as you would for any surface integral.
• Krok 2: Insert the expression for the unit normal vector $\stackrel{^}{\mathbf{\text{n}}}\left(x,y,z\right)$. It's best to do this before you actually compute the unit normal vector since part of it cancels out with a term from the surface integral.
• Krok 3: Simplify the terms inside the integral.
• Krok 4: Compute the double integral.

## The problem

Consider the paraboloid graph defined by the following equation:
$z=4-{x}^{2}-{y}^{2}$
Let $S$ be the portion of this paraboloid which sits above the $xy$-plane:
Whoa, that ended up looking way more like a nuclear warhead than I intended. Ah well, at least it makes clear what surface we're talking about.
For flux integrals, we must specify the orientation of this surface. Let's orient it with outward-facing normal vectors, in the sense that the vectors $\stackrel{^}{\mathbf{\text{i}}}$, $\stackrel{^}{\mathbf{\text{j}}}$ and $\stackrel{^}{\mathbf{\text{k}}}$ are all outward facing from the region under the paraboloid.
Now imagine there is a fluid flowing around in three-dimensional space. Let's say it flows along the vector field defined by the function
$\begin{array}{r}\mathbf{\text{F}}\left(x,y,z\right)=\left[\begin{array}{c}xy\\ xz\\ yz\end{array}\right]\end{array}$
Key question: What is the flux of this flowing fluid through the surface $S$?

## Step 1: Rewrite the flux integral using a parameterization

Right now, the surface $S$ has been defined as a graph, subject to a constraint on $z$.
Graph: $z=4-{x}^{2}-{y}^{2}$
Constraint: $z\ge 0$
But for computing surface integrals, we need to describe this surface parametrically. Luckily, this conversion is not to hard. You basically let one parameter play the role of $x$, while the other parameter plays the role of $y$:
$\begin{array}{r}\mathbf{\text{v}}\left(t,s\right)=\left[\begin{array}{c}t\\ s\\ 4-{t}^{2}-{s}^{2}\end{array}\right]\end{array}$
After writing this function, you still need to specify what region of the $ts$-plane corresponds with our surface $S$. This requires translating the constraint $z\ge 0$ into a constraint on $t$ and $s$.
Concept check: What is the constraint on $t$ and $s$ which will ensure that the $z$-component of $\stackrel{\to }{\mathbf{\text{v}}}\left(t,s\right)$ is greater than or equal to $0$? Write your answer as an inequality.
Since this region is a disk with radius $2$, let's name it ${D}_{2}$.
Later down the road, we'll expand this fully as a set of bounds for $t$ and $s$, but while we work on all the innards of the integral it helps to just have a symbolic representation.
Writing our flux surface integral as a double integral in the parameter space, here's what we get:
$\begin{array}{rl}& \phantom{\rule{1em}{0ex}}{\iint }_{S}\mathbf{\text{F}}\cdot \stackrel{^}{\mathbf{\text{n}}}\phantom{\rule{0.167em}{0ex}}d\mathrm{\Sigma }\\ \\ & =\underset{\begin{array}{c}\text{Double integral in}\\ \text{flat parameter space}\end{array}}{\underset{⏟}{{\iint }_{{D}_{2}}}}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\mathbf{\text{F}}\left(\mathbf{\text{v}}\left(t,s\right)\right)\cdot \stackrel{^}{\mathbf{\text{n}}}\left(\mathbf{\text{v}}\left(t,s\right)\right)\phantom{\rule{0.278em}{0ex}}\underset{d\mathrm{\Sigma }}{\underset{⏟}{|\frac{\partial \mathbf{\text{v}}}{\partial t}×\frac{\partial \mathbf{\text{v}}}{\partial s}|\phantom{\rule{0.167em}{0ex}}dA}}\end{array}$
If this transition to a double integral in the parameter space seems unfamiliar, consider reviewing the article on surface integrals, or the one on surface area.

## Step 2: Insert the expression for a unit normal vector

In a previous article, I talked about how you can find a function which gives the unit normal vector to a surface based on its parameterization $\mathbf{\text{v}}\left(t,s\right)$. Basically, you normalize the cross product of the partial derivatives of $\mathbf{\text{v}}\left(t,s\right)$ (boy is that a mouthful to say):
$\begin{array}{r}\frac{\frac{\partial \mathbf{\text{v}}}{\partial t}×\frac{\partial \mathbf{\text{v}}}{\partial s}}{|\frac{\partial \mathbf{\text{v}}}{\partial t}×\frac{\partial \mathbf{\text{v}}}{\partial s}|}\end{array}$
Now, for those of you who just love computing the magnitude of cross products of partial derivative vectors, hold off for a moment. When we insert this into the flux integral, that magnitude term cancels out:
$\begin{array}{rl}& \phantom{\rule{1em}{0ex}}{\iint }_{{D}_{2}}\mathbf{\text{F}}\left(\mathbf{\text{v}}\left(t,s\right)\right)\cdot \stackrel{^}{\mathbf{\text{n}}}\left(\mathbf{\text{v}}\left(t,s\right)\right)|\frac{\partial \mathbf{\text{v}}}{\partial t}×\frac{\partial \mathbf{\text{v}}}{\partial s}|\phantom{\rule{0.167em}{0ex}}dA\\ \\ & ={\iint }_{{D}_{2}}\mathbf{\text{F}}\left(\mathbf{\text{v}}\left(t,s\right)\right)\cdot \left(\frac{\frac{\partial \mathbf{\text{v}}}{\partial t}×\frac{\partial \mathbf{\text{v}}}{\partial s}}{|\frac{\partial \mathbf{\text{v}}}{\partial t}×\frac{\partial \mathbf{\text{v}}}{\partial s}|}\right)|\frac{\partial \mathbf{\text{v}}}{\partial t}×\frac{\partial \mathbf{\text{v}}}{\partial s}|\phantom{\rule{0.167em}{0ex}}dA\\ \\ & ={\iint }_{{D}_{2}}\mathbf{\text{F}}\left(\mathbf{\text{v}}\left(t,s\right)\right)\cdot \left(\frac{\frac{\partial \mathbf{\text{v}}}{\partial t}×\frac{\partial \mathbf{\text{v}}}{\partial s}}{\overline{)|\frac{\partial \mathbf{\text{v}}}{\partial t}×\frac{\partial \mathbf{\text{v}}}{\partial s}|}}\right)\overline{)|\frac{\partial \mathbf{\text{v}}}{\partial t}×\frac{\partial \mathbf{\text{v}}}{\partial s}|}\phantom{\rule{0.167em}{0ex}}dA\\ \\ & ={\iint }_{{D}_{2}}\mathbf{\text{F}}\left(\mathbf{\text{v}}\left(t,s\right)\right)\cdot \left(\frac{\partial \mathbf{\text{v}}}{\partial t}×\frac{\partial \mathbf{\text{v}}}{\partial s}\right)\phantom{\rule{0.167em}{0ex}}dA\\ \end{array}$

## Step 3: Expand the integrand

Let's start by working out the cross product term:
$\frac{\partial \mathbf{\text{v}}}{\partial t}×\frac{\partial \mathbf{\text{v}}}{\partial s}$
For reference, this was how I defined $\mathbf{\text{v}}\left(t,s\right)$:
$\begin{array}{r}\mathbf{\text{v}}\left(t,s\right)=\left[\begin{array}{c}t\\ s\\ 4-{t}^{2}-{s}^{2}\end{array}\right]\end{array}$
Now compute each partial derivative, then find their cross product.
$\frac{\partial \mathbf{\text{v}}}{\partial t}=$
$\stackrel{^}{\mathbf{\text{i}}}+$
$\stackrel{^}{\mathbf{\text{j}}}+$
$\stackrel{^}{\mathbf{\text{k}}}$

$\frac{\partial \mathbf{\text{v}}}{\partial s}=$
$\stackrel{^}{\mathbf{\text{i}}}+$
$\stackrel{^}{\mathbf{\text{j}}}+$
$\stackrel{^}{\mathbf{\text{k}}}$

$\frac{\partial \mathbf{\text{v}}}{\partial t}×\frac{\partial \mathbf{\text{v}}}{\partial s}=$
$\stackrel{^}{\mathbf{\text{i}}}+$
$\stackrel{^}{\mathbf{\text{j}}}+$
$\stackrel{^}{\mathbf{\text{k}}}$

Concept check: Does the expression for $\frac{\partial \mathbf{\text{v}}}{\partial t}×\frac{\partial \mathbf{\text{v}}}{\partial s}$ that you just found give an outward-facing or inward-facing normal vector?
Wybierz 1 odpowiedź:

Next, write out the term $\mathbf{\text{F}}\left(\mathbf{\text{v}}\left(t,s\right)\right)$ in terms of just $t$ and $s$. For reference, this is how $\mathbf{\text{F}}$ and $\stackrel{\to }{\mathbf{\text{v}}}$ are defined:
$\begin{array}{rlr}\mathbf{\text{F}}\left(x,y,z\right)=\left[\begin{array}{c}xy\\ xz\\ yz\end{array}\right]& & \mathbf{\text{v}}\left(t,s\right)=\left[\begin{array}{c}t\\ s\\ 4-{t}^{2}-{s}^{2}\end{array}\right]\end{array}$
$\mathbf{\text{F}}\left(\mathbf{\text{v}}\left(t,s\right)\right)=$
$\stackrel{^}{\mathbf{\text{i}}}+$
$\stackrel{^}{\mathbf{\text{j}}}+$
$\stackrel{^}{\mathbf{\text{k}}}$

Great! Now we have all the pieces for the innards of our integral.
$\begin{array}{r}{\iint }_{{D}_{2}}\mathbf{\text{F}}\left(\mathbf{\text{v}}\left(t,s\right)\right)\cdot \left(\frac{\partial \mathbf{\text{v}}}{\partial t}×\frac{\partial \mathbf{\text{v}}}{\partial s}\right)\phantom{\rule{0.167em}{0ex}}dA\\ \end{array}$
By taking the dot product of the previous two answers, write the inside of this integral purely in terms of the parameters $t$ and $s$. It will help the integral computation in the next section if you simplify your answer as much as possible.
${\iint }_{{D}_{2}}$
$dA$

## Step 4: Compute the integral

Up until this point, we have been writing a little ${D}_{2}$ under the double integral to indicate that the region of the $ts$-plane we will be integrating over is the disk with radius $2$. Now, as we turn to computing the integral itself, we need to spell this out into concrete bounds on the parameters $t$ and $s$.
To do this, draw yourself a picture of ${D}_{2}$, and imagine cutting it into vertical stripes:
The value $t$ ranges from $-2$ to $2$. The range for $s$ depends on the value of $t$, which you can find using the pythagorean theorem.
From the diagram, you can see that $s$ ranges from $-\sqrt{4-{t}^{2}}$ to $+\sqrt{4-{t}^{2}}$. Applying these bounds to our double integral, here's what we get:
$\begin{array}{r}{\int }_{0}^{2}{\int }_{-\sqrt{4-{t}^{2}}}^{+\sqrt{4-{t}^{2}}}s\left(2{t}^{2}+\left(2t+1\right)\left(4-{t}^{2}-{s}^{2}\right)\right)\phantom{\rule{0.167em}{0ex}}ds\phantom{\rule{0.167em}{0ex}}dt\end{array}$
From here, there are a few ways you might finish the problem
1. Painfully: Compute this double integral in full by hand (ugh!).
2. Pragmatically: Plug this into a calculator, or a computer algebra tool like Wolfram Alpha.
3. Cleverly: You can recognize that the integrand is an odd function with respect to $s$. Distribute the $s$, and notice that all terms either have an $s$ or an ${s}^{3}$. This means the inner integral on the portion from $-\sqrt{4-{t}^{2}}$ to $0$ will cancel out with the portion from $0$ to $\sqrt{4-{t}^{2}}$. Therefore, the integral as a whole is $0$.

## Podsumowanie

A flux integral starts its life looking something like this:
$\begin{array}{r}{\iint }_{S}\mathbf{\text{F}}\cdot \stackrel{^}{\mathbf{\text{n}}}\phantom{\rule{0.167em}{0ex}}d\mathrm{\Sigma }\end{array}$
Solving this involves the following four steps:
• Krok 1: Parameterize the surface, and translate this surface integral to a double integral over the parameter space.
• Krok 2: Apply the formula for a unit normal vector.
• Krok 3: Simplify the integrand, which involves two vector-valued partial derivatives, a cross product, and a dot product.
• Krok 4: Compute the double integral (in practice a computer can handle it from here).

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