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Analiza matematyczna funkcji wielu zmiennych

Kurs: Analiza matematyczna funkcji wielu zmiennych > Rozdział 4

Lekcja 13: Strumień pola w 3D (artykuły)
Matematyka
Analiza matematyczna funkcji wielu zmiennych
Całkowanie funkcji wielu zmiennych
Strumień pola w 3D (artykuły)
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Flux in 3D example

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

Kontekst

Flux in three dimensions
Unit normal vector
Surface integral

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 start color #0c7f99, F, left parenthesis, x, comma, y, comma, z, right parenthesis, end color #0c7f99, and if start color #bc2612, S, end color #bc2612 represents the surface itself, the flux is computed with the following surface integral:
∬SF⋅n^ dΣ\begin{aligned} \iint_\redE{S} \blueE{\textbf{F}} \cdot \greenE{\hat{\textbf{n}}}\, \redE{d\Sigma} \end{aligned}
The vector-valued function start color #0d923f, start bold text, n, end bold text, with, hat, on top, left parenthesis, x, comma, y, comma, z, right parenthesis, end color #0d923f gives the unit normal vector to start color #bc2612, S, end color #bc2612. 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 start color #bc2612, S, end color #bc2612, as you would for any surface integral.
  • Krok 2: Insert the expression for the unit normal vector start color #0d923f, start bold text, n, end bold text, with, hat, on top, left parenthesis, x, comma, y, comma, z, right parenthesis, end color #0d923f. 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, equals, 4, minus, x, squared, minus, y, squared
Let start color #bc2612, S, end color #bc2612 be the portion of this paraboloid which sits above the x, y-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 start bold text, i, end bold text, with, hat, on top, start bold text, j, end bold text, with, hat, on top and start bold text, k, end bold text, with, hat, on top 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
F(x,y,z)=[xyxzyz]\begin{aligned} \blueE{\textbf{F}}(x, y, z) = \left[ \begin{array}{c} xy \\ xz \\ yz \end{array} \right] \end{aligned}
Key question: What is the flux of this flowing fluid through the surface start color #bc2612, S, end color #bc2612?

Step 1: Rewrite the flux integral using a parameterization

Right now, the surface start color #bc2612, S, end color #bc2612 has been defined as a graph, subject to a constraint on z.
Graph: z, equals, 4, minus, x, squared, minus, y, squared
Constraint: z, is greater than or equal to, 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:
v(t,s)=[ts4−t2−s2]\begin{aligned} \textbf{v}(t, s) = \left[ \begin{array}{c} t \\ s \\ 4-t^2-s^2 \end{array} \right] \end{aligned}
After writing this function, you still need to specify what region of the t, s-plane corresponds with our surface start color #bc2612, S, end color #bc2612. This requires translating the constraint z, is greater than or equal to, 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 start bold text, v, end bold text, with, vector, on top, left parenthesis, t, comma, s, right parenthesis 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, start subscript, 2, end subscript.
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:
∬SF⋅n^ dΣ=∬D2⏟Double integral inflat parameter space ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣F(v(t,s))⋅n^(v(t,s))  ∣∂v∂t×∂v∂s∣ dA⏟dΣ\begin{aligned} &\quad \iint_\redE{S} \blueE{\textbf{F}} \cdot \greenE{\hat{\textbf{n}}} \,\redE{d\Sigma} \\\\ &= \underbrace{ \iint_{D_2} }_{\substack{ \text{Double integral in}\\ \text{flat parameter space} }} \!\!\!\!\!\!\! \blueE{\textbf{F}}(\textbf{v}(t, s)) \cdot \greenE{\hat{\textbf{n}}}(\textbf{v}(t, s)) \; \underbrace{ \left| \dfrac{\partial \textbf{v}}{\partial t} \times \dfrac{\partial \textbf{v}}{\partial s} \right| \,dA }_{\redE{d\Sigma}} \end{aligned}
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 start bold text, v, end bold text, left parenthesis, t, comma, s, right parenthesis. Basically, you normalize the cross product of the partial derivatives of start bold text, v, end bold text, left parenthesis, t, comma, s, right parenthesis (boy is that a mouthful to say):
∂v∂t×∂v∂s∣∂v∂t×∂v∂s∣\begin{aligned} \dfrac{ \dfrac{\partial \textbf{v}}{\partial t} \times \dfrac{\partial \textbf{v}}{\partial s} }{ \left| \dfrac{\partial \textbf{v}}{\partial t} \times \dfrac{\partial \textbf{v}}{\partial s} \right| } \end{aligned}
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:
∬D2F(v(t,s))⋅n^(v(t,s))∣∂v∂t×∂v∂s∣ dA=∬D2F(v(t,s))⋅(∂v∂t×∂v∂s∣∂v∂t×∂v∂s∣)∣∂v∂t×∂v∂s∣ dA=∬D2F(v(t,s))⋅(∂v∂t×∂v∂s∣∂v∂t×∂v∂s∣)∣∂v∂t×∂v∂s∣ dA=∬D2F(v(t,s))⋅(∂v∂t×∂v∂s) dA\begin{aligned} &\quad \iint_{D_2} \blueE{\textbf{F}}(\textbf{v}(t, s)) \cdot \greenE{\hat{\textbf{n}}}(\textbf{v}(t, s)) \left| \dfrac{\partial \textbf{v}}{\partial t} \times \dfrac{\partial \textbf{v}}{\partial s} \right| \,dA \\\\ &= \iint_{D_2} \blueE{\textbf{F}}(\textbf{v}(t, s)) \cdot \greenE{\left( \dfrac{ \dfrac{\partial \textbf{v}}{\partial t} \times \dfrac{\partial \textbf{v}}{\partial s} }{ \left| \dfrac{\partial \textbf{v}}{\partial t} \times \dfrac{\partial \textbf{v}}{\partial s} \right| } \right)} \left| \dfrac{\partial \textbf{v}}{\partial t} \times \dfrac{\partial \textbf{v}}{\partial s} \right| \,dA \\\\ &= \iint_{D_2} \blueE{\textbf{F}}(\textbf{v}(t, s)) \cdot \greenE{\left( \dfrac{ \dfrac{\partial \textbf{v}}{\partial t} \times \dfrac{\partial \textbf{v}}{\partial s} }{\cancel{ \left| \dfrac{\partial \textbf{v}}{\partial t} \times \dfrac{\partial \textbf{v}}{\partial s} \right| }} \right)} \cancel{ \left| \dfrac{\partial \textbf{v}}{\partial t} \times \dfrac{\partial \textbf{v}}{\partial s} \right| } \,dA \\\\ &= \iint_{D_2} \blueE{\textbf{F}}(\textbf{v}(t, s)) \cdot \left( \dfrac{\partial \textbf{v}}{\partial t} \times \dfrac{\partial \textbf{v}}{\partial s} \right) \,dA \\\\ \end{aligned}

Step 3: Expand the integrand

Let's start by working out the cross product term:
start fraction, \partial, start bold text, v, end bold text, divided by, \partial, t, end fraction, times, start fraction, \partial, start bold text, v, end bold text, divided by, \partial, s, end fraction
For reference, this was how I defined start bold text, v, end bold text, left parenthesis, t, comma, s, right parenthesis:
v(t,s)=[ts4−t2−s2]\begin{aligned} \textbf{v}(t, s) = \left[ \begin{array}{c} t \\ s \\ 4-t^2-s^2 \end{array} \right] \end{aligned}
Now compute each partial derivative, then find their cross product.
start fraction, \partial, start bold text, v, end bold text, divided by, \partial, t, end fraction, equals
start bold text, i, end bold text, with, hat, on top, plus
start bold text, j, end bold text, with, hat, on top, plus
start bold text, k, end bold text, with, hat, on top

start fraction, \partial, start bold text, v, end bold text, divided by, \partial, s, end fraction, equals
start bold text, i, end bold text, with, hat, on top, plus
start bold text, j, end bold text, with, hat, on top, plus
start bold text, k, end bold text, with, hat, on top

start fraction, \partial, start bold text, v, end bold text, divided by, \partial, t, end fraction, times, start fraction, \partial, start bold text, v, end bold text, divided by, \partial, s, end fraction, equals
start bold text, i, end bold text, with, hat, on top, plus
start bold text, j, end bold text, with, hat, on top, plus
start bold text, k, end bold text, with, hat, on top

Concept check: Does the expression for start fraction, \partial, start bold text, v, end bold text, divided by, \partial, t, end fraction, times, start fraction, \partial, start bold text, v, end bold text, divided by, \partial, s, end fraction that you just found give an outward-facing or inward-facing normal vector?
Wybierz 1 odpowiedź:

Next, write out the term start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, left parenthesis, start bold text, v, end bold text, left parenthesis, t, comma, s, right parenthesis, right parenthesis in terms of just t and s. For reference, this is how start color #0c7f99, start bold text, F, end bold text, end color #0c7f99 and start bold text, v, end bold text, with, vector, on top are defined:
F(x,y,z)=[xyxzyz]v(t,s)=[ts4−t2−s2]\begin{aligned} \blueE{\textbf{F}}(x, y, z) = \left[ \begin{array}{c} xy \\ xz \\ yz \end{array} \right] & & \textbf{v}(t, s) = \left[ \begin{array}{c} t \\ s \\ 4-t^2-s^2 \end{array} \right] \end{aligned}
start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, left parenthesis, start bold text, v, end bold text, left parenthesis, t, comma, s, right parenthesis, right parenthesis, equals
start bold text, i, end bold text, with, hat, on top, plus
start bold text, j, end bold text, with, hat, on top, plus
start bold text, k, end bold text, with, hat, on top

Great! Now we have all the pieces for the innards of our integral.
∬D2F(v(t,s))⋅(∂v∂t×∂v∂s) dA\begin{aligned} \iint_{D_2} \blueE{\textbf{F}}(\textbf{v}(t, s)) \cdot \left( \dfrac{\partial \textbf{v}}{\partial t} \times \dfrac{\partial \textbf{v}}{\partial s} \right) \,dA \\\\ \end{aligned}
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, start subscript, D, start subscript, 2, end subscript, end subscript
d, A

Step 4: Compute the integral

Up until this point, we have been writing a little D, start subscript, 2, end subscript under the double integral to indicate that the region of the t, s-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, start subscript, 2, end subscript, and imagine cutting it into vertical stripes:
The value t ranges from minus, 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 minus, square root of, 4, minus, t, squared, end square root to plus, square root of, 4, minus, t, squared, end square root. Applying these bounds to our double integral, here's what we get:
∫02∫−4−t2+4−t2s(2t2+(2t+1)(4−t2−s2)) ds dt\begin{aligned} \int_0^2 \int_{-\sqrt{4-t^2}}^{+\sqrt{4-t^2}} s\Big(2t^2+(2t+1)(4-t^2-s^2)\Big) \,ds \,dt \end{aligned}
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, cubed. This means the inner integral on the portion from minus, square root of, 4, minus, t, squared, end square root to 0 will cancel out with the portion from 0 to square root of, 4, minus, t, squared, end square root. Therefore, the integral as a whole is 0.

Podsumowanie

A flux integral starts its life looking something like this:
∬SF⋅n^ dΣ\begin{aligned} \iint_\redE{S} \blueE{\textbf{F}} \cdot \greenE{\hat{\textbf{n}}}\, \redE{d\Sigma} \end{aligned}
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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