Would you be able to use cylindrical coordinates in order to make it a nicer integral? And if so where/when would you switch coordinate systems?

I parameterized using r and angle theta where r ranged from 0 to 2 and theta ranged from 0 to 2pi setting x = rcos theta, y = r sin theta z= 4-r^2 and also got a flux of 0. Perhaps just coincidental? Would appreciate comments from anyone more knowledgeable than I.

Why did we have to parameterize this using s & t? wouldn't using x & y give us the same outcome?

Parameterizing means writing x,y, and z as functions of s and t. This means we will have the three functions x(s,t), y(s,t) and z(s,t). If we used x&y instead of s&t, we would have x(x,y) and y(x,y) which doesn't make sense. Of course we will usually be sloppy and just parameterize wrt x&y, but this doesn't make sense "formally", ie. you couldn't program a computer to understand this.

So the integral is 0, but what does that mean?Mass of whatever within stays constant?

Flux out from the surface is zero. If the vector field is really velocity of fluid (I think it is in this problem), then the mass of fluid going into the "region" is exactly equal to the mass of fluid going out. However, we cannot assume that the mass of fluid inside the region is unchanging yet. I mean, what region? The surface is not closed. There is a "hole" on the bottom of the parabolic surface. If you just seal the "hole" with a flat plane, then, no, you cannot be sure yet that mass inside is constant. There might be net fluid escaping or entering through that plane. To make sure, you would need to compute the flux of fluid through that region of plane. If it also equals zero, then yes, the mass of fluid inside is constant. But if the vector field doesn't represent a fluid's velocity, it might mean something else. It might be heat transfer, and then there is no net heat entering from the sides (we still don't know about the bottom). It might be an electric field, and then perhaps there is no net charge inside the parabola.

For the last step, would both s and t be from -2 to 2 since we are considering the parametric plane T mentioned in the surface integral where t and s form a square in t-s plane?

Yes this also works - i did this in mathematica and it gives the same answer. I think for this vector field, the flux is zero everywhere in this area.

Main content

Course: Multivariable calculus > Unit 4

Lesson 14: Flux in 3D (articles)

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.

Background

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

F (x, y, z)

, and if

S

represents the surface itself, the flux is computed with the following surface integral:

$\begin{array}{r} \iint_{S} F \cdot \hat{n} d Σ \end{array}$ ‍

The vector-valued function

\hat{n} (x, y, z)

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.

Step 1: Rewrite the integral in terms of a parameterization of $S$ ‍, as you would for any surface integral.
Step 2: Insert the expression for the unit normal vector $\hat{n} (x, y, z)$ ‍. 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.
Step 3: Simplify the terms inside the integral.
Step 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 $x y$ ‍-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

\hat{i}

\hat{j}

and

\hat{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

$\begin{array}{r} F (x, y, z) = [\begin{array}{c} x y \\ x z \\ y z \end{array}] \end{array}$ ‍

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: $z \geq 0$ ‍

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

$\begin{array}{r} v (t, s) = [\begin{array}{c} t \\ s \\ 4 - t^{2} - s^{2} \end{array}] \end{array}$ ‍

After writing this function, you still need to specify what region of the

t s

-plane corresponds with our surface

S

. This requires translating the constraint

z \geq 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

\vec{v} (t, s)

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:

$\begin{aligned} \iint_{S} F \cdot \hat{n} d Σ \\ = \underset{\begin{array}{c} Double integral in \\ flat parameter space \end{array}}{\underset{⏟}{\iint_{D_{2}}}} F (v (t, s)) \cdot \hat{n} (v (t, s)) \underset{d Σ}{\underset{⏟}{| \frac{\partial v}{\partial t} \times \frac{\partial v}{\partial s} | d A}} \end{aligned}$ ‍

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

v (t, s)

. Basically, you normalize the cross product of the partial derivatives of $v (t, s)$ ‍ (boy is that a mouthful to say):

$\begin{array}{r} \frac{\frac{\partial v}{\partial t} \times \frac{\partial v}{\partial s}}{| \frac{\partial v}{\partial t} \times \frac{\partial v}{\partial s} |} \end{array}$ ‍

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:

$\begin{aligned} \iint_{D_{2}} F (v (t, s)) \cdot \hat{n} (v (t, s)) | \frac{\partial v}{\partial t} \times \frac{\partial v}{\partial s} | d A \\ = \iint_{D_{2}} F (v (t, s)) \cdot (\frac{\frac{\partial v}{\partial t} \times \frac{\partial v}{\partial s}}{| \frac{\partial v}{\partial t} \times \frac{\partial v}{\partial s} |}) | \frac{\partial v}{\partial t} \times \frac{\partial v}{\partial s} | d A \\ = \iint_{D_{2}} F (v (t, s)) \cdot (\frac{\frac{\partial v}{\partial t} \times \frac{\partial v}{\partial s}}{| \frac{\partial v}{\partial t} \times \frac{\partial v}{\partial s} |}) | \frac{\partial v}{\partial t} \times \frac{\partial v}{\partial s} | d A \\ = \iint_{D_{2}} F (v (t, s)) \cdot (\frac{\partial v}{\partial t} \times \frac{\partial v}{\partial s}) d A \end{aligned}$ ‍

This is a bit detailed, but it's a very satisfying intuition once you wrap your mind around it.

Remember the intuition behind a flux integral: We divide the surface

S

into many tiny pieces, and look at how much fluid flows through each one of them per unit time:

Khan Academy video wrapper

See video transcript

Typically, we think of one of these tiny pieces of area as being a parallelogram, spanned by the two infinitesimal vectors

(\frac{\partial v}{\partial t} d t)

and

(\frac{\partial v}{\partial s} d s)

Because this piece is so small, we can assume that all the fluid particles passing through it move with basically the same velocity vector,

F (\vec{v} (t, s))

for some point

(t, s)

which lands you inside the piece. This means the volume formed by fluid passing through this piece forms a parallelepiped (my favorite word!), which is the three-dimensional analog of a parallelogram.

Let's assume the parallelepiped represents the fluid passing through the parallelogram over one unit of time. Then to find its volume, you want to multiply the area of the parallelogram by the component of the vector $F (\vec{v} (t, s))$ ‍ which is perpendicular to that parallelogram. But think about what this cross product represents:

$(\frac{\partial v}{\partial t} d t) \times (\frac{\partial v}{\partial s} d s)$ ‍

It gives a vector perpendicular to the parallelogram, whose magnitude is the area of that parallelogram. So what is the meaning of this dot product:

$F (\vec{v} (t, s)) \cdot (\frac{\partial v}{\partial t} d t \times \frac{\partial v}{\partial s} d s)$ ‍

It will give the component of

F (\vec{v} (t, s))

that is perpendicular to the parallelogram, multiplied by the area of that parallelogram. But this is exactly the volume we want!

When you pull out the

d t

and

d s

terms, whose product is

d A

, you get the integral that we found above after canceling out the norm of the cross product.

$\begin{array}{r} \iint_{D_{2}} F (v (t, s)) \cdot (\frac{\partial v}{\partial t} \times \frac{\partial v}{\partial s}) d A \end{array}$ ‍

Step 3: Expand the integrand

Let's start by working out the cross product term:

$\frac{\partial v}{\partial t} \times \frac{\partial v}{\partial s}$ ‍

For reference, this was how I defined

v (t, s)

$\begin{array}{r} v (t, s) = [\begin{array}{c} t \\ s \\ 4 - t^{2} - s^{2} \end{array}] \end{array}$ ‍

Now compute each partial derivative, then find their cross product.

$\frac{\partial v}{\partial t} =$ ‍
$\hat{i} +$ ‍
$\hat{j} +$ ‍
$\hat{k}$ ‍

$\frac{\partial v}{\partial s} =$ ‍
$\hat{i} +$ ‍
$\hat{j} +$ ‍
$\hat{k}$ ‍

$\frac{\partial v}{\partial t} \times \frac{\partial v}{\partial s} =$ ‍
$\hat{i} +$ ‍
$\hat{j} +$ ‍
$\hat{k}$ ‍

To take the cross product, use the usual determinant trick, where the $2^{nd}$ ‍ and $3^{rd}$ ‍ rows are the answers to the previous two questions:
$\begin{array}{r} det ([\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & - 2 t \\ 0 & 1 & - 2 s \end{array}]) \end{array}$ ‍
For the $\hat{i}$ ‍ component, cross out the top row and left column, then take the determinant:
$\begin{array}{r} det ([\begin{array}{cc} 0 & - 2 t \\ 1 & - 2 s \end{array}]) = (0) (- 2 s) - (- 2 t) (1) = 2 t \end{array}$ ‍
For the $\hat{j}$ ‍ component, cross out the top row and middle column, then take the negative determinant:
$\begin{array}{r} - det ([\begin{array}{cc} 1 & - 2 t \\ 0 & - 2 s \end{array}]) = - ((- 2 s) (1) - (- 2 t) (0)) = 2 s \end{array}$ ‍
Finally, for the $\hat{k}$ ‍ component, cross out the top row and last column, then take the determinant:
$\begin{array}{r} det ([\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]) = 1 \end{array}$ ‍

Concept check: Does the expression for

\frac{\partial v}{\partial t} \times \frac{\partial v}{\partial s}

that you just found give an outward-facing or inward-facing normal vector?

Choose 1 answer:
(Choice A)
Outward-facing
(Choice B)
Inward-facing

The best way to find out is to just plug in a specific input. The easiest input is probably $(t, s) = (0, 0)$ ‍, which corresponds to the very top of our paraboloid surface:
$\begin{array}{r} [\begin{array}{c} 2 t \\ 2 s \\ 1 \end{array}] \to [\begin{array}{c} 2 (0) \\ 2 (0) \\ 1 \end{array}] \to [\begin{array}{c} 0 \\ 0 \\ 1 \end{array}] \end{array}$ ‍
This is a vector pointing straight up, which is outside the region enclosed by the paraboloid. Therefore, it is an outward facing vector. If it was not, we would need to flip the sign of our surface integral.

Next, write out the term

F (v (t, s))

in terms of just

t

and

s

. For reference, this is how

F

and

\vec{v}

are defined:

$\begin{aligned} F (x, y, z) = [\begin{array}{c} x y \\ x z \\ y z \end{array}] & v (t, s) = [\begin{array}{c} t \\ s \\ 4 - t^{2} - s^{2} \end{array}] \end{aligned}$ ‍

$F (v (t, s)) =$ ‍
$\hat{i} +$ ‍
$\hat{j} +$ ‍
$\hat{k}$ ‍

Great! Now we have all the pieces for the innards of our integral.

$\begin{array}{r} \iint_{D_{2}} F (v (t, s)) \cdot (\frac{\partial v}{\partial t} \times \frac{\partial v}{\partial s}) d A \end{array}$ ‍

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.

$\iint_{D_{2}}$ ‍
$d A$ ‍

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

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_{2}

, and imagine cutting it into vertical stripes:

The value

t

ranges from

- 2

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}}

+ \sqrt{4 - t^{2}}

. Applying these bounds to our double integral, here's what we get:

$\begin{array}{r} \int_{0}^{2} \int_{- \sqrt{4 - t^{2}}}^{+ \sqrt{4 - t^{2}}} s (2 t^{2} + (2 t + 1) (4 - t^{2} - s^{2})) d s d t \end{array}$ ‍

From here, there are a few ways you might finish the problem

Painfully: Compute this double integral in full by hand (ugh!).
Pragmatically: Plug this into a calculator, or a computer algebra tool like Wolfram Alpha.
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 $0$ ‍.

Summary

A flux integral starts its life looking something like this:

$\begin{array}{r} \iint_{S} F \cdot \hat{n} d Σ \end{array}$ ‍

Solving this involves the following four steps:

Step 1: Parameterize the surface, and translate this surface integral to a double integral over the parameter space.
Step 2: Apply the formula for a unit normal vector.
Step 3: Simplify the integrand, which involves two vector-valued partial derivatives, a cross product, and a dot product.
Step 4: Compute the double integral (in practice a computer can handle it from here).

Want to join the conversation?

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Xingxing Yang
Posted 7 years ago. Direct link to Xingxing Yang's post “For the last formula, sho...”
For the last formula, shouldn't t range from -2 to 2?
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(33 votes)
Answer
- bean
  Posted 6 months ago. Direct link to bean's post “I'm late, but yeah I beli...”
  I'm late, but yeah I believe so.
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  (1 vote)
James Taylor
Posted 8 years ago. Direct link to James Taylor's post “So the integral is 0, but...”
So the integral is 0, but what does that mean?Mass of whatever within stays constant?
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(2 votes)
Answer
- Alexander Wu
  Posted 8 years ago. Direct link to Alexander Wu's post “Flux out from the surface...”
  Flux out from the surface is zero. If the vector field is really velocity of fluid (I think it is in this problem), then the mass of fluid going into the "region" is exactly equal to the mass of fluid going out. However, we cannot assume that the mass of fluid inside the region is unchanging yet. I mean, what region? The surface is not closed. There is a "hole" on the bottom of the parabolic surface. If you just seal the "hole" with a flat plane, then, no, you cannot be sure yet that mass inside is constant. There might be net fluid escaping or entering through that plane. To make sure, you would need to compute the flux of fluid through that region of plane. If it also equals zero, then yes, the mass of fluid inside is constant.
  
  But if the vector field doesn't represent a fluid's velocity, it might mean something else. It might be heat transfer, and then there is no net heat entering from the sides (we still don't know about the bottom). It might be an electric field, and then perhaps there is no net charge inside the parabola.
  Button navigates to signup page
  (16 votes)
Jenna Varcak
Posted 7 years ago. Direct link to Jenna Varcak's post “Would you be able to use ...”
Would you be able to use cylindrical coordinates in order to make it a nicer integral? And if so where/when would you switch coordinate systems?
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(4 votes)
Answer
- luminasea
  Posted 7 years ago. Direct link to luminasea's post “I parameterized using r a...”
  I parameterized using r and angle theta where r ranged from 0 to 2 and theta ranged from 0 to 2pi setting x = rcos theta, y = r sin theta z= 4-r^2 and also got a flux of 0. Perhaps just coincidental? Would appreciate comments from anyone more knowledgeable than I.
  Comment on luminasea's post “I parameterized using r a...”
  (5 votes)
kennygoldman
Posted 7 years ago. Direct link to kennygoldman's post “Why did we have to parame...”
Why did we have to parameterize this using s & t? wouldn't using x & y give us the same outcome?
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(3 votes)
Answer
- Samuel Laferriere
  Posted 7 years ago. Direct link to Samuel Laferriere's post “Parameterizing means writ...”
  Parameterizing means writing x,y, and z as functions of s and t.
  This means we will have the three functions x(s,t), y(s,t) and z(s,t).
  If we used x&y instead of s&t, we would have x(x,y) and y(x,y) which doesn't make sense.
  Of course we will usually be sloppy and just parameterize wrt x&y, but this doesn't make sense "formally", ie. you couldn't program a computer to understand this.
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  (7 votes)
Alexander Wu
Posted 8 years ago. Direct link to Alexander Wu's post “Is there some better way ...”
Is there some better way to find out where the normal vector points? Other than by plugging in some value. Sometimes we might not know what the surface looks like so we cannot just observe where the vector points and say, "Oh, that's outward." Also, we might be solving abstractly. So maybe v(t,s) = [t, s, a - b t^2 - b t^2], and we want to see how flux changes as we vary a and b. In this case, as a and b change, the directions of everything change, so a plug-in at one value doesn't work for a plug-in for another. We need a systematic procedure to find which way the vector really points (and maybe the expression we get using this procedure contains a and b, so we can generalize without having to get an absolute answer).

As for this problem, I did find a way around plugging in values, with intuition. As we see, t is equivalent to x, and s to y. We used ∂v/∂t × ∂v/∂s, so t cross s is x cross y which is in the positive z direction. But positive z direction is out of the parabola. So we are pointing out.

Of course, this intuitive approach is quite weak, with no "proof." And it probably doesn't work for many other problems. But obviously there must be one that does. What is it?
Button navigates to signup pageComment on Alexander Wu's post “Is there some better way ...”
(5 votes)
Answer
Ruimin Zhang
Posted 5 years ago. Direct link to Ruimin Zhang's post “For the last step, would ...”
For the last step, would both s and t be from -2 to 2 since we are considering the parametric plane T mentioned in the surface integral where t and s form a square in t-s plane?
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(1 vote)
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- yunki.yau
  Posted 3 years ago. Direct link to yunki.yau's post “Yes this also works - i d...”
  Yes this also works - i did this in mathematica and it gives the same answer. I think for this vector field, the flux is zero everywhere in this area.
  Button navigates to signup page
  (2 votes)
festavarian2
Posted 2 years ago. Direct link to festavarian2's post “Why does the S-bound depe...”
Why does the S-bound depend on T? Why can't the S-bounds be from -2 to +2?
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(3 votes)
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- VideoGameAddict
  Posted a year ago. Direct link to VideoGameAddict's post “You are correct. This is ...”
  You are correct. This is a mistake in the article.
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  (0 votes)
ricoruotongjia
Posted 6 years ago. Direct link to ricoruotongjia's post “Anyone tried polar coordi...”
Anyone tried polar coordinates? I set x = r*cos(t), y - r* sin(t), z = z= 4-r^2. And the normal vector I got was <2r^2*cos(t), 2r^2*sin(t), r>. At the point (0, 0, 4), the normal vector is (0, 0, 0). By comparison, the normal vector above, which is < 2s, 2t, 1>, at the point (0, 0, 4), the normal vector is (0, 0, 1).

I know there is a factor r and when we normalize it, it will be gone. But I still can't explain why the normal vectors are different.

Thank you for your explanation!
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(1 vote)
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festavarian2
Posted 2 years ago. Direct link to festavarian2's post “Why are we going through ...”
Why are we going through this whole UNIT normal business, when all we need for the flux equation is the AREA vector which IS a normal vector and is simply the the cross products of the partial derivatives times dA?
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(1 vote)
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