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We are going to continue to look at stuff in space.
We have been working with triple integrals and seeing how
to set them up in all sorts of coordinate systems.
And the next topic we will be looking at are vector fields in
space. And so, in particular,
we will be learning about flux and work.
So, just for a change, we will be starting with flux
first. And we will do work,
actually, after Thanksgiving. Just to remind you,
a vector field in space is just the same thing as in the plane.
At every point you have a vector, and the components of
this vector depend on the coordinates x,
y and z. Let's say the components might
be P, Q, R, or your favorite three letters,
where each of these things is a function of coordinates x,
y, z. You have seen that in the plane
it is already pretty hard to draw a vector field.
Usually, in space, we won't really try too hard.
But it is still useful to try to have a general idea for what
the vectors in there are doing, whether they are all going in
the same direction, whether they may be all
vertical or horizontal, pointing away from the origin,
towards it, things like that.
But, generally-speaking, we won't really bother with
trying to draw a picture because that is going to be quite hard.
Just to give you examples, well, the same kinds of
examples as the plane, you can think of force fields.
For example, the gravitational attraction --
-- of a solid mass, let's call this mass big M,
at the origin on a mass M at point x, y, z.
That would be given by a vector field that points toward the
origin and whose magnitude is inversely proportional to the
square of a distance from the origin.
Such a field would be directed towards the origin and its
magnitude would be of the order of a constant over pho squared
where pho is the distance from the origin.
The picture, if I really wanted to draw a
picture, would be everywhere it is a field that points towards
the origin. And if I am further away then
it gets smaller. And, of course,
I am not going to try to draw all these vectors in there.
If I wanted to give a formula for that -- A formula for that
might be something of a form minus c times x,
y, z over pho cubed. Let's see.
Well, the direction of this vector, this vector is
proportional to negative x, y, z.
is the vector that goes from the origin to your point.
The negative goes towards the origin.
Then the magnitude of this guy, well, the magnitude of x,
y, z is just the distance from the origin rho.
So the magnitude of this thing is one over rho cubed times some
constant factor. That would be an example of a
vector field that comes up in physics.
Well, other examples would be electric fields.
Actually, if you look at the electric field generated by a
charged particle at the origin, it is given by exactly the same
kind of formula, and there are magnetic fields
and so on. Another example comes from
velocity fields. If you have a fluid flow,
for example, if you want to study wind
patterns in the atmosphere. Well, wind, most of the time,
is kind of horizontal, but maybe it depends on the
altitude. At high altitude you have jet
streams, and the wind velocity is not the same at all
altitudes. And, just to give you more
examples, in math we have seen that the gradient of a function
of three variables gives you a vector field.
If you have a function u of x, y, z then its gradient field
has just components, u sub x, u sub y and u sub z.
And, of course, the cases are not mutually
exclusive. For example,
the electric field or gravitational field is given by
the gradient of the gravitational or electric
potential. So, these are not like
different cases. There is overlap.
Anyway, hopefully, you are kind of convinced that
you should learn about vector fields.
What are we going to do with them?
Well, let's start with flux. Remember not so long ago we
looked at flux of a two-dimensional field of a
curve. We had a curve in the plane and
we had a vector field. And we looked at the component
of a vector field in the direction that was normal to the
curve. We formed the flux integral
that was a line integral F dot n ds.
And that measured how much the vector field was going across
the curve. If you were thinking of a
velocity field, that would measure how much
fluid is passing through the curve in unit time.
Now let's say that we were in space.
Well, we cannot really think of flux as a line integral.
Because, if you have a curve in space and say that you have wind
or something like that, you cannot really ask how much
air is flowing through the curve.
See, to have a flow through something you need a surface.
If you have a net maybe then you can ask how much stuff is
passing through that surface. There is going to be a big
difference here. In the three-dimensional space,
flux will be measured through a surface.
And so it will be a surface integral, not a line integral
anymore. That means we will be
integrating, we will be summing over all the pieces of a surface
in space. Because a surface is a
two-dimensional object, that will end up being a double
integral. But, of course,
we will have to set it up properly because the surface
that is in space, and we will probably have x,
y and z to deal with at the same time,
and we will have to somehow get rid of one variable so that we
can set up and evaluate a double integral.
So conceptually it is very similar to line integrals.
In the line integral in the plane, you had two variables
that you reduced to one by figuring out what the curve was.
Here you have three variables that you will reduce to two by
figuring out what the surface is.
Let me give you a definition of flux in 3D.
Let's say that we have a vector field and s, a surface in space.
Let me draw some kind of a picture.
I have my surface and I have my vector field F.
Well, at every point it changes with a point.
Well, I want to figure out how much my vector field is going
across that surface. That means I want to figure out
the normal component of my vector field,
so I will use, as in the plane case,
the unit normal vector to s. I take my point on the surface
and build a unit vector that is standing on it perpendicularly.
Now, we have to decide which way it is standing.
We can build our normal vector to go this way or to go the
other way around. There are two choices.
Basically, whenever you want to set up a flux integral you have
to choose one side of the surface.
And you will count positively what flows toward that side and
negatively what flows towards the other side.
There are two choices for n. We need to choose a side of the
surface. In the case of curves,
we made that choice by deciding that because we were going along
some direction on the curve we could choose one side by saying
let's rotate clockwise from the tangent vector.
And, in a way, what we were doing was really
it was a recipe to choose for us one of the two sides.
Here we don't have a notion of orienting the surface other than
by precisely choosing one of the two possible normal vectors.
So, in fact, this is called choosing an
orientation of a surface. When you are saying you are
orienting the surface that really means you are deciding
which side is which. Let's call that orientation.
Now, there is no set convention that will work forever.
But the usually traditional settings would be to take your
normal vector pointing maybe out of the solid region because then
you will be looking at flux that is coming out of that region of
space. Or, if you have a surface that
is not like closed or anything but maybe you will want the flux
going up through the region. Or, there are various
conventions. Concretely, on problem sets it
will either say which choice you have to make or you get to
choose which one you want to make.
And, of course, if you choose the other one
then the sign becomes the opposite.
Now, once we have made a choice then we can define the flux
integral. It will just be the double
integral over a surface of F dot n dS.
Now I am using a big dS. That stands for the surface
area element on this surface. I am using dS rather than dA
because I still want to think of dA as maybe the area in one of
the coordinate planes like the one we had in double integrals.
You will see later where this comes in.
But conceptually it is very similar.
Concretely what this means is I cut my surface into little
pieces. Each of them has area delta S.
And, for each piece, I take my vector field,
I take my normal vector, I dot them and I multiply by
this surface area and sum all these things together.
That is what a double integral means.
In particular, an easy case where you know you
can get away without computing anything is, of course,
if your vector field is tangent to the surface because then you
know that there is no flux. Flux is going to be zero
because nothing passes through the surface.
Otherwise, we have to figure out how to compute these things.
That is what we are going to learn now.
Well, maybe I should box this formula.
I have noticed that some of you seem to like it when I box the
important formulas. (APPLAUSE) By the way,
a piece of notation before I move on, sometimes you will also
see the notation vector dS. What is vector dS?
Vector dS is this guy n dS put together.
Vector dS is a vector which points perpendicular to the
surface and whose length corresponds to the surface
element. And the reason for having this
shortcut notation, well, it is not only laziness
like saving one n, but it is because this guy is
very often easier to compute than it is to set up n and dS
separately. Actually, if you remember in
the plane, we have seen that vector n little ds can be
written directly as dy, - dx.
That was easier than finding n and ds separately.
And here the same is going to be true in many cases.
Well, any questions before we do examples?
No. OK.
Let's do examples. The first example for today is
we are going to look at the flux of vector field xi yj xk through
the sphere of radius a -- -- centered at the origin.
What does the picture look like? We have a sphere of radius a.
I have my vector field. Well, , see,
that is a vector field that is equal to the vector from the
origin to the point where I am, so it is pointing radially away
from the origin. My vector field is really
sticking out everywhere away from the origin.
Now I have to find the normal vector to the sphere if I want
to set up double integral over the sphere of F dot vector ds,
or if you want F dot n dS. What does the normal vector to
the sphere look like? Well, it depends,
of course, whether I choose it pointing out or in.
Let's say I am choosing it pointing out then it will be
sticking straight out of a sphere as well.
Hopefully, you can see that if I take a normal vector to the
sphere it is actually pointing radially out away from the
origin. In fact, our vector field and
our normal vector are parallel to each other.
Let's think a bit more about what a normal vector looks like.
I said it is sticking straight out.
It is proportional to this vector field.
Maybe I should start by writing
because that is the vector that goes from the origin to my point
so it points radially away from the origin.
Now there is a small problem with that.
It is not a unit vector. So what is its length?
Well, its length is square root of x^2 y^2 z^2.
But, if I am on the sphere, then that length is just equal
to a because distance from the origin is a.
In fact, I get my normal vector by scaling this guy down by a
factor of a. And let me write it down just
in case you are still unsure. This is unit because square
root of x^2 y^2 z^2 is equal to a on the sphere.
OK. Any questions about this?
No. It looks OK? I see a lot of blank faces.
That physics test must have been hard.
Yes? I could have put a rho but I
want to emphasize the fact that here it is going to be a
constant. I mean rho has this connotation
of being a variable that I will need to then maybe integrate
over or do something with. Yes, it would be correct to put
rho but I then later will want to replace it by its actual
value which is a number. And the number is a.
It is not going to actually change from point to point.
For example, if this was the unit sphere
then I would just put x, y, z.
I wouldn't divide by anything. Now let's figure out F dot n.
Let's do things one at a time. Well, F and n are parallel to
each other. F dot n, the normal component
of F, is actually equal to the length of F.
Well, times the length of n if you want, but that is going to
be a one since F and n are parallel to each other.
And what is the magnitude of F if I am on the sphere?
Well, the magnitude of F in general is square root of x^2
y^2 z^2 on the sphere that is going be a.
The other way to do it, if you don't want to think
geometrically like that, is to just to do the dot
product x, y, z doted with x over a,
y over a, z over a. You will be x^2 y^2 z^2 divided
by a. That will simplify to a because
we are on the sphere. See, we are already using here
the relation between x, y and z.
We are not letting x, y and z be completely
arbitrary. But the slogan is everything
happens on the surface where we are doing the integral.
We are not looking at anything inside or outside.
We are just on the surface.
Now what do I do with that? Well, I have turned my integral
into the double integral of a dS.
And a is just a constant, so I am very lucky here.
I can just say this will be a times the double integral of dS.
And, of course, some day I will have to learn
how to tackle that beast, but for now I don't actually
need to because the double integral of dS just means I am
summing the area of each little piece of the sphere.
I am just going to get the total area of the sphere which I
know to be 4pi a2. This guy here is going to be
the area of S. I know that to be 4pi a^2.
So I will get 4pi a^3. That one was relatively
painless. That was too easy.
Let's do a second example with the same sphere.
But now my vector field is going to be just z times k.
Well, let me give it a different name.
Let me call it H instead of f or something like that just so
that it is not called F anymore. Well, the initial part of the
setup is still the same. The normal vector is still the
same. What changes is,
of course, my vector field is no longer sticking straight out
so I cannot use this easy geometric argument.
It looks like I will have to compute F dot n and then figure
out how to integrate that with dS.
Let's do that. We still have that n is <x,
y, z>/a. That tells us that H dot n will
be dot <x, y,
z> / a. It looks like I will be left
with z^2 over a. H dot n is z^2 over a.
The double integral for flux now becomes double integral on
the sphere of z^2 over a dS. Well, we can take out one over
a, that is fine, but it looks like we will have
to integrate z^2 on the surface of the sphere.
How do we do that? Well, we have to figure out
what is dS in terms of our favorite set of two variables
that we will use to integrate. Now, what is the best way to
figure out where you are on the sphere?
Well, you could try to use maybe theta and z.
If you know how high you are and where you are around,
in principle you know where you are on the sphere.
But since spherical coordinates we have actually learned about
something much more interesting, namely spherical coordinates.
It looks like longitude / latitude is the way to go when
trying to figure out where you are on a sphere.
We are going to use phi and theta.
And, of course, we have to figure out how to
express dS in terms of d phi and d theta.
Well, if you were paying really, really close attention
last time, you will notice that we have actually already done
that. Last time we saw that if I have
a sphere of radius a and I take a little piece of it that
corresponds to small changes in phi and theta then we said that
-- Well, we argued that this side here,
the one that is going east-west was a piece of the circle that
has a radius a sin phi because that is r,
so that side is a sin phi delta theta.
And the side that goes north-south is a piece of the
circle of radius a corresponding to angle delta phi,
so it is a delta phi. And so, just to get to the
answer, we got dS equals a^2 sin phi d phi d theta.
When we set up a surface integral on the surface of a
sphere, most likely we will be using
phi and theta as our two variables of integration and dS
will become this. Now, it is OK to think of them
as spherical coordinates, but I would like to encourage
you not to think of them as spherical coordinates.
Spherical coordinates are a way of describing points in space in
terms of three variables. Here it is more like we are
parameterizing the sphere. We are finding a parametric
equation for the sphere using two variables phi and theta
which happen to be part of the spherical coordinate system.
But, see, there is no rho involved in here.
I am not using any rho ever, and I am not going to in this
calculation. I have two variable phi and
theta. That is it.
It is basically in the same way as when you parameterize a line
integral in the circle, we use theta as the parameter
variable and never think about r.
That being said, well, we are going to use phi
and theta. We know what dS is.
We still need to figure out what z is.
There we want to think a tiny bit about spherical coordinates
again. And we will know that z is just
a cos phi. In case you don't quite see it,
let me draw a diagram. Phi is the angle down from the
positive z axes, this distance is a,
so this distance here is a cos phi.
Now I have everything I need to compute my double integral.
z^2 over a dS will become a double integral.
z^2 becomes a^2 cos^2 phi over a times, ds becomes,
a^2 sin phi d phi d theta. Now I need to set up bounds.
Well, what are the bounds? Phi goes all the way from zero
to pi because we go all the way from the north pole to the south
pole, and theta goes from zero to 2pi.
And, of course, I can get rid of some a's in
there and take them out. Let's look at what number we
get. First of all,
we can take out all those a's and get a^3.
Second, in the inner integral, we are integrating cos^2 phi
sin phi d phi. I claim that integrates to cos3
up to some factor, and that factor should be
negative one-third. If you look at cos3 phi and you
take its derivative, you will get that guy with a
negative three in front between zero and pi.
And, while integrating over theta, we will just multiply
things by 2pi. Let me add the 2pi in front.
Now, if I evaluate this guy between zero and pi,
well, at pi cos^3 is negative one, at zero it is one,
I will get two-thirds out of this.
I end up with four-thirds pi a^3.
Sorry I didn't write very much because I am trying to save
blackboard space. Yes?
That is a very natural question. That looks a lot like somebody
we know, like the volume of a sphere.
And ultimately it will be. Wait until next class when we
talk about the divergence theorem.
I mean the question was is this related to the volume of a
sphere, and ultimately it is, but for now it is just some
coincidence. Yes?
The question is there is a way to do it M dx plus N dy plus
stuff like that? The answer is unfortunately no
because it is not a line integral.
It is a surface integral, so we need to have to variables
in there. In a way you would end up with
things like some dx dy maybe and so on.
I mean it is not practical to do it directly that way because
you would have then to compute Jacobians to switch from dx dy
to something else. We are going to see various
ways of computing it. Unfortunately,
it is not quite as simple as with line integrals.
But it is not much harder. It is the same spirit.
We just use two variables and set up everything in terms of
these two variables. Any other questions?
No. OK.
By the way, just some food for thought.
Never mind. Conclusion of looking at these
two examples is that sometimes we can use geometric.
The first example, we didn't actually have to
compute an integral. But most of the time we need to
learn how to set up double integrals.
Use geometry or you need to set up for double integral of a
surface. And so we are going to learn
how to do that in general. As I said, we need to have two
parameters on the surface and express everything in terms of
these. Let's look at various examples.
We are going to see various situations where we can do
things. Well, let's start with an easy
one. Let's call that number zero.
Say that my surface S is a horizontal plane,
say z equals a. When I say a horizontal plane,
it doesn't have to be the entire horizontal plane.
It could be a small piece of it. It could even be,
to trick you, maybe an ellipse in there or a
triangle in there or something like that.
What you have to recognize is my surface is a piece of just a
flat plane, so I shouldn't worry too much about what part of a
plane it is. Well, it will become important
when I set up bounds for integration.
But, when it comes to looking for the normal vector,
be rest assured that the normal vector to a horizontal plane is
just vertical. It is going to be either k or
negative k depending on whether I have chosen to orient it
pointing up or down. And which one I choose might
depend on what I am going to try to do.
The normal vector is just sticking straight up or straight
down. Now, what about dS?
Well, it is just going to be the area element in a horizontal
plane. It just looks like it should be
dx dy. I mean if I am moving on a
horizontal plane, to know where I am,
I should know x and y. So dS will be dx dy.
If I play the game that way, I have my vector field F.
I do F dot n. That just gives me the z
component which might involve x, y and z.
x and y I am very happy with. They will stay as my variables.
Whenever I see z, well, I want to get rid of it.
That is easy because z is just equal to a.
I just plug that value and I am left with only x and y,
and I am integrating that dx dy.
It is actually ending up being just a usual double integral in
x, y coordinates. And, of course,
once it is set up anything is fair game.
I might want to switch to polar coordinates or something like
that. Or, I can set it up dx dy or dy
dx. All the usual stuff applies.
But, for the initial setup, we are just going to use these
and express everything in terms of x and y.
A small variation on that. Let's say that we take vertical
planes that are parallel to maybe the blackboard plane,
so parallel to the yz plane. That might be something like x
equals some constant. Well, what would I do then?
It could be pretty much the same.
The normal vector for this guy would be sticking straight out
towards me or away from me. Let's say I am having it come
to the front. The normal vector would be
plus/minus i. And the variables that I would
be using, to find out my position on this guy,
would be y and z. In terms of those,
the surface element is just dy dz.
Similarly for planes parallel to the xz plane.
You can figure that one out. These are somehow the easiest
ones, because those we already know how to compute without too
much trouble. What if it is a more
complicated plane? We will come back to that next
time. Let's explore some other
situations first. Number one on the list.
Let's say that I gave you a sphere of radius a centered at
the origin, or maybe just half of that sphere or some portion
of it. Well, we have already seen how
to do things. Namely, we will be saying the
normal vector is x, y, z over a,
plus or minus depending on whether we want it pointing in
or out. And dS will be a^2 sin phi d
phi d theta. In fact, we will express
everything in terms of phi and theta.
If I wanted to I could tell you what the formulas are for x,
y, z in terms of phi and theta. You know them.
But it is actually better to wait a little bit.
It is better to do F dot n, because F is also going to have
a bunch of x's, y's and z's.
And if there is any kind of symmetry to the problem then you
might end up with things like x^2 y^2 z^2 or things that have
more symmetry that are easier to express in terms of phi and
theta. The advice would be first do
the dot product with F, and then see what you get and
then turn it into phi and theta. That is one we have seen.
Let's say that I have -- It is a close cousin.
Let's say I have a cylinder of radius a centered on the z-axis.
What does that look like? And, again, when I say
cylinder, it could be a piece of cylinder.
First of all, what does the normal vector to
a cylinder look like? Well, it is sticking straight
out, but sticking straight out in a slightly different way from
what happens with a sphere. See, the sides of a cylinder
are vertical. If you imagine that you have
this big cylindrical type in front of you,
hopefully you can see that a normal vector is going to always
be horizontal. It is sticking straight out in
the horizontal directions. It doesn't have any z component.
I claim the normal vector for the cylinder,
if you have a point here at (x, y, z), it would be pointing
straight out away from the central axis.
My normal vector, well, if I am taking it two
points outwards, will be going straight away
from the central axis. If I look at it from above,
maybe it is easier if I look at it from above,
look at x, y, then my cylinder looks like a
circle and the normal vector just points straight out.
It is the same situation as when we had a circle in the 2D
case. The normal vector for that is
just going to be x, y and 0 in the z component.
Well, plus/minus, depending on whether you want
it sticking in or out.
We said in our cylinder normal vector is plus or minus x,
y, zero over a. What about the surface element?
Before we ask that, maybe we should first figure
out what coordinates are we going to use to locate ourselves
in a cylinder. Well, yes,
we probably want to use part of a cylindrical coordinate,
except for, well, we don't want r because r
doesn't change, it is not a variable here.
Indeed, you probably want to use z to tell how high you are
and theta to tell you where you are around.
dS should be in terms of dz d theta.
Now, what is the constant? Well, let's look at a small
piece of our cylinder corresponding to a small angle
delta theta and a small height delta z.
Well, the height, as I said, is going to be delta
z. What about the width?
It is going to be a piece of a circle of radius a corresponding
to the angle delta theta, so this side will be a delta
theta. Delta S is a delta theta delta
z. DS is just a dz d theta or d
theta dz. It doesn't matter which way you
do it. And so when we set up the flux
integral, we will take first the dot product of f with this
normal vector. Then we will stick in this dS.
And then, of course, we will get rid of any x and y
that are left by expressing them in terms of theta.
Maybe x becomes a cos theta, y becomes a sin theta.
These various formulas, you should try to remember them
because they are really useful, for the sphere,
for the cylinder. And, hopefully,
those for the planes you kind of know already intuitively.
What about marginals or faces? Not everything in life is made
out of cylinders and spheres. I mean it is a good try.
Let's look at a marginal kind of surface.
Let's say I give you a graph of a function z equals f of x,
y. This guy has nothing to do with
the integrand. It is not what we are
integrating. We are just integrating a
vector field that has nothing to do with that.
This is how I want to describe the surface on which I will be
integrating. My surface is given by z as a
function of x, y.
Well, I would need to tell you what n is and what dS is.
That is going to be slightly annoying.
I mean, I don't want to tell them separately because you see
they are pretty hard. Instead, I am going to tell you
that in this case, well, let's see.
What variables do we want? I am going to tell you a
formula for n dS. What variables do we want to
express this in terms of? Well, most likely x and y
because we know how to express z in terms of x and y.
This is an invitation to get rid of any z that might be left
and set everything up in terms of dx dy.
The formula that we are going to see,
I think we are going to see the details of why it works
tomorrow, is that you can take negative
partial f partial x, negative partial f partial y,
one, dx dy.
Plus/minus depending on which way you want it to go.
If you really want to know what dS is, well, dS is the magnitude
of this vector times dx dy. There will be a square root and
some squares and some stuff. What is the normal vector?
Well, you take this vector and you scale it down to unit
length. Just to emphasize it,
this guy here is not n and this guy here is not dS.
Each of them is more complicated than that,
but the combination somehow simplifies nicely.
And that is good news for us. Now, concretely,
one way you can think about it is this tells you how to reduce
things to an integral of x and y.
And, of course, you will have to figure out
what are the bounds on x and y. That means you will need to
know what does the shadow of your surface look like in the x,
y plane. To set up bounds on whatever
you will get dx dy, well, of course you can switch
to dy dx or anything you would like,
but you will need to look at the shadow of S in the x y
plane. But only do that after you
gotten rid of all the z. When you no longer have z then
you can figure out what the bounds are for x and y.
Any questions about that? Yes?
For the cylinder. OK.
Let me re-explain quickly how I got a normal vector for the
cylinder. If you know what a cylinder
looks like, you probably can see that the normal vector sticks
straight out of it horizontally. That means the z component of n
is going to be zero. And then the x,
y components you get by looking at it from above.
One last thing I want to say. What about the geometric
interpretation and how to prove it?
Well, if your vector field F is a velocity field then the flux
is the amount of matter that crosses the surface that passes
through S per unit time. And the way that you would
prove it would be similar to the picture that I drew when we did
it in the plane. Namely, you would consider a
small element of a surface delta S.
And you would try to figure out what is the stuff that flows
through it in a second. Well, it is the stuff that
lives in a small box whose base is that piece of surface and
whose other side is given by the vector field.
And then the volume of that is given by base times height,
and the height is F dot n. It is the same argument as what
we saw in the plane. OK.
Next time we will see more formulas.
We will first see how to prove this, more ways to do it,
more examples. And then we will get to the
divergence theorem.