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PROFESSOR: Hi.
In the reading assignment for today, you'll notice in the
textbook that the topic is double integrals in terms of
polar coordinates.
Now, the trouble with polar coordinates is that aside from
straight lines, they're perhaps the only coordinate
system that we've studied in terms of Euclidean geometry in
a high school class.
And consequently, if we were to have a change of variables
other than strict linear changes of variables or polar
coordinates, it might be difficult to geometrically try
to determine what the new double integral looks like
with respect to the new variables.
The text, you may recall, when you-- or not you may recall,
you haven't read it yet-- but when you read the text, you'll
notice at the end that Professor Thomas says there is
a technique called the Jacobian, multiplying by the
Jacobian determinant, that tells you how to transfer a
double integral from x- and y-coordinates into another
coordinate system.
At any rate, with that prolog as background, our aim in
today's lecture is to show more generally how the
Jacobian sneaks into the study of multiple integrals.
In particular, we call today's lecture multiple integration
and the Jacobian.
And by way of review, let me pick a problem that we've
solved in the past in great detail, but perhaps from a
slightly different perspective, that will lead
into where the Jacobian matrix and the Jacobian determinant
comes from.
Recall that when we want to compute the definite integral
1 to 3 2x squared root of x squared plus 1 dx, we make the
substitution u equals x squared plus 1.
Or inverting this, x equals positive square
root of u minus 1.
And I emphasize the positive to point out that in general
the inverse of a squaring function is not one to one.
See a square root is usually double valued.
But notice that with the restriction that x must be on
the integral from 1 through 3, x cannot be negative.
And therefore, we certainly can assume that locally,
meaning in the region in which we're interested in, that x is
the positive square root of u minus 1.
From this we saw that du was 2xdx.
We then went back to this equation here.
We replaced 2xdx by its value du.
We replaced the square root of x squared plus 1 by the
square root of u.
And then noticing that when x equaled 1, u equaled 2 and
when x equaled 3, u equaled 10, we wound up with the fact
that the number named by this definite integral was the same
as the number named by this definite integral.
Now the only thing that I'd like to say here as an aside
is the following, there is sometimes a tendency to think
of dx as just being a symbol over here, that we think of it
as saying all we want is a function whose derivative with
respect to x is this.
And that other than that, it makes no difference what we
put in here.
What I would like you to see at this time and--
to review at this time because we know it happens--
is that if we made the substitutions mentioned here,
forgetting about the dx-- in other words, if we replaced x
by the positive square root of u minus 1, if we replace x
squared plus 1 by u, and if we replace the limits 1 to 3 by 2
to 10 and then just tacked on the du to indicate that we
were doing this problem with respect to u, the resulting
definite integral would not be equivalent to
the original one.
That is not to say that this couldn't be computed.
What I mean is this number is incorrect if by this number
you mean the value of this definite integral here.
And notice that from a pictorial point of view, all
we're really saying is that the integral 1 to 3 2x squared
root of x squared plus 1 dx is the area of the region R when
R is that region in the xy-plane bounded between the
lines x equals 1, x equals 3, below by the x-axis and above
by the curve y equals 2x times the square root of x
squared plus 1.
And I've simply put the values of these endpoints in here--
namely when x is 1, y is 2 square roots of 2, when x is
3, y is 6 square roots of 10--
to give you sort of an orientation of
this particular curve.
On the other hand, that other integral that was incorrect--
integral from 2 to 10, et cetera--
is the area of the region S where S is the region that's
obtained by taking that integral from 1 to 3 along the
x-axis, mapping it by u equals x squared plus 1, in other
words it goes from 2 to 10.
And in fact, you don't even have to know that.
All I'm saying is if you just read this thing mechanically
in the yu-plane, this would be the area of the region S where
S is bounded vertically by the lines u equal 2 and u equal
10, below by the u-axis and above by the curve y equals
twice square root u minus 1 square root of u.
And it should be clear by inspection that there is no
reason to expect that the area of the region R is the same as
the area of the region S even though both
R and S have areas.
Now, what the whole geometrical impact is on this
technique of integration, techniques of integration,
what the whole geometric impact is is this.
This is a difficult integral to evaluate to find the area.
Hopefully, one would hope that we could find a way of scaling
an element of area here to correspond to an element of
area here which was easier to compute.
And that since the mapping was one to one, by adding up the
appropriately scaled pieces here, we equivalently add up
the pieces to find the area of the region R.
Now, because I know that sounds vague to you, I am
going to do that in much more detail.
For the time being let me point out, though, that if I
want to view this as a mapping, the interesting thing
is that any point on the x-axis maps into the
corresponding point on the u-axis by the mapping what?
u equals x squared plus 1.
But it's important to point out that the values of x and u
were not independent.
They were chosen to obey the identity u equals x squared
plus 1, or x is the positive square root of u minus 1.
So what that means geometrically is that whatever
height was above a point in the region R along the
x-axis-- whatever height was here--
that height is the same when that point is moved to the
region S.
Because that again sounds like a difficult mouthful, let me
write that.
All I'm saying is notice that for the u corresponding to a
given x, 2x times the square root of x squared plus 1 is
equal to 2 times the square root of u minus 1 times the
square root of u.
How do I know that?
Well, I know that because I picked x to be the square root
of u minus 1.
Or equivalently what? u equals x squared plus 1.
This says that I can replace x by the square
root of u minus 1.
This says I can replace x squared plus 1 by u.
Consequently, as long as the x matches with the image u, this
number is the same as this number.
In other words again, in terms of a picture, if I start with
the point on the x-axis x equals 2 and I'm looking at
the point p being the point on the region R directly above x
equals 2, since 2 gets mapped into 5 by the mapping u equals
x squared plus 1-- see 2 squared plus 1 is 5--
what was the height that went with the point 2 over here?
The height that went with the point 2 over
here was simply what?
4 square roots of 5.
And I claim that that's the same as this, because when x
was 2, u is 5.
This is 2 square roots of 5.
5 minus 1 is 4.
Square root of 4 is 2.
This is 4, therefore--
2 times 2--
4 square roots of 5.
It makes no difference whether your using the x or the u, but
that the point keeps the same height.
It shifted laterally, but it does not distort the height.
Which means now if we want to view this not as a mapping
from xy-plane into the uy-plane but more
traditionally in terms of the xy-plane into the uv-plane,
that's what this v is in parentheses here.
It means that I might want to name the y-axis the v-axis
just so that I can use my identification established in
the previous block in our course when we talked about
mapping the xy-plane into the uv-plane.
That is not an accent mark over the v. That just happens
to be an interruption of this arrow that connects 2 to 5.
But at any rate, in terms of mappings, notice that the
region R is mapped onto the region S by the mapping u
equals x squared plus 1 and v equals y.
In other words, u equals x squared plus 1.
But the y-coordinate is the same as the v-coordinate.
Just changing the name of the axis here to
correspond to the uv-plane.
And now, the idea is simply this.
Since--
at least in the domain that we're interested in-- the
mapping u equals x squared plus 1 v equals y maps R onto
S in a one to one manner, each increment of area delta A sub
S matches one and only one delta A sub R.
Let me give you an example of what I mean by this.
Let's suppose I start with the region S and let me
arbitrarily divide it up into a number--
the interval from 2 to 10--
into a number of pieces here.
Oh, I guess in the diagram that I've used here, I've
divided this up into four pieces, all right.
So I get these four little rectangles.
An approximation for the area of the region S would be the
sum of the areas of these four rectangles.
What I'm saying is that under our mapping, these four
regions induce four mutually exclusive regions that cover
all of R. In fact, since v equals y, the way we do this
mechanically is--
for example, let's focus on just one of these little
elements over here.
Let's suppose I want to find how to match this shaded
rectangle with a suitable rectangle of R. What I said is
whatever the v-value is over here, it must be the same as
the y-value of the point that mapped into this
point on the axis.
In other words, if I call this point here u0, that comes from
some point here which I'll call x0.
See u0 comes from x0.
What must the height above this point be?
Since the transformation does not change the y-value at all
since v equals y, what this means is I can now draw a line
parallel to the u-axis here, come over to here, and that
locates where on the curve I'm going to locate the point x0.
See, in other words, I just come across like this, either
of these two ways.
This is how I locate the x0 that matches the u0.
I take its height, take that same height over to this
curve, and project down.
Notice, of course, that the delta x that measures the
difference between these two points on the x-axis is not
the same as the delta u that measures the distance between
these two points.
But what I want you to see is that these four rectangles
here induce four rectangles here.
And what I would like to be able to do is somehow be able,
hopefully, to find out how to express a typical rectangle
here as a scaled version of one of these rectangles,
hoping that when I then take the sum the resulting
summation leads to an integral which is easy to evaluate.
And before I get into that to show you
what does happen here--
I think I'm making this longer than it may really seem--
but let me just get onto the next step.
What I want to do next is to blow up these two shaded area
so I can look at them in more detail.
What I have is a region which I'll call delta A sub S in the
uv-plane and a delta A sub R in the xy-plane where the
mapping is, again, what? u equals x squared
plus 1 v equals y.
This piece matches with this piece.
All right.
Now, for small delta u notice that by definition of
derivative, delta x divided by delta u is approximately dxdu.
Consequently, I can say that delta x is
approximately dxdu times du.
Now the thing that I really want is not the area of a
piece of S, I want a portion of the area of R. I want delta
A sub R. Notice that delta A sub R has as its height y sub
0 and as it's width delta x.
Notice also that since v sub 0 equals y sub 0-- see y equals
v in this transformation--
notice that the area delta AR is v0 times delta x.
OK.
We also know that delta x is approximately
dxdu times delta u.
So making this substitution, I see that my little increment
of area in the region R is precisely what?
It's v0 times the replacement for delta x,
dxdu times delta u.
Let me also notice that the region delta AS over here has
as its height v0, as its base delta u.
So the area of this rectangle is v0du.
Let me, therefore, group these two factors together, rewrite
this term in this fashion, noticing that v0 delta u is
delta A sub S. And I now have the very interesting result
that delta A sub R is not delta A sub S. But the
correction factor is what?
It's dxdu multiplying delta A sub S where, for the sake of
argument over a small enough strip here, let we assume that
I've chosen dxdu to be evaluated at u equals u0, say.
At any rate, what we are saying is to find all of the
area of region R, we want to add up all of these delta A
sub Rs as the maximum delta x sub k goes to 0.
But from what we've just seen, a typical delta A sub R is a
dxdu evaluated at u equals u0 times delta A sub S. And
therefore, to find delta A sub R, this limit is precisely the
same as this limit.
Now the point is that delta A sub S is certainly just as
messy as delta A sub R, in general.
It may also happened that when I scale delta A sub S by
multiplying it by dxdu, the result is even more messy than
the original expression.
But it's also possible that dxdu happens to be the factor
that wipes out the nasty part of delta A sub S. You see,
what I'm saying is if there is a one to one correspondence--
which there is--
between the delta ASes and the delta ARs, if this expression
here happens to be convenient, I can find this sum simply by
computing this sum.
And that's why in techniques of integration, that's
precisely what we look for.
We look for the change of variable, the substitution,
that makes this thing simplify.
And that's precisely what happened in
this particular example.
Keep in mind that what I've written down over here is true
for any substitution when x is some function of u, not just
for x equals u squared plus 1.
I could've done this any time.
But what I claim is that if x weren't equal to--
what was it-- the square root of u minus 1, this wouldn't
have turned out to be a very nice expression.
And you see this is going to be called the one-dimensional
Jacobian later on.
This is the correction factor, the scaling factor, you see.
Let's see how that worked out.
Notice that delta A sub S was v0 times delta u.
And notice that by definition of what the curve looked like
in the uv-plane, v0 is twice the square root of u0 minus 1
times the square root of u0.
On the other hand, what is dxdu?
Since u is equal to x squared plus 1, it's easy to show that
dxdu is 1 over twice the square root of u minus 1.
So in particular when u equals u0, dxdu is 1 over 2 square
root of u0 minus 1.
Notice now, even though delta A sub S is quite messy, when I
multiply it by this particular scaling factor, look at what
that scaling factor wipes out.
The 2 square root of u0 minus 1 wipes this out.
All I have left is the square root of u0 times delta u.
When I form the definite integral summing this thing
up, it's trivial, you see, to see that that simply comes out
to be what?
The definite integral from 2 to 10 square root of u du.
And since this particular sum was equal to A sub R, that is
the mapping interpretation of why the area of the region R
can be evaluated by this particular integral.
Now in a sense, all of this has been review even though
the pitch has been slightly different.
Let me now generalized this to a bona fide mapping of
two-dimensional space into two-dimensional space.
And the reason I use the word bona fide is that when you say
let u equal x squared plus 1 and let v equal y, you really
haven't got a general mapping of two-dimensional space into
two-dimensional space.
You've essentially let the y-axis remain fixed.
So let me talk about something more general.
Let's suppose I have an arbitrary region R and an
arbitrary function f bar which maps R onto S in
a one to one manner.
By the way, notice the whole idea is this.
When I want the area of the region R, it's going to
involve a dxdy inside the double integral.
When I want the area of the region S, that's what's going
to involve a delta u times delta v. Now the reason that
delta u times delta v is indeed a bona fide element of
area when we're breaking up S lies in the fact that in the
us-plane delta u times delta v is the actual area of an
increment of area in S when we break up S by lines parallel
to the u and v axis.
On the other hand, notice that if we see what the line u
equals a constant comes from, back in the xy-plane u is a
function of x and y.
And to say that u of xy equals a constant does not mean that
you have a straight line.
You could have some pretty squiggly lines over here.
In other words, it might be a very funny looking line that
maps into a straight line, a straight vertical line, in the
uv-plane with respect to the mapping f bar.
And in a similar way, the lines v equal a constant--
the lines v equal a constant--
in the xy-plane I read what? v of xy equals a constant.
That's a general curve in the xy-plane.
What we're saying is that since this mapping is one to
one, when I break up this region into elementary
elements of rectangles, that will induce a breaking up of
this region into little elements here.
But notice that the resulting elements-- say we take a piece
like this, and we see where that piece comes from.
Let's say that particular piece happened
to come from here.
Say that that was the one to one correspondence.
You can take delta u times delta v here.
But notice that delta u times delta v simply means
multiplying two edges which aren't straight lines, which
aren't necessarily perpendicular, and hence in no
way should represent what the area of this little
element here is.
The key point is we do not want the area of the region S.
We want the area of the region R. And what we're hoping that
we can do is that by making the change of variables that
maps the region R in the xy-plane into the region S in
the uv-plane that we somehow find a convenient way of
scaling an individual element of area here with respect to
one here, and define the area of this region by adding up
the appropriate pieces here.
And, again, let me show you what that means in terms of
just an enlargement again.
You see, in the same as I did before, let me take this
little piece over here and really blow it up.
Let me take this little piece that is the backmap of this--
in other words, the piece that maps into this-- let
me blow that up.
And the idea is this.
If I pick delta u and delta v sufficiently small, notice
that the area of the backmap off delta A sub S is
approximately the area of a parallelogram.
You see, come back to this statement after I've explained
the picture.
What I'm saying is I start with an element of area delta
A sub S in the uv-plane.
You see I pick its vertices.
I'll call them A bar, B bar, D bar, C bar.
That's a little rectangle over here.
Because the mapping is one to one, I know that there is one
and only one point in the xy-plane that maps into A bar.
See that?
Let's call that point A. There is one and only one point in
the xy-plane that maps into B bar.
Let's call that B. One and only one point in the xy-plane
that maps into C bar.
Let's call that point C. And let me leave the point D out
for a moment.
Now the idea is if we call the coordinates of A bar u0 comma
v0, then because this is a line parallel to the axis--
call this dimension delta u--
b bar is u0 plus delta u comma v0.
C bar--
call this dimension delta v--
is u0 comma v0 plus delta v. Now the point is there is no
reason why the image of B bar-- name of the point B--
has to be on the line that joins A
parallel to the x-axis.
In other words, B is up here someplace, C is
up here some place.
In other words, again, there's no reason why these backmaps
give me a rectangle over here.
The point is that B has some coordinates.
It's x0 plus some increment involving x-- let me
call that delta x1--
and its y-coordinate is y0 plus delta y1.
C is the point x0 plus some increment delta x2 comma y0
plus delta y2.
And what I'm saying is now, if I were to just look at the
parallelogram which had AB and AC as adjacent sides, it's
very easy for me to find the area of that parallelogram.
Namely to find the area of a parallelogram in vector form,
I just take the magnitude of the cross-product of the two
vectors AB and AC.
You see what's wrong with this is that that particular
parallelogram is not the exact image of the
backmap of delta AS.
Sure, A bar maps exactly into A. C bar maps exactly into C.
B bar maps exactly into B. But the point is that these points
along A bar C bar do not map into the straight line from A
to C, in general.
In other words, what characterizes this?
This is characterized by delta u equals 0.
And when u is written in terms of x and y that doesn't say
that delta x or delta y is 0.
So, the true image of this might be what I've represented
with this dotted array here.
In other words, what the true backmap of delta A is this
dotted thing where D is now this vertex here.
You see, there's no guarantee that the backmap of D bar is
going to be the fourth vertex of this parallelogram.
But what the key point is-- and this is where that
familiarity is so important--
is that if the transformation is smooth enough, continuously
differentiable, then what it does mean is that as long as
delta u and delta v are sufficiently small, the true
area of the region delta AR that we're looking for is
approximately equal to the area of this parallelogram.
And by approximately equal I mean what?
That the arrow goes to 0 as we take the limit in forming the
double sum.
In other words, again, the key point is this.
The backmap of delta A sub S yields delta AR, but that
delta AR is approximately this parallelogram.
And the area of this parallelogram is exactly the
magnitude of AB--
the vector AB--
crossed with AC.
In other words, the approximation comes in because
this is exactly the area of the parallelogram.
But delta A sub R is only approximately equal to the
area of the parallelogram.
At any rate, notice in terms of i and j components how easy
it is to compute AB and AC.
You see, where are the components of the
vector from A to B?
The i component is delta x1.
See, this minus this.
The j component is this minus this, right.
In other words, AC has what as its components?
It's this minus this, namely delta x2.
This minus this is the y-component.
That's delta y2.
In other words-- to write this out so you don't have to
listen to how fast I'm talking--
AB is this particular vector.
AC is this particular vector.
Remembering that when we take a cross-product, i crossed j
is 0, j cross j is 0, i cross j is k, and j
cross i is minus k.
Remember, for the cross-product we can't change
the order of the terms in the order in which they appear.
We then determined what?
That AB cross AC is delta x1 delta y2 minus delta x2 delta
y1 times the vector k, the unit
vector in the z direction.
Now, delta x1 is exactly--
see, remember, we're looking at the parallelogram, which is
a straight line.
If you want to think of it in terms of the region R, by
differentials delta x1 would be what?
It's approximately the partial of x with respect to u times
delta u plus the partial of x with respect to v times delta
v. Similarly, delta y1 is y sub u times delta u plus y sub
v times delta v.
But now keep in mind where delta x1 and
delta y1 come from.
Delta x1 and delta y1 come from the back mapping from A
bar B bar back to AB.
And along A bar B bar, noticed that v is always equal to v0.
So that means that delta v is 0.
That means, therefore, that because delta v is 0, delta x1
and delta y1 are simply this.
See this drops out.
Similarly, delta x2 and delta y2 come from the
backmap of A bar A bar.
Along A bar C bar, u is equal to u0.
So delta u is zero.
So writing this out, these drop out.
And now with these terms being 0, with these terms being 0, I
could very simply compute the product delta x1 delta y2
minus delta x2 delta y1.
And if I do that, you see, right away
what I obtain is what?
It's this times this.
I now, OK, collect the terms here.
In other words, I want the delta u
and the delta v together.
The multiplier out front here is x sub u times y sub v. In a
similar way, delta x2 times delta y1 is x sub v y sub u
times delta u delta v. Therefore, putting that into
here, the magnitude of AB cross AC is simply this
expression here, noticing that the k vector drops out because
its magnitude is 1.
Notice that delta u times delta v is precisely delta A
sub S and that x sub u times y sub v minus x sub v times y
sub u is precisely the determinant of our old friend
the Jacobian matrix, the Jacobian matrix of x and y
with respect to u and v. In other words, delta A sub R is
approximately equal to the determinant of the Jacobian
matrix of x and y with respect to u and v times delta AS.
And if I now perform this double sum, you see this
becomes what?
The area of the region R--
just write that in, that's really the area
of the region R--
is a double integral over R dxdy.
And that's the same as what?
The double integral over S times dudv multiplied by the
scaling factor of the Jacobian determinant.
By the way, notice I dropped the
determinant symbol over here.
The reason for that is that many textbooks, including our
own, use this notation not to name the Jacobian matrix but
to name the Jacobian determinant.
I have been using this to name the Jacobian matrix.
From this point on, I will now switch to become
uniform with the text.
And unless otherwise specified, I will write this
rather than put the determinant symbol in.
From now on in our course, when I write this I am
referring to the Jacobian determinant, OK.
But the thing is this.
Notice that the given mapping might straighten out the
region R into a nicer looking region S. But to offset this,
it may also turn out the resulting integrand here is
much worse than the integrand here.
Here the multiplier of dxdy was the simple
number 1, wasn't it?
Here, what's multiplying dudv, no matter how nice S is, is
this expression here, which may be quite messy.
And therefore, in most practical applications where
one solves multiple integrals by a change of variable, one
not only wants a change of variables that straightens out
the region into a nice looking one, he wants a combination of
two things.
He would like a nice looking region.
And more importantly, even if he can't get a nicer looking
region, at least if he gets a correction factor, a Jacobian
determinant, that gives him something that's easy to
integrate, he'll settle for that.
And what that means, hopefully, will become clearer
as we go through the exercises and the reading material.
At any rate, I think that's all I want to say about a
supplement to Professor Thomas' treatment of polar
coordinates at this time.
And until next time, then, good bye.
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