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5.4 More on Differential Stretching

Back

Remark 5.4.1. Recall my discussing the change of variable formula for

integration of functions of a single variable. I explained how it is useful to

think of the derivative as a differential length stretching term. I elaborated

on this because it works also for higher dimensions.

76 CHAPTER 5. GREEN’S THEOREM

Remark 5.4.2. It works in particular for cases where the curve is in R2 or

Example 5.4.1. Find an expression for the arc length of the graph of y = x2

from the origin to the point [1, 1]T . Using Mathematica or otherwise, find

the length of the curve.

Solution We can write the problem as

c d` where d` is an ‘infinitesimal’ bit

of the curve. It is reasonable to write

d` =

(dx)2 + (dy)2

Parametrise the curve by _

Then note that the quantity d` becomes just the norm of the differential

term so we get: Z 1

x˙ (t)

y˙(t)

Z 1

1 + 4t2dt

Writing

Integrate[Sqrt[1 + 4t^2], {t, 0, 1}]

in Mathematica (or by making the substitution 2t = sinh(x)) we get:

5 + sinh−1(2)

Writing

N[%]

NIntegrate[Sqrt[1 + 4t^2], {t, 0, 1}]

5.4. MORE ON DIFFERENTIAL STRETCHING 77

to get a numerical evaluation, we get 1.47894 which looks believable.

Note that all arc length problems, in R2 or R3, can be seen as

kx˙ k dt

where

x : I −! Rn

t x(t)

is the parametrisation of the curve.

Exercise 5.4.1. Find the length of the arc of y = cosh(x) between x = −1

and x = 1.

Remark 5.4.3. As well as doing it for curves, it also works just as well for

parametrisation of surfaces.

Let g : I2 −! R2 be a map of the unit square into R2. Recall:

Definition 5.10.

I2 ,

: 0 _ x _ 1, 0 _ y _ 1

Remark 5.4.4. The picture corresponding to figure 5.3 which did it for

curves is figure 5.9

I have drawn a cubical box at the front over g(I2) and pulled it back to a

box over I2 to show what happens when you integrate the (blue coloured)

function f. I have chosen g to be 1-1 and smooth.

Remark 5.4.5. It is plain that the idea of the one-dimensional case still

works here, but the cubical boxes standing on a tiny square base in I2, will

be taken by g to cubical boxes standing on deformed squares in g(I2). We

get these deformed squares in the limit by taking the derivative of g at the

point in I2 to see what the deformation is, and in particular what it does to

the area of those tiny squares. The area stretching factor will then go into

the formula for changing the variable by composing with g.

The derivative of g at a point a in I2 is going to be 2 × 2 matrix

78 CHAPTER 5. GREEN’S THEOREM

Figure 5.9: Change of variables in Integration for a function of two variables

where

g =

x(s, t)

y(s, t)

When I evaluate those partial derivatives at the p[oint a I shall get a matrix

of numbers which is the linear map which best approximates g at the point

Remark 5.4.6. We need to see what linear maps from R2 to R2 do to area.

I have drawn in figure 5.10 the image of the unit square by the linear map

given by the matrix _

a c

b d

Since the area of the unit square is one, and since a linear map will take

smaller squares to proportionately smaller parallelograms, the area stretching

factor for the linear map is simply the area of the parallelogram with vertices

at the origin, _

and

a + c

b + d

Now we can calculate the area of this parallelogram by chopping off the

triangle at the top and moving it down to the bottom. This is now gives a

new parallelogram which is the image of I2 by the matrix

a − bc/d c

0 d

5.4. MORE ON DIFFERENTIAL STRETCHING 79

Figure 5.10: Image by a linear map of the unit square

These two maps have the same area stretching factor. I hope you can see

that I merely subtracted the proper multiple of

from

to get the second matrix.

The diagram in figure 5.11 shows the new parallelogram. It can also be

skewed, without changing the area, (chop out a triangle and move it) so that

it is rectangular, and has vertices at the origin, at

a − bc

and

a − bc

This rectangle would arise from the linear map represented by the matrix:

_ ad−bc

d 0

0 d

which has the same area stretch as the original matrix, and takes the unit

square to the rectangle. It is easy to see that the area of this rectangle is

80 CHAPTER 5. GREEN’S THEOREM

Figure 5.11: Image by a linear map of the unit square

ad − bc. Since none of the transformations have changed the area, this is

the area of the original parallelogram and is the area stretching factor of the

linear map. Note that it is negative when the image of i has been moved past

the image of j so that the original square has been turned upside down. So we

actually get the oriented area stretching term out of what is the determinant

of the matrix _

a c

b d

Remark 5.4.7. It is a good idea to take g to be 1-1 and smooth and to

have a smooth inverse except perhaps on a set of area zero.4 In this case,

the Jacobian Determinant _____

_____

is either always positive or always negative since it cannot be zero. We get

the right answer for the area stretch either way if we take its absolute value.

4 I was rather dismissive of curves which failed to be differentiable at only a finite set of

points, and I propose to be equally dismissive of functions from I2 which fail to be smooth

or 1-1 at a finite number of lines; the reason is the same, they won’t make any difference

to the integral.

5.4. MORE ON DIFFERENTIAL STRETCHING 81

Proposition 5.4.1. The change of variable formula for integrating a function

over a surface is:

I2 g

−! R2 f

−! R

g(I2)

f =

(f _ g)| det(D(g))|

where D(g) is the derivative of g. You should be able to see that | detD(g)|

is the “differential area stretching factor”.

Proof: We have done all the hard work in messing about with parallelograms.

We note then that we can get an approximation to a Riemann

double sum over little regions in g(I2) by taking squares in I2 and their images

by g. In the neighbourhood of a point of I2, g can be approximated by

the (linear) derivative (together with a shift to put the image in the right

place). The height of the function f _ g over a point a is, by definition, the

height of f over g(a) and the base of the cuboidal box over g(a) has had its

area stretched by an amount

_____

So the volume of the little box has been stretched by the same amount,

and so the Riemann Sum of the little cuboids over g(I2) is the sum of the

corresponding cuboids over I2 with the area stretching factor taken into

account. The approximation improves as the boxes are made smaller and so

the formula comes out of the limit of the Rieman sums. _

Remark 5.4.8. This lacks the careful rigour that Pure Mathematicians prefer,

but making it rigorous is not very difficult. The idea is the main thing.

The guys who invented these ideas were happy with proofs like this one.

Example 5.4.2. Find the area enclosed by the ellipse

= 1

Typing:

<< Graphics‘ImplicitPlot‘

ImplicitPlot[x^2/9 + y^2/4 == 1, {x, -4, 4}]

82 CHAPTER 5. GREEN’S THEOREM

Figure 5.12: The ellipse x2/9 + y2/4 = 1

into Mathematica and running it gives the picture of figure 5.12

Now we write the ellipse as a transform of the unit circle:

T : R2 −! R2

It is easy to see that this stretches the circle by a factor of 3 in the x direction

and 2 in the y direction. The derivative of this map is the diagonal matrix

with diagonal entries 3 and 2 and determinant 6. So the area enclosed by

the ellipse is six times the area of the unit disc, that is, 6_.

Remark 5.4.9. A little thought suggests that if you take a map g : I2 −! R3

then there ought to be a formula for the area stretching in this case. This

would be nice, for then we could calculate areas of tori and spheres and

ellipsoids.

Remark 5.4.10. All of these things are taken care of automatically by using

differential forms.

This strikes me as a good argument in favour of them. I shall now show how

easy it is to take care of the differential area stretch without having to do

much.

Example 5.4.3. Let I2 g

−! R2 f

−! R be a pair of smooth maps.

I shall regard f as a 2-form

Pdx ^ dy

5.4. MORE ON DIFFERENTIAL STRETCHING 83

I write g as

so g =

x(s, t)

y(s, t)

I want

g(I2) f, which means

g(I2) Pdx ^ dy

I want to transform this to be

I2 f _ g ds ^ dt. I write down the chain rule:

_ @x

_ _

dx =

ds +

dy =

ds +

dx ^ dy =

ds +

ds ^ dt +

dt ^ ds

−

ds ^ dt

= det(Dg) ds ^ dt.

Remark 5.4.11. And out comes the differential area stretch without me

having to do any sweating. The good news is that this works for maps which

take the square and parametrise some surface in R3.

Remark 5.4.12. I shall refer to this process as ‘composing with g on the

functional part and composing with g0 on the differential part of the form.’

Example 5.4.4. Parametise S2 in R3 and integrate the function

over S2 to obtain the area of the sphere

Solution:

Recall that

g : [0, 2_] × [−1, 1] −! R3

(s , t)

1 − t2 p cos(s)

1 − t2 sin(s)

84 CHAPTER 5. GREEN’S THEOREM

parametrised the 2-sphere by wrapping the rectangle around the sphere in a

cylinder and then pushing the cylinder in horizontally. (Check that x2+y2+

z2 = 1, which shows that we do finish up on the sphere, then check that for

the point on the sphere with cylindrical coordinates (1, _, z) there is a point

which gets sent to it.) Writing

g(s, t) =

x(s, t)

y(s, t)

z(s, t)

The derivative of g is therefore:

g0(s, t) =

664

775

and

5 =

664

775

This leads to:

dx ^ dy =

that is:

dx ^ dy =

−

ds ^ dt

and similarly:

dx ^ dz =

−

ds ^ dt

and

dy ^ dz =

−

ds ^ dt

This gives the transformation on the differential part using g0.

5.4. MORE ON DIFFERENTIAL STRETCHING 85

Evaluating them for my map g I get

dx ^ dy = (t sin2(s) + t cos2(s)) ds ^ dt = t ds ^ dt

and

dx ^ dz = −

1 − t2 sin(s) ds ^ dt

and

dy ^ dz =

1 − t2 cos(s) ds ^ dt

Finding the area in R3 is like finding the length of a curve in R2; there, recall,

we had Z

kx˙ k

for the length. Similarly, here we have

[0,2_]×[−1,1]

dx ^ dy

dx ^ dz

dy ^ dz

The differential area-stretching factor is the norm of the 2-form

1( dx ^ dy + dx ^ dz + dy ^ dz)

with a corresponding result for maps g embedding I2 (or any other rectangle

or two dimensional region) in Rn for any n > 2.

This gives the area of the sphere as

[0,2_]×[−1,1]

p t

p1 − t2 sin(s)

1 − t2 cos(s)

ds ^ dt

Z 1

−1

Z 2_

1 ds ^ dt

Now we have it in the final form we can leave the wedge out and the answer

is 4_. Note that this is the same as the area of the circumscribing cylinder:

the projection onto the sphere does not change the area.

Remark 5.4.13. You would naturally like to know what the formula is in

Old Fashioned Language. The answer is rather natural.

86 CHAPTER 5. GREEN’S THEOREM

Figure 5.13: Curves and tangents

Proposition 5.4.2. If g : I2 −! R3 is a smooth 1-1 map with smooth

inverse, and

g(s, t) =

x(s, t)

y(s, t)

z(s, t)

then

and

are two tangents to curves in g(I2) and are linearly independent in R3.

Figure 5.13 shows one tangent to a curve.

Proof:

Going back to the definition of the partial derivative at a point of I2 we see

that if we keep t = b and look at the curve in R3 obtained by letting s vary,

then the first partial derivative vector is just the tangent to this curve at the

point (a, b)T when we evaluate it at that point. Similarly for the other. If

the derivative of g at the point is non singular then the two vectors in I2 are

taken to independent vectors in R3. But the derivative of g is never singular

because we have a smooth inverse. _

Remark 5.4.14. In R3, the cross product of two vectors is orthogonal to the

pair and has length the product of the two lengths times the sine of the angle

5.4. MORE ON DIFFERENTIAL STRETCHING 87

between them. This length is in fact the area of the parallelogram defined

by the two vectors, the origin and their sum.

The area stretch done by g at a neighbourhood of a point is going to be given

by the length stretch in the s direction multiplied by the length stretch in

the t direction, multiplied by the sine of the angle between the image of the

unit vectors i and j by the derivative. In other words, the cross product of

the above partial derivatives.

We have therefore the change of variables and area stretching formula:

g(I2)

f =

f _ g

This is equivalent to the formula obtained from calculating the differential

part of the 2-form. As it had better be.

I keep using the following idea:

Definition 5.11. Smooth embedding A map g : Ik −! Rn is said to be

a smooth embedding of Ik in Rn if it is a map which is smooth, 1-1, and has

a smooth inverse from the image.

Definition 5.12. Smooth Embedding a.e A map g : Ik −! Rn is said to

be a smooth embedding almost everywhere (a.e.) of Ik in Rn if it is continuous

and is a smooth embedding except on a subset of Ik having Lebesgue measure

zero. Lebesgue measure is the natural generalisation of length in R, area in

R2 and volume in R3. In particular the measure of the cube Ik is one, whereas

its boundary has measure zero in Rk.

Remark 5.4.15. Since we are doing a certain amount of integration, we can

usually be dismissive about things going wrong on sets of zero length, area,

volume, whatever. So g can fail to be smooth or 1-1 on such negligible sets

and we can neglect them.

Remark 5.4.16. Ik is a subset of Rk so differentiability makes sense. You

can only embed Ik in Rn if n _ k.

Proposition 5.4.3. If f : Rn −! R is a map that is integrable and g : I2 −!

Rn is a map which is a smooth embedding of I2 in Rn almost everywhere,

then Z

g(I2)

f =

(f _ g)(s, t) · k!k

88 CHAPTER 5. GREEN’S THEOREM

where ! is the constant 2-form

1_i<j_n

dxi ^ dxj

and

g(t) =

6664

x1(s, t)

x2(s, t)

...

xn(s, t)

7775

on Rn.

Proof: I shan’t prove it, it requires more (multi-)linear algebra than you have

covered. The result should be intuitively appealing if you think about it. It

is obviously consistent with my claim about the differential area stretching_

Remark 5.4.17. This is the general change of variable formula for maps

from I2 into Rn and we are now integrating f over some surface sitting in

Rn.

Exercise 5.4.2. Write down what you feel ought to be the formula for finding

the length of a curve embedded in Rn. Test it out on particlar curves where

you can make some estimate of the result.

Example 5.4.5. The region T2 _ R4 is defined by:

T2 =

8>><

>>:

664

775

2 R4 : w2 + x2 = 1 and y2 + z2 = 1

9>>=

>>;

Find its area.

Solution Parametrise T2 by

g : [0, 2_] × [0, 2_] −! R4

(s , t)

664

w(s, t) = cos(s)

x(s, t) = sin(s)

y(s, t) = cos(t)

z(s, t) = sin(t)

775

Then

dw =

ds +

dt = −sin(s) ds

5.5. GREEN’S THEOREM AGAIN 89

Similarly, dx = cos(s) ds, dy = −sin(t) dt, dz = cos(t) dt Hence

dw ^ dx = (−sin(s) cos(s)) ds ^ ds = 0

Similarly

dw ^ dy = sin(s) sin(t) ds ^ dt, dw ^ dy = sin(s) sin(t) ds ^ dt,

dw ^ dz = −sin(s) cos(t) ds ^ dt, dx ^ dy = cos(s)(−sin(t)) ds ^ dt

and

dx ^ dz = cos(s) cos(t) ds ^ dt, dy ^ dz = 0

The norm of this vector of six components is:

02 + (sin(s) sin(t))2 + (−sin(s) cos(t))2

+((cos(s)(−sin(t))2 + (cos(s) cos(t))2 + 02

This is just 1. So the area of T2 is

Z 2_

1 ds dt = 4_2

Remark 5.4.18. Most of the old fashioned guys wouldn’t have the faintest

idea how to start on this. A modern mathematician is someone who can

do this in his head in a few minutes. An old fashioned mathematician is

someone who can’t see any reason why T2 should have an area, let alone

know how to compute it.

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