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Lecture 1: AMATH 231-F14-001 September 8, 2014

0 Preliminary Material

0.1 Review Vector Algebra and Introduce Tensor Notation

Motivation for vectors and tensors:

Previously learned vector notation and this makes things more compact Easier than writing out each component However, proving vector identities can still be very lengthy Here we review some vector algebra and present tensor (indicial) notation

Basic Notation

Small letters with half vectors for vectors:~a = (a1, a2).

Capital letters denote matrices:A =

(a11 a12a21 a22

).

One free index denotes a vector: ai where i = 1, 2. Two free indices denotes a matrix: Aij where i = 1, 2 and j = 1, 2. Einstein summation convention: A repeated index means we sum up over all the indices

2i=1

aibi is reduced to aibi

Can work for vectors of any size 2, 3, 4, !Review of Vector Algebra

We will assume that ~a,~b,~c are all vectors in three-space, s is a scalar andA is a 3 3 matrix.(a) Addition and Subtraction of Vectors: These operations are commutative and associative.

~a+~b =3i=1

(ai + bi) in tensor notation becomes ai + bi

(b) Multiplication of Vectors by Numbers:

s~a =3i=1

sai in tensor notation becomes sai

1

(c) Scalar Products: Dot

~a ~b =3i=1

aibi in tensor notation becomes aibi

(d) Matrix Multiplication:

A~b =3j=1

Aijbj in tensor notation becomes Aijbi

(e) Vector Products: Curl

~a~b = (a2b3 a3b2, a3b1 a1b3, a1b2 a2b1)

(f) Triple Scalar Products:

~a (~b ~c

)= a1 (b2c3 b3c2) + a2 (b3c1 b1c3) + a3 (b1c2 b2c1)

~a(~b ~c

)= ???

At the moment we dont see any clever way to write some of these vectors in tensorial notationbut we will figure out a way of doing that when needed. Now, we begin with the course notesand start our discussion of Vector Calculus.

Before we get to talking about curves lets discuss some motivation to know why we mightwant to learn vector calculus. To do this we visit the course notes for MATH 212.

0.2 Motivation to study Vector Calculus

Often we can translate fundamental laws of nature into the language of mathematics and VectorCalculus is a useful way to accomplish this.

example 0.1 For example, consider water flowing into a cylindrical pipe. How can we determine howthe mass of water in a particular section is changing in time? (Units of kilograms/second)

1) Measure the concentration of water in that section at every instant and find the rate of change ofmass with time.

2) Find out how much water is coming in and going out through the boundaries per second.

By the Conservation of Mass both must give us the same answer. But in the first we are consideringthe whole volume whereas in the second we are only looking at the boundaries. This simple idea is used agreat deal to determine the Differential Equations (DE) that describe a given system.

In physics we have four laws of electromagnetism (EM) that can be stated in words as follows:

1) Gauss Law [Flux of electric field

through any closed surface

]=

[Net change

inside the volume

]

2

2) Faraday [circulation of an electric field

around the perimeter of a surface

]=

[flux of the magnetic

field through the surface

]3) No magnetic dipoles [

flux of a magnetic fieldthrough any closed surface

]= 0

4) Ampe`re

circulation of a magneticfield around the perimeterof a surface

= ddt

flux of theelectric fieldthrough the surface

+

flux of theelectric currentdensity through

the surface

Please dont be scared off by these equations. In this course you will not be tested in yourknowledge of physics. However, by the end of the course if I give you these principles youshould be able to deriving the governing system of Partial Differential Equations (PDEs) thatdescribes the motion.

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1 Curves and Vector Fields

This is mostly taken from sections 1.2 and 1.3 from the MATH 212 notes or section 1.1.1 of theAMATH 231 course notes.

Definition 1.1 A scalar-valued function is of the form,

f : [a, b] R or x = f(t).Definition 1.2 A vector-valued function is of the form,

~g : [a, b] R3 or ~g(t) = (g1(t), g2(t), g3(t)) tensor= gi(t).

1.1 Curves in Rn

1.1.1 Curves as vector-valued functions

Both functions produce curves in 1D space.

The first maps a 1D domain ([a, b]) to a 1D domain from f(a) to f(b). The second maps a 1D domain ([a, b]) to a 3D domain from ~g(a) to ~g(b).

Next we need to define what we mean by a field.

Definition 1.3 A scalar field is a function of 2, 3, , n variables that yields a scalar. For example,z = f(x, y),

u = g(x, y, z),

w = h(x1, , xn).The input can be a vector but the output is always a scalar. Examples of scalar fields include chargedensity, temperature, height of a mountain and speed of the winds.

Definition 1.4 A vector field is a function of 2, 3, , n variables that yields a vector. For example,~f(x, y) = (f1(x, y), f2(x, y)),

~g(x, y, z) = (g1(x, y, z), g2(x, y, z), g2(x, y, z)).

The input can be a vector but the output is always a vector. Examples of vector fields include velocity.

example 1.5 Recall that we can parameterize a curve as,

~x(t) = (r cos(t), r sin(t)), if 0 t 2pi.

To make things concrete we can think of t as time, is the frequency (units of radians/second) and the ~x(t)is the position of a particle (object) in the plane. How does the path change if we change the frequency? Thevector function, ~x(t), (position), is determined by the parameter, t (time). We say that the vector-valuedfunction is parameterized with respect to t.

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example 1.6 We can parameterize a line using a linear relationship between the parameter and thevector-valued function

~x(t) = ~r0 + t~u, if 0 t a,where ~r0 and ~u are constant vectors and a is a scalar. This is the equation of a line that passes throughposition ~r0 at t = 0 and has a slope of ~u.

Definition 1.7 Let ~g : [a, b] R2 be a vector function given by

~g(t) = (x(t), y(t)).

We call ~g a path and its image C, the curve of the path. The input t is a parameter and the above is theparametric representation of the path.

example 1.8 Consider the path ~g : [0, 2] R2 given by

~g(t) = (t, t2) = (x(t), y(t)).

The input is t [0, 2] and the output is x(t) = t and y(t) = t2. We can graph this for different values of tor find out what the relation is between x and y:

y = t2 = (t)2 = x2.

Therefore, this curve is nothing more that a parabolic centred at the origin opening upwards.

example 1.9 Consider the path ~g : [0, a] R2 given by

~g(t) = (a cos t, b sin t) with 0 t 2pi.

We can find the following relations,

x

a= cos t, and

y

b= sin t.

From our knowledge of trigonometry we know that cos2 t+sin2 t = 1 and therefore we deduce the followingalgebraic relation between x and y:

x2

a2+y2

b2= 1,

which is of course the equation of an ellipse.

See section 1.1.1 of the AMATH 231 course notes for more details on the examples and alsoone involving a helix.

1.1.2 Representation of a curve

Not only is there not a unique representation of a curve but there is always an infinite number ofparameterizations. For example in the previous example with the ellipse, we can replace t with2s and as long as we change the bounds appropriately we get the same curve. We say that thetwo paths are different parameterizations of the same curve C.

5

To make this concrete leth : [, ] [a, b]

be a continuous and increasing and therefore one-to-one function between these two intervals.Then, given the path ~g : [a, b] Rn we define, ~g : [, ] Rn by

~g() = ~g(h()), on .

The two paths ~g(t) and ~g() follow the same points and thus yield the same curve.How do we interpret these two paths? If we assume that the paths denote the position of

a particle and we want the independent parameters t and to both be time in seconds, thenthat means the two particles follow the same paths but at different rate. That is to say differentspeeds.

example 1.10 Define the following path,

~g(t) = (cos t, sin t).

This is clearly a circular motion of radius one and frequency equal to one.If we pick t = h() = 3 then on possible path is

~g() = ~g(h()) = (cos 3, sin 3).

This does not have a constant frequency and indeed, the frequency increases quadratically in time.If instead we pick t = h() = a then another path is

~g() = ~g(h()) = (cos a, sin a).

This rotates with a constant frequency of a.

6


1.1.3 Limits

Next, we begin to discuss the derivative of ~g(t). This is build up by looking at the derivative ofeach component. First we discuss the definition of a limit of ~g(t).

Definition 1.11 Given a vector-valued function (path)

~g(t) = (g1(t), , gn(t)),

defined in a neighbourhood of t0 and a constant vector

~L = (L1, , Ln),

thenlimtt0

~g(t) = ~L limtt0

gi(t) = Li, for i = 1, , n.

Definition 1.12 The vector-field ~g is continuous at t0 means that

i) ~g(t0) is defined

ii) limtt0 ~g(t) exists

iii) limtt0 ~g(t) = ~g(t0).

Note the following properties:

i) ~g(t) is continuous at t0 gi(t) is continuous at t0 for i = 1, 2, , n.ii) Physical curves are necessarily continuous functions.

iii) In general paths need not be continuous.

1.1.4 Derivatives

Definition 1.13 Given ~g : [a, b] Rn, the derivative of ~g at t, denoted by, d~gdt

(t) or ~g(t), is defined by

~g(t) = limt0

1

t[~g(t+ t) ~g(t)] ,

provided the limit exists.

Note:

i) If ~g(t) exists then ~g(t) is differentiable at t.

ii) ~g(t) exists ~gi(t) exists for i = 1, 2, n.iii) The derivative is computed component-wise, ~g(t) = (g1(t), , gn(t).

7

Physical Interpretation of the derivative:Consider a curve C in R3 and the path (position) of the particle described by ~x = ~g(t). From

the definition of the derivative we see that it is a change in distance divided by the change intime. That is to say it is the displacement vector over time t divided by the time interval. Thatis why if ~g(t) is the position then the velocity is

~v(t) = ~g(t) = (x(t), y(t), z(t)) ,

and the speed is simply,

~v(t) =x(t)2 + y(t)2 + z(t)2 .

Following the same rationale, the second derivative yields the acceleration vector of the par-ticle,

~a(t) = ~g(t) = (x(t), y(t), z(t)) .

Geometrical Interpretation of the derivative:Suppose a curve is defined by ~x = ~g(t) with ~g(t0) 6= 0, then ~g(t0) is a vector in the direction of

the tangent line at the point ~x = ~g(t0). Draw a picture like that of Figure 1.10 showing the secondand tangent lines.

Question: Why do we need ~g(t0) 6= 0?Equation of the tangent line:

Consider a curve ~x = ~g(t) with ~g(t0) 6= 0. Since we know that ~g(t0) is the direction of thetangent line and it must go through the point ~g(t0) at t = t0 we can write down the equation ofthe tangent line as,

~Lt0 = ~g(t0) + (t t0)~g(t0).Note that we used the variable ~Lt0 to denote the tangent line at the point t0 since clearly it willchange in general.

Question: For what curve are all the tangent lines the same?

Definition 1.14 We say that a vector-value function ~g : [a, b] Rn is of of class C1 or just C1 if ~g(t)is differentiable.

Definition 1.15 We say that a vector-value function ~g : [a, b] Rn is piecewise C1 if there is a partitionof [a, b],

a = t0 < t1 < < tN = b,such that ~g(t), when restricted to each open interval (ti, ti+1) for i = 0, 1, , N1, coincides with thefunction that is C1 on the closed interval [ti, ti+1].

example 1.16 Find the speed of ~x(t) = (r cost, r sint).The velocity is

~x(t) = r( sint, cost).The speed is therefore,

~x(t) = |r|.This means that the linear speed is linear proportional to the radius and the frequency. The units are ofcourse length/time.

8

example 1.17 Show the followingd

dt

(~x(t)2) = 2~x ~x.Expand using the definition and the product rule:

d

dt

(~x(t)2) = ddt

(~x ~x)= ~x(t) ~x+ ~x ~x= 2~x ~x.

example 1.18 Compute the followingd

dt~x(t).

Expand using the power rule:

d

dt

(~x(t)2) = 2~x~x2~x ~x = 2~x~x.

From this we deduce the result,

~x = ~x ~x

~x .

example 1.19 Example 1.5 from AMATH 231 Course Notes.Consider the curves defined by

i) ~x = ~g1(t) = (t|t|, t2), 1 t 1.ii) ~x = ~g2(t) = (t, |t|), 1 t 1.Observe that we can get non-parameteric representation of these curves:

i) y = t2 = |t||t| = |t|t|| = |x|.ii) y = |t| = |x|.

Therefore, both curves are the same.Question: Are the parameterizations differentiable at t = 0?Since we cannot compute the derivative of the absolute value at t = 0 the only hope we have is to use

first principles. First for ~g1(t),

limh0

~x(h) ~x(0)h

= limh0

(h|h|, h2) (0, 0)h

= limh0

(|h|, h) = (0, 0),

and second for ~g2(t),

limh0

~x(h) ~x(0)h

= limh0

(h, |h|) (0, 0)h

= limh0

(1,|h|h

),

which does not exist.Question: What does this mean?

9


To figure this out we can compute the derivatives for t < 0 and t > 0, respectively. For the firstfunction we get,

~g1(t) = (2t, 2t), for t < 0~g1(t) = (2t, 2t), for t > 0,

The speed is the same on either side and is ~g = 22|t|. From this we see that the particle slows downas it approaches t = 0 and stops. For the second function we obtain a different result,

~g2(t) = (1,1), for t < 0~g2(t) = (1, 1), for t > 0.

The speed on either side is ~g = 2. So the speed is continuous but the velocity is discontinuous. We seethat the particle always moves in the positive x-direction at the same speed. But for t < 0 it moves downat a constant speed and then, when it gets at t = 0 it suddenly starts to move upwards at the same speed.Clearly we see that some paths can be rather confusing and non-physical.

See example 1.6 in the AMATH 231 course notes for an example of a cycloid.

1.1.5 Arclength

It is easy to determine the length of a straight lines is easy but how do we find the length of anycurve in Rn? To answer this let us consider a path given by

~x = ~g(t), with a t b.

The mapping is from an interval of the real line to Rn.insert pictureAs in calculus, we use what we know. At present all we know is how to find length of straight

lines. That is why we approximate the curve with straight lines to get an estimate. Then we usemore and more straight lines. This is the same idea used in Riemann sums!

The process boils down to the following steps:

1) To begin, we partition our interval [a, b] into n subintervals,

a = t0 < t1 < t2 < < tn1 < tn = b.

2) We linearly approximate the curve along each subinterval using the tangent line,

~xi+1 ~xi ti~g(ti), with ti = ti+1 ti.

3) Find the length of each segment (si denotes the length of arc i),

si ~xi+1 ~xi = ~g(ti)ti, for all i

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4) We sum over all the segments to get an an exact value of the arc length, for this partition,

sN =N1i=1

~g(ti)ti.

We used sN to denote the arc length of the N -th piecewise constant approximation.

5) Take the limit as N or ti 0 to get the exact value for the arc length of the curve,

s =

ba

~g(t) dt.

If the image C is the trajectory traversed by a particle then s is the distance travelled.

Note: At each point t [a, b] we have that ~g(t) dt computes the infinitesimal distancetravelled in a time dt. Integration then sums over all values of t and therefore yields the arclength (or distance) exactly.

example 1.20 Compute the length of the path ~g1(t) = (cos t, sin t) on t [0, 2pi] and ~g2(t) = (cos 3t, sin 3t)on t [0, 2pi].

s(~g1(t)) =

2pi0

~g1(t) dt = 2pi

0

( sin t, cos t) dt = 2pi

0

sin2 t+ cos2 t dt =

2pi0

1 dt = 2pi

s(~g2(t)) =

2pi0

~g2(t) dt = 2pi

0

3( sin 3t, cos 3t) dt = 3 2pi

0

sin2 3t+ cos2 3t dt = 6pi.

The second is three times bigger because it wraps around the circle three times.

example 1.21 What is the length of the curve y = f(x) on x [a, b]?We parameterize the curve as ~g(t) = (x, f(x)) and x [a, b], which implies ~g(x) = (1, f (x)). Theformula yields,

s(~g(x)) =

ba

1 + (f (x))2 dx.

example 1.22 Prove that the shortest distance between two points is a straight line. Suppose that thepoints are A = (a, 0) and B = (b, 0).

We parameterize any smooth curve connecting A to B with the path ~g(t) = (t, f(t)). Recall that theformula for arc length is

s(~g(t)) =

ba

1 + (f (t))2 dt.

Clearly the distance between the points is |b a|. We must show that any other path has a distance thatis larger than that, i.e. s(~g(t)) > b a, where we assume b > a.

If the path is not exactly a straight line then there must be a point where f (t) 6= 0. Since f(t) isassumed to be C1, then f (t) is continuous and there must be an interval (a, b) around t where f (t) 6= 0.

11

Thus,

1 + (f (t))2 > 1 holds on (a, b). But,

s(~g(t)) =

ba

1 + (f (t))2 dt,

=

aa

1 + (f (t))2 dt+

ba

1 + (f (t))2 dt+

bb

1 + (f (t))2 dt,

>

aa

1 dt+

ba

1 dt+

bb

1 dt,

=

ba

1 dt = b a.

Therefore, if the line is ever not straight, we necessarily get a length that is larger than that of a line andwe are done.


example 1.23 Let ~g : [a, b] R2 by a C1 path. The arc length function of ~g(t) is given by

s(t) =

ta

~g() d.

Observe that,

a) s(a) = 0.

b) s(b) = s(~g).

c) s(t) is a monotonically decreasing function.

d) dsdt

= ~g().This turns out to be a very powerful tool. Based on our previous work we can parameterize anypath with its arc length. Unlike our previous parameterizations, we have that there is as uniquepath that does this. But to do this we need to find t(s).

example 1.24 Compute the arc length function for the path ~g(t) = (3 cos t, 3 sin t) with t [0, 2pi] andparameterize it by its arc length.Solution:Compute the derivative of the path,

~g(t) = (3 sin t, 3 cos t)Therefor the norm of this is ~g(t) = 3. Therefore, the arc length function is easily computed,

s(t) =

t0

~g(t) dt = t

0

3 dt = 3t.

In this case the inversion is easy, t = s/3.Therefore, the parameterization by arc length is,

~g(s) = (3 coss

3, 3 sin

s

3)

and note that ~g(s) = 1 by design, as we will prove shortly.

12

example 1.25 Compute the arc length function for the path ~g(t) = (a cos t, a sin t, bt) with t [0, 2pi]with a, b > 0 and parameterize it by its arc length.Solution:ELFS (Exercise Left for the Student)

1.1.6 Acceleration and Curvature

Now we have a method to parameterize any path, say ~g(t), by its arc length, ~g(s) (at least inprinciple). Observe that if we differentiate with respect to arc length we get, after using thechain rule,

d

ds~g(s) =

d

dt~g(t)

dt

ds=

ddt~g(t)dsdt

=~g(t)~g(t)

Therefore, the length of this is unity, dds~g(s) = ~g(t)~g(t)

= 1.Definition 1.26 For any path ~g(t) we can define the unit tangent vector to ~g(t) as

~T (t) =~g(t)~g(t) .

This is the unique choice of tangent vectors that has a length of one for any value of t.Note that ~T (t) = ~g(s).

example 1.27 Let ~g(t) = (2t3/2, 2t) from (0, 0) to (2, 2)

a) Find the arc length function s and then ~g(s).

b) Check dsdt

= ~g(t)c) Find ~T (t) = ~g(s).

Solution:

a) The derivative of the path is ~g(t) = (3t1/2, 2) and thus ~g(t) = 9t+ 4.With this we can calculate the arc length function,

s =

t0

9t+ 4 dt =

2

27(9t+ 4)3/2

t0

=2

27

[(9t+ 4)3/2 8] .

Next, we invert to solve for t in terms of s,

(9t+ 4)3/2 =27

2s+ 8,

t =1

9

[27

2s+ 8

]2/3 4

9.

We can substitute this into the path to obtain our unique parameterization in terms of the arc length,

~g(s) = (2

{2

27

[(9t+ 4)3/2 8]}3/2 , 2

9

[27

2s+ 8

]2/3 8

9)

13

b) We differentiate the arc length with respect to t to verify that it is equal to ~g(t) = 9t+ 4,ds

dt=

2

27

27

2(9t+ 4)1/2 =

9t+ 4.

c) We compute the unit tangent vector,

~T (t) =~g(t)~g(t) =

(3t1/2, 2)9t+ 4

.

Physical Quantities

Let ~g(t) be a C2 path describing the trajectory of the motion. Speed is ds

dt= ~g(t) is the rate of change of arc length with time.

Velocity is the speed times the unit tangent vector.

~v(t) = ~g(t) = ~g(t) ~g(t)

~g(t) = ~g(t)~T (t) = ds

dt~T (t).

Note that ds/dt is the magnitude and ~T is the unit tangent vector.

Acceleration: sum of two components and directions.

~a(t) =d

dt~v(t) =

d

dt

(ds

dt~T (t)

),

~a(t) =d

dt

(ds

dt

)~T (t) +

ds

dt

d

dt~T (t),

~a(t) =d2s

dt2~T (t) +

ds

dt

d

dt~T (t).

Orthogonality property: Since we know that ~T (t)2 = ~T ~T = 1 we can differentiate thisto show that the unit tangent vector is orthogonal to its derivative,

~T ~T = 0.

The acceleration has a part parallel to ~T and a second parallel to ~T

Tangential acceleration (speeding up in the same direction)

~aT =d2s

dt2~T (t) .

Normal acceleration (tendency to turn)

~aN =ds

dt

d

dt~T (t).

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Can simplify the derivative of the unit tangent vector,d

dt~T (t) =

d

ds~T (t)

ds

dt= ~g(t) d

ds~T (t).

Normal component of acceleration can be rewritten as,

~aN =

(ds

dt

)2d

ds~T (t) .

Acceleration can then be written as,

~a =d2s

dt2~T (t) +

(ds

dt

)2d

ds~T (t).


Definition 1.28 ~g(t) is a C2 path on R2 (or R3) parameterized by arc length. The curvature (s) of ~g(s)is defined as,

(s) dds ~T

=~T (t)~g(t) ,

using the above identity. Geometrically, that means that the curvature is the magnitude of the rate ofchange of the unit tangent vector with respect to arc length.

example 1.29 Find the curvature of the line ~g(t) = (at, bt).Solution:

~T (t) =~g(t)~g(t) =

(a, b)a2 + b2

.

Since this is a constant we then deduce that the curve is zero, as it should be for a straight line,

(s) =

~T (t)~g(t) = 0.

example 1.30 Find the curvature of the circle ~g(t) = (a cos t, a sin t).Solution:

~T (t) =~g(t)~g(t) =

(a sin t, a cos t)a

= ( sin t, cos t).

Next, we can compute the curvature from the formula

(s) =

~T (t)~g(t) =

( cos t, sin t)a

=1

a.

The curvature of a circle of radius a is 1/a. In the limit as a we have a straight line and thecurvature is zero. This confirms what we verified in the previous example.

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example 1.31 Find the curvature of the helix ~g(t) = (a cos t, a sin t, bt) with a, b > 0 and t 0.Solution:

~T (t) =~g(t)~g(t) =

(a sin t, a cos t, b)a2 + b2

=1

a2 + b2(a sin t, a cos t, b).

Next, we can compute the curvature from the formula

(s) =

~T (t)~g(t) =

(a cos t,a sin t, 0)a2 + b2

=a

a2 + b2.

In the limit where b = 0 we recover the result for a circle. But in general we find that if b 6= 0 then thecurvature of the helix tends to be smaller than that of a circle.

exercise 1.32 Find the curvature of the parabola y = x2. [Solution: (t) = 2(1+4t2)3/2

]

1.2 Vector Fields

1.2.1 Examples from physics

example 1.33 Gravitational field due to a spherical body of mass M is a vector field

~F (~x) = GM ~rr3, with r = ~r ,

where G is the gravitational constant and ~F is the force exerted on a test body of unit mass.

example 1.34 Velocity of a fluid is a vector field, such as the atmosphere or the oceans.

example 1.35 Density of a fluid is a scalar field.

1.2.2 Field lines of a vector field

Assume that ~F (~x) determines the velocity of a fluid at position ~x.

Definition 1.36 Let ~F (~x) be a continuous vector field defined on a subset R2 or R3. A flow (field) lineof F is a path ~g(t) such that

d

dt~g(s) = ~F (~g(t)).

Note: Flow lines are called integral curves if a path. That is to say they are the curves that satisfy thisabove Differential Equation (DE).

Calculating the flow lines is equivalent to answering the following question: given a velocityfield ~F , what trajectory do particles follow? For example, consider a river with a winding coast-line. If we drop a light nerf ball in the river, what path does it traverse?

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example 1.37 The field lines ~x = ~g(t) satisfies ~g(t) = ~F (~g(t)). Then we deduce,

d

dt~x = (x(t), y(t)) = (y(t),x(t)),

that goes through the point (, ). The flow lines yield two DE,

dx

dt= y, and

dy

dt= x.

If we differentiate the first we can then substitute in the second to get,

d2x

dt2=dy

dt= x, or d

2x

dt2+ x = 0.

This is the equation for the simple harmonic oscillator that we can solve exactly.We look for a solution of the form et and find that the characteristic equation is,

2 + 1 = 0,

and so the roots are = 1. Therefore, the solution can be written as,x = aeit + beit.

But using Eulers formula and redefining our constants we can write this in terms of real functions,

x = a cos t+ b sin t.

To satisfy the initial conditions we pick a = and b = and therefore our solution is,

x = cos t+ sin t.

Crash Course in ODEsFor second order DE with constant coefficients of the form,

ax + bx + c = 0,

we look for solutions of the form x = et to obtain our characteristic equation (after dividingthrough by the exponential)

a2 + b+ c = 0.

The roots are,

1,2 =bb2 4ac

2a.

example 1.38 The field lines ~x = ~g(t) satisfies ~g(t) = ~F (~g(t)). Then we deduce,

d

dt~x = (x(t), y(t)) = (x(t), y(t)),

that goes through the point (, ). The flow lines yield two DE,

dx

dt= x, and

dy

dt= y.

These equations separate and we can solve each independently. The exact solution is,

x = et, and y = et.

17


2 Line Integrals and Greens Theorem

2.1 Line Integral of a Scalar-Field

2.1.1 Motivation and Definitions

So far we are able to compute the arclength of a curve given its path ~g(t). What if we want to domore? Suppose that f(x, y) is the density of a hanging wire and t parameterizes the wire. Whatis the mass of the wire?

Definition 2.1 The curve C in Rn is defined by ~x = ~g(t) on a t b where ~g is C1 and f(~x) is C0. Theintegral of f along path C is,

~g

f ds =

Cf ds =

ba

f(~g(t)) ~g(t) dt.

In worlds we say, f is evaluated on the curve C giving f(~g(t)) and the symbol ds is the arc length andtherefore yields ~g(t) dt. The left-hand side is new but the right-hand side is a standard Riemannintegral. Note that in the special case where f = 1 we recover the formula for arc length,

s =

~g

ds =

Cds =

ba

~g(t) dt.

Consider a curve like in Figure 2.2 of the course notes.

1) We divide our interval [a, b] into N subintervals.

2) Linearly approximate the curve on each segment.

3) For the i-th segment we can write the approximate arc length as,

s(~g(ti)) ~g(ti)ti.

4) The mass of a small segment (that is nearly linear) is equal to the density times the arclength. If ti is sufficiently small then the density is nearly constant and therefore themass is

mi f(~g(ti)) ~g(ti)ti.This says that locally mass = density length.

5) To find the mass of the entire curve we sum up over all the segments and then take thelimit as N . This results in the above equation.

example 2.2 Compute the line integral~gf ds of f(x, y, z) = xyz along the curve

~g(t) = ( sin t,

2 cos t, sin t), on t [0, pi/2].

18

Solution:Must compute the derivative,

~g(t) = ( cos t,

2 sin t, cos t),

then the magnitude of the derivative,

~g(t) =

cos2 t+ 2 sin2 t+ cos2 t =

2.

Now we can unravel the line integral,~g

f ds =

ba

f(~g(t)) ~g(t) dt,

=

pi/20

(

2 sin2 t cos t)

2 dt,

= 2 sin t cos t|pi/20 ,= 2

3sin3 t

pi/20

,

= 23.

Is it bad to have a negative answer? It depends on what were computing. Here, no.

example 2.3 Compute the line integral~gf ds of f(x, y) = 2x+ y along the paths:

a) Clockwise quarter circle: ~g1(t) = (cos t, sin t) on t [0, pi/2].b) Along a line segment: ~g2(t) = (1, 0) + t(1, 1) = (1 t, t) on t [0, 1].c) Along two line segments: ~g3(t) = (1 t, 0) on t [0, 1] and ~g4(t) = (0, t) on t [0, 1].

Solution:

a) Compute the derivative and its magnitude,

~g1(t) = ( sin t, cos t), = ~g1 = 1.

Therefore, the line integral becomes,~g1

f ds =

ba

f(~g(t)) ~g(t) dt,

=

pi/20

(2 cos t+ sin t)1 dt,

= 2 sin t cos t|pi/20 ,= (2 0) (0 1) = 3.

b) Compute the derivative and its magnitude

~g2(t) = (1, 1), = ~g2(t) =

2.

19

Now we compute the line integral~g2

f ds =

2

10

(2 t) dt,

=

2 (2t t2/2)10,

=

2(2 1/2) = 3/

2.

c) The same thing but now for two line segments:

~g3(t) = (1, 0), = ~g3(t) = 1.

~g4(t) = (0, 1), = ~g4(t) = 1.Sum the two line integrals,

~g3

f ds+

~g4

f ds =

10

2(1 t) dt+ 1

0

t dt,

=[2t t2 + t2/2]1

0dt,

= 2 1 + 1/2 = 3/2.

Even though the starting and ending points are the same the line integral depends on the path, andso the answers are different.

Definition 2.4 The average value of a scalar field is,

f = 1s(~g)

~g

f ds =1

s(~g)

ba

f(~g(t)) ~g(t) dt.

Consistency Property:If ~g(t) : a t b is C1 and ~g() : is C1 and they parameterize the same curve C thenthe line integral of any function f over the two paths must be equal:

Cf ds =

ba

f(~g1(t)) ~g1(t) dt =

f(~g2()) ~g2() d.

example 2.5 Compute the average temperature of a wire in the shape

~g(t) = (cos t, t/10, sin t), on t [0, 10pi],

if the temperature is T (~x) = x2 + y + z2.Solution:

1 +pi

2.

20


2.2 Line Integral of Vector Field

2.2.1 Motivation and Definition

Suppose a force field ~F (~x) moves a particle along a curve C given by ~g : [a, b] R2 from ~g(a) to~g(b). What is the work done by ~F ? (Recall that work done is equal to the force dotted with thedisplacement.)

Definition 2.6 The curve C in Rn is given by ~x = ~g(t), a t b,~g is C1 and ~F is continuous. The lineintegral of ~F along C is,

~g

~F d~x =C

~F d~x = ba

~F (~g(t)) ~g(t) dt.

Note: ~F (~x) d~x = ~F (~g(t)) ~g(t) dt is the work (force times distance) done by ~F along the segment denotedby dx.

Method proceeds as before but with one major difference:

1) We divide our interval [a, b] into N subintervals.

2) Linearly approximate the curve on each segment and compute ~xi = ~xi+1 ~xi.3) The line integral (work done) on the i-th segment is

~F (~xi) ~xi.

This says that locally work = force displacement.4) Rewrite the displacement in terms of the path ~xi ~g(ti)ti and substitute into the above

~F (~xi) ~xi = ~F (~xi) ~g(ti)ti.

5) Sum over all the segments and then take the limit as N .

example 2.7 Compute the work done by ~F acting upon a particle that moves along the trajectory ~g(t) :[a, b] R3 is given by the path integral,

W =

ba

~F (~g(t)) ~g(t) dt.

where ~F = (y, x, 1) if we consider the following curves:a) ~g(t) = (cos t, sin t, 0), t [0, pi].b) ~g(t) = (t, t, t), t [0, 1].c) ~g(t) = (cos t, sin t, t), t [0, 2pi].

Solution:

21

a) We must evaluate the vector field along the path and compute the derivative of the path,

~F (~g(t)) = (y(t), x(t), 1) = ( sin t, cos t, 1),~g(t) = ( sin t, cos t, 0).

Now we can evaluate the integral,

W =

pi0

~F (~g(t)) ~g(t) dt = pi

0

sin2 t+ cos2 t dt = pi.

b) Similarly,

~F (~g(t)) = (y(t), x(t), 1) = (t, t, 1),~g(t) = (1, 1, 1),

and the work becomes

W =

10

~F (~g(t)) ~g(t) dt = 1

0

t+ t+ 1 dt = 1.

~g(t) = (t, t, t), t [0, 1].c) Finally,

~F (~g(t)) = (y(t), x(t), 1) = ( sin t, cos t, 1),~g(t) = ( sin t, cos t, 1),

and the work becomes

W =

2pi0

~F (~g(t)) ~g(t) dt = 2pi

0

sin2 t+ cos2 t+ 1 dt = 4pi.

Geometric Interpretation:The path integral can be rewritten using the dot product

~g

~F d~x = ba

~F (~g(t)) ~g(t) dt. = ba

~F (~g(t)) ~g(t) cos (t) dt.The integrand, and therefore the integral, is largest for paths such that is closest to zero. Phys-ically, that means the most work is done by a vector field that along a path that is parallel to thevector field. These optimal paths are the flow lines of ~F .Consistency requirement:If ~g and ~g are two parameterizations of the same curve then b

a

~F (~g(t)) ~g(t) dt. =

~F (~g(t)) ~g(t) dt,

where ~g : [a, b] Rn and ~g : [, ] Rn.Alternative Notation:If ~F = (F1, F2) and ~g(t) dt = (x(t), y(t)) dt = (dx, dy), then we could write,

C

~F d~x =C

(F1, F2) (dx, dy) =C

F1 dx+ F2 dy.

22

example 2.8 Evaluate the line integral,

I =

C(3x+ 4y) dx+ (2x+ 3y2) dy,

around the circle x2 + y2 = 4.Solution:We can parameterize the circle as ~g(t) = (2 cos t, 2 sin t) and t [0, 2pi]. We compute the derivative to be,

~g(t) = (2 sin t, 2 cos t),

then substitute the resultant into the integral,

I =

C(3x+ 4y) dx+ (2x+ 3y2) dy,

=

2pi0

[2(3 cos t+ 4 sin t)(2 sin t) + 2(2 cos t+ 6 sin2 t)(2 cos t)] dt,

=

2pi0

[12 cos t sin t 16 sin2 t+ 8 cos2 t+ 24 sin2 t cos t] dt,=

2pi0

[12 cos t sin t 8 (1 cos 2t) + 4 (1 + cos 2t) + 24 sin2 t cos t] dt,=

[6 sin2 t 8

(t 1

2sin 2t

)+ 4

(t+

1

2sin 2t

)+ 8 sin3 t

]2pi0

,

= 8pi.

Properties of Line Integrals:

(i) Linearity: C

(~F + ~G) d~x =C

~F d~x+ C

~G d~x.

(ii) Additivity: C is a C1 curve and C = C1 C2 and C1 C2 = thenC

~F d~x =C1

~F d~x+C2

~F d~x.

(iii) Reversal of Orientation: If you reverse the orientation of a path C we denote that with Cand we have that

C

~F d~x = C

~F d~x.

Note that each curve has two possible orientations.

Proof of Reversal of OrientationSee proposition 2.2 of the course notes. Not too difficult but technical.

23


example 2.9 Work equals gain in kinetic energySuppose that ~F (~x) is the force acting on a particle m that moves along the path ~g(t) on the intervalt [a, b]. Newtons second law implies that,

~F (~g(t)) = m~a(t) = m~v(t) = m~g(t).

The work done by the force of gravity is,

W =

~g

~F d~x = ba

~F (~g(t)) ~g(t) dt = ba

m~v(t) ~v(t) dt.

But the product rule yields,d

dt

(~v2) = ddt

(~v ~v) = 2~v ~v.When we substitute this into the line integral we obtain,

W =

ba

m~v(t) ~v(t) dt = m2

ba

d

dt

(~v2) dt.By a fundamental theorem of calculus then we can evaluate this easily,

W =m

2~v(b)2 m

2~v(a)2 .

That is to say the work done on an object is equal to its final (kinetic) energy minus the initial (kinetic)energy. This is completely independent on what path it takes. This leads to our next topic.

2.3 Path Independent Line Integrals (Conservative Fields)

Definition 2.10 The image of a one to one, piecewise C1 mapping (path) ~g : [a, b] R2 is said to be asimple curve. That is to say there are no intersections.

Definition 2.11 The image of a piecewise C1 mapping (path) ~g : [a, b] R2 that is one to one on [a, b]and ~g(a) = ~g(b) is called a simple closed curve. These curves can be clockwise or counterclockwise.

Definition 2.12 A domain D is connected if every pair of points in D can be joined by a continuouscurve lying strictly in D.

Definition 2.13 A simply connected domainD is a connected domain in which every simple closed curvecan be continuously shrunk to a point in D without ever leaving D.

Definition 2.14 A positive orientation is one that is counter-clockwise.

We have see many examples of line integrals that depend on path and that is very typical.However, there is a special family of line integrals that are path independent. These occur whenthe vector field is said to be conservative. Gravitational fields are conservative whereas air resis-tance is non-conservative (path dependent).

24

Definition 2.15 If ~F (~x) = ~(~x) in a domain D, then ~F is a conservative vector field in D and is apotential for ~F for D.example 2.16 Say,

~F (~x) = km(

xx2 + y2

,y

x2 + y2

)= km ~r~r ,

is a conservative (gradient) vector field on R2 {(0, 0)} since if we define

= kmx2 + y2, = ~F = ~.

Check:Suppose the definition from above and evaluate

~ = ~(km ~r) = km~~r .How do we evaluate the above? We can use the definition,

~~r =(

x

~r ~r,

y

~r ~r

),

=

(~r ~r

x~r ~r ,

~r ~ry

~r ~r

).

But since ~r = (x, y) we have that the partial derivatives are the unit vectors in each direction,

~~r = 1~r (~r ex, ~r ey) =~r

~r .

Therefore, returning to our above calculations,

~ = km~~r = km ~r~r .

Thus, we have demonstrated that ~F (~x) = ~.example 2.17 Show that the velocity field ~u = (y,x) of rigid body motion about the z-axis is non-conservative if 6= 0.Solution:If the field is conservative then ~u = (y, x) = and therefore,

x= y, and

y= x.

Method 1:Integrate the first equation with respect to x to get,

= xy + f(y).Then, differentiate with respect to y and match with the second equation,

y= x+ f (y) = x.

25

But there is no function f (y) that can satisfy this equation. Therefore it is impossible and so our fieldmust be non-conservative.Method 2:If ~u is C1 and exists then must be C2 and therefore we require that,

2

xy=

2

yx.

But, from the above equations we have,

2

xy=

y

(

x

)=

y(y) = ,

2

yx=

x

(

y

)=

x(x) = .

Since these are not equal a function cannot exist and therefore a conservation function does not exist.

Theorem 2.18 Necessary condition for a conservative field in R2 to exist.If ~F = (F1(x, y), F2(x, y)) is a conservative vector field in the domain D then,

yF1(x, y) =

xF2(x, y).

Proof:If ~F = then, by definition,

~F = (F1, F2) =

(

x,

y

)But if ~F is C1 then must be C2 and thus,

F1y

=

yx=

xy=F2x

.

26


Theorem 2.19 Necessary condition for a conservative vector field in R3.If ~F = (F1, F2, F3)(x, y, z) is a conservative vector field in D in R3 then we must have everywhere in Dthat,

yF1 =

xF2,

zF2 =

yF3,

xF3 =

zF1.

Or, this can be more easily stated as saying that the curl of the vector field is zero,

~ ~F = ~0.

The proof is not very difficult and is analogous to the 2D result from before.

example 2.20 Show that ~F (x, y) = (x,y) is a conservative vector field in R2 and find the potential .Solution:From a previous theorem,

x= x, and

y= y.

Integrate the first equation,

=1

2x2 + f(y).

Differentiate with respect to y and then use the second equation,

y= f (y) = y,

and therefore f(y) = 12y2 + c. Therefore, the vector field is conservative and the potential function is,

=x2

2 y

2

2+ c.

example 2.21 Find the potential function of ~F (x, y, z) = (xy sin z, x22 ey

2, e

y

z2 x cos z)

Solution:From a previous theorem we surmise that

x= xy sin z,

y=x2

2 e

y

z,

z=ey

z2 x cos z.

Integrate the first equation,

=x2y

2 x sin z + f(y, z).

Differentiate with respect to y and then use the second equation,

y=x2

2+f

y=x2

2 e

y

z.

27

Therefore, f = ey

z+ g(z). We plug this int our expression for , differentiate with respect to z and then

use our final equation to obtain that

x cos z ey

z2+ g(z) =

ey

z2 x cos z.

Therefore, we can pick g(z) = 0 and so the potential function is

=x2y

2 x sin z + e

y

z+ constant.

Theorem 2.22 Necessary and sufficient conditions for a vector field to be a gradient vector.Let ~F be a C1 vector field defined on an open simply connected set U , a subset of R2. Then ~F = ~f forsome function f if and only if

F2x

=F1y

.

In R3 we require that ~ ~F = ~0.Theorem 2.23 Generalization of the Fundamental Theorem of Calculus.Let f : Rn R be a C1 function and let ~g : [a, b] Rn be a piecewise C1 path. Then

~g

~f d~x = f(~g(b)) f(~g(a)).

Proof:By the definition of a line integral we have

~g

~f d~x = ba

~f(~g(t)) ~g(t) dt.

But observe that the chain rule yields,

d

dt(f(~g(t))) = ~f(~g(t)) ~g(t).

Therefore, we obtain our desired result,~g

~f d~x = ba

~f(~g(t)) ~g(t) dt.

=

ba

d

dt(f(~g(t))) dt.

= f(~g(b)) f(~g(a)).Observe that to prove the above we needed to use the Fundamental Theorem of Calculus for one variableand the chain rule for paths.

Corollary 2.24 As above but if ~g is a closed path then~g

~f d~x = 0.

Note: we typically denote the line integral over a closed path with a circle on the integral, as used above.

28

example 2.25 Compute the work of the electrostatic force,

~F =Qq

4pi0

~r

r3 ,

acting on a change q that moves from point A to point B along the path ~g.Solution:Easy to show if we realize that

~F = ~V (~r), with V (~r) = Qq4pi0

1

~r .

Then the work is computed as follows:

W =

~g

~F d~x,

= ~g

~V (~r) d~x,

= V (~r)|~r(b)~r(a) ,= V (~r(a)) V (~r(b)),=

Qq

4pi0

(1

~r(a) 1

~r(b)).

Lecture 11: AMATH 231-F14-001 October 1, 2014

Theorem 2.26 Properties of a gradient vector fieldLet ~F be a C1 vector field defined on an open, connected set U , a subset of Rn. Then the following areequivalent:

a) ~F is a gradient field (~F = ~)b) For any oriented simple closed path ~g,

~g

~F d~x = 0.

c) For any two orientated simple curves ~g1, ~g2, having the same initial and final positions,~g1

~F d~x =~g2

~F d~x

If U is simply connected then the above are equivalent to

d)F2x

=F1y

and the analogue for R3.

29

The equivalence of a) and c) is the First Fundamental Theorem of Line Integrals.Proof:a) = b)This is equivalent to a previous Corollary. Assume ~F = ~ in D and say C is parameterized by ~g(t) fora t b and then evaluate the line integral,

~g

~F d~x = ba

~ ~g(t) dt,

=

ba

d

dt((~g(t))) dt,

= (~g(b)) (~g(a)),= 0,

since the curve is assumed to be closed.b) = c)Suppose we have any orientated simple closed curve C. We can split it up into C = C1 (C2). Then

0 =

~g

~F d~x,

0 =

C1

~F d~xC2

~F d~x,C1

~F d~x =C2

~F d~x.

Therefore, the line integral is path independent.c) = a)Suppose that the line integral is path independent. Call the initial and final points P0 = (x0, y0) andP = (x, y), respectively. Then decompose the curve C into C1 C2 where C1 is the line from (x0, y0) to(x0, y) and C2 is from (x0, y) to (x, y). The two paths can be written as,

~g1(t) = (x0, t), y0 t y,~g2(t) = (t, y), x0 t x.

We can then define the following potential function and also decompose the line integral into the lineintegral over each subpath

(x, y) =

C

~F d~x =C1

~F d~x+C2

~F d~x.

We can evaluate the integrals to obtain,

(x, y) =

yy0

(F1, F2) (0, 1)dt+ xx0

(F1, F2) (1, 0)dt,

(x, y) =

yy0

F2(x0, t) dt+

xx0

F1(t, y) dt.

If we then compute the x partial derivative we get,

x= F1(x, y).

30

To show this for the y partial derivative we need to construct a path that first goes along x and then alongy. Please try it and convince yourself that you understand how this works.

Therefore, we can show that ~F is conservative since we can write it as the gradient of a function.Part d) follows from the definition of a conservative function if you make the additional assumption.

2.4 Greens Theorem

Theorem 2.27 Let D be a bounded subset of R2 whose boundary D is a piecewise C1 simple closedcurve oriented counter clockwise (positive). If ~F = (F1, F2) is C1 on D D then

D~F d~x =

D

(F2x F1

y

)dxdy

Note: Greens Theorem relates a line integral around the perimeter of the domain to an areaintegral within the domain. What are these quantities that we are summing up?

Suppose we can describe D in the following form,

f(x) y g(x),a x b,

and H(x, y) is continuous function on D D then the double integral over the domain can bewritten as,

DH(x, y) dxdy =

bx=a

[ y=g(x)y=f(x)

H(x, y) dy

]dx.

This will be useful in the proof.Proof of Greens Theorem:

First approach that Greens Theorem has special cases for (F1, 0) and (0, F2). If we can proveeach of them independently, then we can add them to get the total result.

To do this we need to evaluate the line integral. Let us suppose that it can be bounded intwo different ways. The first is as mentioned before. That is to say we assume that C1 and C2parameterize the bottom and top of the region, respectively, then we get,

~g1 = (x, f(x)), = ~g1 = (1, f (x)), on a x b~g2 = (x, g(x)), = ~g2 = (1, g(x)), on a x b

where on both curves we have a x b. This parameterization implies that D = C1 (C2).Alternatively, we could assume that we have left and right bounds on the functions

~g3 = (h(y), y), = ~g3 = (h(y), 1), on c y d,~g4 = (k(y), y), = ~g2 = (k(y), 1), on c y d.

31

For the first part, consider the special case with ~F = (F1, 0):D

~F d~x =C1~F d~x

C2~F d~x,

=

ba

(F1(x, f(x)), 0) (1, f (x))dx ba

(F1(x, g(x)), 0) (1, g(x))dx,

=

ba

F1(x, f(x))dx ba

F1(x, g(x))dx,

=

ba

[F1(x, f(x)) F1(x, g(x))] dx.

But observe that the following double integral takes on the same form, after some manipulation,DF1y

dxdy = ba

[ g(x)y=f(x)

F1y

dy

]dx,

= ba

[F1(x, g(x)) F1(x, f(x))] dx,

=

ba

[F1(x, f(x)) F1(x, g(x))] dx,

=

D

~F d~x.

In the above equation we used the previous result just before to obtain the desired result.The other case we consider is ~F = (0, F2):

D~F d~x =

C1~F d~x

C2~F d~x,

=

dc

(0, F2(k(y), y)) (k(y), 1)dy dc

(0, F2(h(y), y)) (h(y), 1)dy,

=

dc

[F2(k(y), y)) F2(h(y), y)] dy.

But we can rewrite the other double integral asD

F2x

dydx =

dc

[ k(y)x=h(y)

F2x

dx

]dy,

=

dc

[F2(k(y), y) F2(h(y), y)] dy,

=

D

~F d~x.

In the above equation we used the previous result just before to obtain the desired result.By linearity we can sum these two results and get the theorem as stated above.

32


example 2.28 Verify Greens Theorem if ~F = (x2 xy, xy y2) if D is the triangular region withvertices (0, 0), (1, 1), (2, 0).

Solution:Label the curves on the bottom, right and left as C1, C2, C3, respectively. Note that the boundaryis the union of the three, D = C1 C2 C3.LHS:

D~F d~x =

C1

~F d~x+C2

~F d~x+C3

~F d~x

We must parameterize each path,

C1 : ~g1(t) = (t, 0) 0 t 2,C2 : ~g2(t) = (2 t, t) 0 t 1,C3 : ~g3(t) = (1 t, 1 t) 0 t 1.

We evaluate each integral,~g1

~F d~x = 2

0

(t2, 0) (1, 0)dt = 13t320

=8

3,

~g2

~F d~x = 1

0

((2 t)2 (2 t)t, (2 t)t t2) (1, 1)dt = 43,

~g3

~F d~x = 1

0

(0, 0) (1,1)dt = 0.

Therefore, the final answer is the sum of the three integrals,D

~F d~x = 43.

RHS:Since the boundary is piecewise C1 and a simple closed curve and ~F is C1 on D D we can

use Greens theorem,D

~F d~x =D

(F2x F1

y

)dxdy

D

(y + x) dxdy.

We can parameterize the area with,

y x 2 y, on 0 y 1.

33

This allows us to set up the double integer,D

(y + x) dxdy =

10

( 2yy

(y + x) dx

)dy,

=

10

[xy +

1

2x2]2yy

dy,

=

10

[(2 y)y + 1

2(2 y)2 (y2 + 1

2y2)

]dy,

= 2

10

[1 y2] dy,

= 2

[y 1

3y3]1

0

=4

3.


Corollary 2.29 If we make all the assumptions of Greens theorem and further assume that,

F2x F1

y= 0,

or ~F is a conservative field for all points in D then,D

~F d~x = 0.

As we have already shown but now we verify with Greens Theorem.

example 2.30 Define U = R2 {(0, 0)} and

~F (~x) =

( yx2 + y2

,x

x2 + y2

).

Show thatF2x F1

y= 0,

but that D

~F d~x 6= 0,if we define D to be the perimeter of the unit circle.Solution:First we verify that the difference of partial derivatives is zero,

F2x F1

y=

(x2 + y2) 2x2(x2 + y2)2

+(x2 + y2) 2y2

(x2 + y2)2=

0

(x2 + y2)2= 0.

To compute the line integral we define the path of the unit circle ~g(t) = (cos t, sin t) on 0 t 2pi and evaluate the line integral,

~F d~x = 2pi

0

( sin t, cos t) ( sin t, cos t) dt = 2pi

0

dt = 2pi 6= 0.

34

Question: Is Greens Theorem wrong?No! The conditions require that the boundary is a piecewise C1 simple closed curve and that

~F is C1 on the boundary and interior. The problem here is that the origin is not included andtherefore the domain is not simply connected.

If C is any simple closed curve that has (0, 0) in its interior than one can show thatD

~F d~x = 2pi.

If C is any simple closed curve that does not contain (0, 0) then,D

~F d~x = 0.

At all points except (0, 0) we have that ~F is conservative and therefore ~F = ~ for the potentialfunction

= arctan(yx

).

We can only apply Greens Theorem to curves that do not contain the origin.

example 2.31 Define ~F (~x) = (2xy, 1 + x2) and

a) Test that ~F is conservative

b) If yes, find the potential function.

Solution:

a) ~F is C1 everywhere because it consists of polynomials.

For any open simply connected set U a subset of R2,F2x F1

y= 2x 2x = 0.

Thus, ~F is a conservative field.

b) Define the potential function ~F = ~ and therefore,

x= 2xy, and

y= 1 + x2.

We integrate the first equation and get = x2y + f(y). When we compute the partialderivative with respect to y we then deduce that x2 + f (y) = 1 + x2. Therefore we requiref(y) = y and a potential function is,

= x2y + y.

The level curves of (curves where is constant) are called equipotential lines.

35

2.5 Vorticity and Circulation

Definition 2.32 Given a C1 field ~F : R2 R2 and a simple closed path C defined by ~g : [a, b] R2, theline integral,

=

~g

~F d~x,

is called the circulation of ~F around ~g.

Physical Interpretation of Circulation:Suppose that we pick ~F = ~u to be the velocity of an object. The circulation is the line integral ofthe velocity dotted with the tangent vector. The quantity that is integrated is

~F d~x~g

= ~F (~g(t)) ~g(t)dt.

If the velocity is always orthogonal to the path then the circulation is zero. If the velocity istangent to the path then the circulation is largest. Therefore, the circulation is the tendency for thevelocity field to flow (circulate) counter clockwise around the path.

The three possible cases are:

a) If > 0 fluid moves counter clockwise.

b) If < 0 the fluid moves clockwise.

c) If = 0 there is no net circulation.

If the circulation is zero that does not imply there is no movement along the boundary. It simplymeans that there is no net movement. There could be regions of clockwise movement but theyare essentially balanced out by regions of counter clockwise movement.

Definition 2.33 The vorticity of a vector field ~F = (u(x, y), v(x, y)) is

=

(v

x uy

),

which, if ~F is the velocity, is the tendency for objects in the flow to rotation counter clockwise. We willshow that the vorticity is interpreted to be the circulation per unit area.

Physical Interpretation of Vorticity:The vorticity (or rotation rate) of a fluid is defined as the sum of the rotation rate of two or-

thogonal lines. By convention we choose the vorticity to be positive if the rotation counter (anti)clockwise. Figure 1 illustrates two initially orthogonal lines each of which is deformed due tothe orthogonal component of velocity. If we consider the vertical line for instance, and if u1 isuniform on this line there will be a translation but no rotation. Its only due to the vertical shearof u1 do you get a component of rotation from this axis. A similar argument applies to the hor-izontal line which implies that it is the horizontal gradient in the vertical velocity, that inducesrotation. Each of these act in opposite directions however and a typical outcome after a shorttime is illustrated in Figure 2.

36

Figure 1: Two orthogonal lines are shown in the xy-plane and how gradients in the velocitiescan give rise to local rotation, and thus vorticity.

Figure 2: This is what could transpire after some short time if the gradients in the previous plotsare both positive.

37

Following this argument, we observe that the vorticity about the z-axis, which we denotewith 3 is

3 = ddt

+d

dt

=1

dt

[1

x2

(u1x2

x2dt

)+

1

x1

(u2x1

x1dt

)]=

u2x1 u1x2

.

Analogously, we can define a vorticity component about the other two orthogonal axes. Thismotives the following definition.

Vorticity exists if the vector field changes in space Suppose that ~F = (u, v) is a velocity andconsider cases with constants , :

(u, v) = (0, x) = = ,(u, v) = (y, 0) = = ,

(u, v) = (y, x) = = + .The vorticity is a sum of the two and is the sum of the tendency for the fluid to rotate counterclockwise due to both components of velocity.

Suppose the velocity field is ~V (x, y) = (u(x, y), v(x, y)). If ~V is C1 in a small disk of radius 1 denoted by D with boundary D, then Greens theorem implies

D

~V d~x =

D

(v

x uy

)dxdy,

=

(v

x uy

)(P )area(D).

In the above we used the Mean Value Theorem to approximate the double integral with thevalue of the integrand at point P , somewhere in the disk. We can divide by the area and invertthis to get, (

v

x uy

)(P ) =

D

~V d~xarea(D)

=pi2

.

Therefore, we have that the circulation is equal to the vorticity times the area contained within.

Physical Interpretation of Greens Theorem:Suppose ~F is the velocity field of a fluid, say the velocity in your sink, then Greens theoremstates,

D~F d~x =

D

(F2x F1

y

)dxdy

The LHS is the line integral of the velocity field along a boundary D. It computes thetangental component of velocity and sums it up over all parts of the boundary. This yieldsthe circulation of ~F along the boundary D. The RHS is the double integral of the vorticity over the region D, which is the area con-

tained within the boundary D. This computes the vorticity at every point in the domainand then integrates (sums) this up over the entire domain.

38

Combining these two ideas yields the following interpretation for Greens theorem: Thetotal vorticity (like rotation) in D is equal to the circulation along the boundary D.

Figure 3: A cartoon to try and illustrate what Greens Theorem states.

example 2.34 Pick ~v = (y, x) with > 0 with units of 1/time. Previously, the field (flow) lines havebeen shown to be x2 + y2 = c2, for some constant Lc.

At (1, 0) the velocity is upwards and therefore the rotation is counter clockwise.The vorticity is easily computed,

=v2x v1

y= + = 2 > 0.

By Greens theorem the circulation is,

=

D

~F d~x =D

(F2x F1

y

)dxdy = 2pic2 > 0.

example 2.35 Pick ~v = ( yx2+y2

, xx2+y2

) on (x, y) 6= (0, 0) with > 0 with units of 1/time. The flowlines are the same as before, x2 + y2 = c2, for some constant c.

39

At (1, 0) the velocity is upwards and therefore the rotation is counter clockwise.The vorticity is computed to be zero,

=v2x v1

y= 0.

We can compute the circulation around a circle of radius c directly from the definition,

=

D~v d~x,

=

2pi0

1

c( sin t, cos t) (c sin t, c cos t) dt,

=

2pi0

1 dt,

= 2pi.

It can be shown that if we compute the circulation around any closed counter that containsthe origin, then we will get a non-zero value. This is in contrast to the circulation around anyclosed contour that does not contain the origin where the result is zero,

=

D

~F d~x =D

(F2x F1

y

)dxdy = 0.

The distinction in the two cases comes form the fact that the velocity is singular at the origin.

40


3 Surfaces and Surface Integrals

3.1 Parameterized Surfaces

Some discussion and Figure 3.1 form the course notes.

3.1.1 Surfaces as Vector-Valued Functions

Definition 3.1 A parameterized surface is a map ~g : D R2 R3 given by,~g(u, v) = (x(u, v), y(u, v), z(u, v))

Surface is denoted by and is the image of ~g. If ~g is C1 we call a C1-surface. Each component is C1.

example 3.2 Given f : D R where D R2 such that z = f(x, y), each (x, y) in R2 has a heightf(x, y). We can parameterize this surface as,

~g(u, v) = (u, v, f(u, v)).

Draw a picture like that of Figure 3.1 that shows the xyz space with D in the xy plane with (x, y)and the surface above with f(x, y) and some curvature to the surface.

example 3.3 Surfaces can also be defined implicitly by h(x, y, z) = 0, such as x2 + y2 + z2 = a2.

3.1.2 The Tangent Plane

We can fix one of the two parameters. For example if we set u = u0, then we get a curve thatpasses through (u0, v0),

~(v) = ~g(u0, v) = (x(u0, v), y(u0, v), z(u0, v)) .

Or, we can set v = v0 and get a different curve that also passes through (u0, v0),

~(u) = ~g(u, v0) = (x(u, v0), y(u, v0), z(u, v0)) .

Given what we already know we can find tangent vectors to the surface:

~Tu : ~(u) =

~g

u(u, v0),

~Tv : ~(v) =

~g

v(u0, v).

These two (usually) nonparallel vectors determine the tangent plane to the surface.A normal vector to the surface at (u0, v0) is

~N(u0, v0) = ~Tu(u0, v0) ~Tv(u0, v0),~N(u0, v0) = ~

(u0) ~(v0).

41

The unit normal vector can be written as,

~n =~Tu ~Tv~Tu ~Tv

Draw a picture of a surface with tangent vectors Tu, Tv.We require that the two tangent vectors are linearly independent otherwise this breaks down.

For a counter example consider the parameterization of a cone ~x = ~g(u, v) = (u cos v, u sin v, u).Compute tangent vectors and consider what happens at u = 0.

example 3.4 Given ~g(u, v) = (u, v, f(u, v)) with (u, v) D find two tangent vectors to the surface anda normal vector.Solution:We use the formulas above,

~g

u=

(1, 0,

f

u

), and

~g

v=

(0, 1,

f

v

).

Finally, the normal vector is the cross product of these two,

~N =~g

u ~gv,

=

i j k1 0 fu

0 1 fv

,=

(fu,f

v, 1

)We can check this with what we learned in MATH 237. If we define h(x, y, z) = z f(x, y) = 0 then anormal vector is found by the 3D gradient operation,

~h =(fu,f

v, 1

)To find the equation of the tangent plane we use the fact that any line in the tangent play is necessarilyorthogonal to the normal. Therefore, if ~g(u0, v0) = (u0, v0, f(u0, v0)) then

0 = ~N (~x ~g(u0, v0))0 =

(fu,f

v, 1

) (u u0, v v0, z f(u0, v0)),

z = f(u0, v0) + (u u0)fu

(u0, v0) + (v v0)fv

(u0, v0).

The equation of the tangent plane can be written as,

z = f(u0, v0) + (u u0)fu

(u0, v0) + (v v0)fv

(u0, v0).

42


3.1.3 Surface Area

Consider a surface parameterized by ~x = ~g(u, v) which is C1. The surface area can be built upusing the following steps:

Partition our domain into small rectanglesDuv using the standard Cartesian grid. This thenpartitions the surface into small surface elements (Draw a Figure like 3.6).

If u and v are sufficiently small then the surface element is very similar to a plane andthe edges are parallel vectors. Therefore, we can approximate the surface element with asmall plane parallelogram.

Recall: the area of a parallelogram defined by the vectors ~A and ~B is ~A ~B, that is to

say the magnitude of the cross product, ~A ~B = ~A B sin ,where is the angle between the two vectors.

Using our linear approximation we can obtain,

~A (u)~gu

(u0, v0), and ~B (v)~gv

(u0, v0),

Therefore, the surface area, S, is approximately,

S ~gu(u0, v0) ~gv (u0, v0)

uv Next we sum over all the surface elements determined by the partition we have chosen

and finally take the limit as N and u,v 0.This motivates the following definition for the surface area.

Definition 3.5 The surface area of the surface ~x = ~g(u, v), (u, v) Duv, where ~g is C1, is defined by

S =

Duv

~gu ~gv dudv.

Locally, this computes the tangent vectors at a point, the surface area of the parallelogram and thenintegrates that over all the points.

example 3.6 Calculate the surface area of the cone of radius b and height h. See example 3.4 of the coursenotes. The surface is given by

~g(r, ) = (br cos , br sin , hr)

with (r, ) Duv = {(r, ) : 0 r 1, 0 2pi}.

43

Solution:We compute the tangent vectors and their cross product:

~g

r= (b cos , b sin , h),

~g

= (br sin , br cos , 0),

~N = r(bh cos ,bh sin , b2).

The norm the unit vector isb2 + h2br. Therefore, the surface area is,

S =

1d,

=

10

2pi0

b2 + h2br ddr,

=b2 + h2bpi[r2]10 = pib

b2 + h2,

which is the surface area of a cone. Note that at r = 0, the point we have that one of the tangentvectors is the zero vector. That does not have a direction, and therefore where we cannot find anormal vector.

44


example 3.7 Suppose ~g : Duv R3 is defined by

~g(u, v) = a(sinu cos v, sinu sin v, cosu),

with Duv = {(u, v)|0 u pi, 0 v 2pi}.Verify that x2 + y2 + z2 = a2. [Parameterization of the sphere of radius a centred at the origin.]Then compute the surface area of the sphere.

Solution:We check the algebraic identity:

x2 + y2 + z2 = a2 sin2 u cos2 v + a2 sin2 u sin2 v + a2 cos2 u,

= a2 sin2 u+ a2 cos2 u,

= a2.

Before we can use the formula for surface area we need to compute the tangent vectors and thenormal vector,

~g

u= a(cosu cos v, cosu sin v, sinu),

~g

v= a( sinu sin v, sinu cos v, 0),

when then yields,

~g

u ~gv

= a2(sin2 u cos v, sin2 u sin v, sinu cosu) = (a sinu)~x.

Therefore, the surface element becomes,~gu ~gv = (a sinu) ~x = a2 sinu.

Finally, we substitute into the formula for the surface are,

d =

2pi0

pi0

a2 sinu dudv,

= 2pia2 [ cosu]pi0 ,= 4pia2.

Of course this confirms with what we already know.

3.1.4 Orientation of Surface

Given a surface described by ~x = ~g(u, v) on (u, v) Duv where ~g is one to one and C1 and thenormal vector is,

~N =~g

u ~gv,

45

is non-zero for all (u, v) Duv. Then the normal vector varies continuously and if it startspointing on one side and move it around it cannot ever point in the other direction at the samepoint. Surfaces where you can have this peculiar behaviour are non-orientable. An example is aMoebius band.

We will assume that our surfaces are orientable.When dealing with a surface with boundary is a piecewise smooth closed curve, we

relate the orientation of the two. In particular, when viewed form the side of on which thenormal vector points, the boundary is orientated counter-clockwise. [Kind of like the righthand rule for vectors.]

3.2 Surface Integrals

3.2.1 Scalar Fields

Suppose we have that the density of a thin fluid on a surface is with units of mass/unit area,then can we figure out what is the mass of the fluid?

We can revisit our derivation of surface area and make one minor modification. Instead ofsimply computing the surface area on each surface element, we would like to compute the mass,which is the product of the local density times the surface area. Following the same rational asbefore motivates the following definition.

Definition 3.8 If a surface R3 is parameterized by

~g(u, v) = (x(u, v), y(u, v), z(u, v)), (u, v) Duv R2,

and f : R is a continuous function. Then, the surface integral of f over is

fd =

Duv

f(~g(u, v))

~gu ~gv dudv.

If f has units of stuff/unit area then the surface integral of f is going to have units of stuff.Compare this to the line integral of a scalar function from before,

~g

fds =

ba

f(~g) ~g(t) dt.

On the LHS, we change the line for a surface and a ds for a d. On the RHS we change the normof the derivative of the path to the norm of the cross product of tangent vectors. Yes, this is morework but it is doable!

example 3.9 Example 3.5 of the course notes considers a sphere and asks to compute the surface integralof z2. There calculations are slightly longer but its a good exercise that I encourage you to go through toget more practice.

example 3.10 Find

x2 + y2 + 1 d where is the helicoidal surface,

~g(u, v) = (u cos v, u sin v, v), 0 u 1, 0 v 2pi.

46

Solution:We find the tangent vectors and normal vector:

~g

u= (cos v, sin v, 0),

~g

v= (u sin v, u cos v, 1),

~N = (sin v, cos v, u).Then we compute the surface integral,

x2 + y2 + 1 d =

10

2pi0

u2 + 1

1 + u2 dvdu,

=

10

(u2 + 1) du

2pi0

dv,

= [1

3u3 + u]102pi,

=8

3pi.

Theorem 3.11 Surface Integrals are independent of ParameterizationLet be a smooth surface and let g : Duv R be a real-value continuous function. Then surface integral

f d does not depend on the parameterization of the surface, proved the parameterization is smooth.


example 3.12 Find the formula for the integral

h d where is determined by the smooth function

z = f(x, y) defined on Duv R2 and h(x, y, z) : R3 R is continuous.Solution:The surface is parameterized by ~g(x, y) = (x, y, f(x, y)). Previously, we found the surface ele-ment for this type of function. Then, all we have to do is evaluate the integrand on the surfaceand multiply by the surface element, to obtain

h d =

Duv

h(u, v, f(u, v))

1 +

(f

u

)2+

(f

v

)2dA.

example 3.13 Find the surface area of a torus denoted by

~g(u, v) = ((R + cos v) cosu, (R + cos v) sinu, sin v), 0 u, v 2pi.where R, are constants. R denotes the radius of the torus from the centre and is the radius along eachsegment.

Solution: We compute the tangent vectors and the normal vector,

~g

u= ((R + cos v) sinu, (R + cos v) cosu, 0),

~g

v= ( sin v cosu, sin v sinu, cos v),

~N = (R + cos v) (cosu cos v, sinu cos v, sin v) .

47

The norm of the unit vector is ~N = (R + cos v)cos2 v + sin2 v = (R + cos v).Therefore, we can now evaluate the surface area,

S =

2pi0

2pi0

(R + cos v) dvdu,

=

2pi0

( 2pi0

(R + cos v) dv

)du,

=

2pi0

[(Rv + sin v)]2pi0 du,

=

2pi0

2piRdu,

= 4pi2R.

3.2.2 Vector Fields

Suppose that ~F is a velocity field defined in an around a surface that has unit outward normal~n. The dot product of these two vectors

~F (~g(u, v)) ~n(~g(u, v)),

projects the velocity field normal to the surface and yield the component of the velocity movingthrough the surface. We call this the flux of ~F through the surface at ~g(u, v). If we sum up thisquantity over an entire surface we get the total (net) outward flux of the velocity field throughthe surface,

~F ~n d.

When would we need this concept? Suppose we have water flowing and we want to find outhow much is flowing through a given region. That would be exactly what we described above.We will considering some applications but first lets figure out how to compute this formally.

Definition 3.14 Consider a C1 orientated surface given by ~g(u, v) with (u, v) D and a vector field~F continuous on the surface. The surface integral of ~F over is

~F ~n d =

Duv

~F (~g(u, v)) (~g

u ~gv

)dudv

Note:

~F ~n d =

Duv

~F (~g(u, v)) (~gu ~g

v

) ~gu ~gv

~gu ~gv dudv

and so the norms cancel and therefore why it doesnt appear in the surface integral of a vector field.

example 3.15 Compute the outward flux of ~F through the surface if ~F = (y,x, z2) and the surfaceis the helicoid ~g(u, v) = (u cos v, u sin v, v) with Duv = {(u, v) : 0 u 1, 0 v pi/2}

48

Solution:From a previous example we have that

~g

u ~gv

= (sin v, cos v, u).

Now, we can compute the flux,

~F ~n d =D

(u sin v,u cos v, v2) (sin v, cos v, u) dvdu,

=

10

pi/20

u+ uv2 dudv,

=

10

[uv +

1

3uv3]pi/2

0

du

=

10

[upi

2+

1

3u(pi

3

)3]du

=

[1

2u2pi

2+

1

6u2(pi

3

)3]10

=pi

4+pi3

48.


example 3.16 Compute the outward flux of ~F through the surface if ~F = (xyz, 0, 0) and the surfaceis the portion of the denoted by x2 + y2 + z2 = 4 in the first quadrant.

Solution:The surface mapping for the sphere is

~g(u, v) = (2 cos v cosu, 2 cos v sinu, 2 sin v)

onDuv = {(u, v) : 0 u pi/2, 0 v pi/2}

We compute the normal vector,

~g

u= (2 cos v sinu, 2 cos v cosu, 0),

~g

v= (2 sin v cosu,2 sin v sinu, 2 cos v),

~N = 4(cos2 v cosu, cos2 v sinu, cos v sin v

),

~N = 4 cos v (cos v cosu, cos v sinu, sin v) .

49

We now can compute the flux,

~F ~n d =D

(8 cos2 v sin v cosu sinu, 0, 0) 4 cos v (cos v cosu, cos v sinu, sin v) dvdu,

= 32

D

cos4 v sin v cos2 u sinu dudv,

= 32

pi/20

cos2 u sinu du

pi/20

cos4 v sin v dv,

= 32

[1

3cos3 u

]pi/20

[1

5cos5 v

]pi/20

,

=32

15.

example 3.17 Compute

~F n d with ~F = (x, 0, z) and is the surface of a cube bounded by

x = 0, y = 0, z = 0 and x = 1, y = 1, z = 1 and n is the outward normal.

Solution:If n is parallel to x, y, z maybe it is simpler to evaluate the flux using ~F n rather than using thedefinition that we presented.

We parameterize each surface,

1 : n1 = (0, 0,1), and ~g1(u, v) = (u, v, 0) 0 u, v 12 : n2 = (0, 0, 1), and ~g2(u, v) = (u, v, 1) 0 u, v 13 : n3 = (0,1, 0), and ~g3(u, v) = (u, 0, v) 0 u, v 14 : n4 = (0,1, 0), and ~g4(u, v) = (u, 1, v) 0 u, v 15 : n5 = (1, 0, 0), and ~g3(u, v) = (0, u, v) 0 u, v 16 : n6 = (1, 0, 0), and ~g4(u, v) = (1, u, v) 0 u, v 1.

With these it is easy enough to evaluate the line integrals over a vector field,1

~F n1 d =

1

(u, 0, 0) (0, 0,1) d = 0,2

~F n2 d =

2

(u, 0, 1) (0, 0, 1) d = 1,3

~F n3 d =

3

(u, 0, v) (0,1, 0) d = 0,4

~F n4 d =

4

(u, 0, v) (0, 1, 0) d = 0,5

~F n5 d =

5

(0, 0, v) (1, 0, 0) d = 0,6

~F n6 d =

6

(1, 0, v) (1, 0, 0) d = 1.

Therefore the line integral over the whole cube is

~F n d = 2.

From the above example we see that we can compute surface integrals of piecewise C1 sur-faces but it is a lot of work. Later, we will see that sometimes this can be rewritten in terms of avolume integral. In that case we only have one integral that can be easier to compute.

50

example 3.18 Compute

~F n d over given by,

~g(u, v) = (u, v, u2 + v2), D ={

(u, v)|u2 + v2 4}and ~F = (y3, x3, 3z2).

Solution:Compute the normal vector,

~g

u= (1, 0, 2u),

~g

v= (0, 1, 2v),

~g

u ~gv

= (2u,2v, 1).

Now, we can compute the surface integral of the vector field,

~F n d =

D

(v3, u3, 3(u2 + v2)2) (2u,2v, 1) dudv,

=

D

(2uv(u2 + v2) + 3(u2 + v2)2) dudv.

To evaluate this integral it us easier to transform to polar co-ordinates,Dr = {(r, )|0 r 2, 0 2pi}

~F n d = 2pi

0

20

2r4 cos sin + 3r4 rdrd,

=

2pi0

[1

3r6 cos sin +

1

2r6]2

0

d,

=26

6

2pi0

2 cos sin + 3 d,

=26

6

[cos2 + 3

]2pi0,

= 64pi.

3.2.3 Properties of surface integrals

Similar to line integrals, surface integrals posses the properties of linearity and additivity.Also, if the surface is piecewise C1 (the surface is a union of several surfaces who are each

C1), the integral of can be decomposed into the integral over each component.

3.2.4 Applications

Definition 3.19 The centre of mass (or centre of gravity) of a surface is a point on which balanceswhen supported at that point.

The point need not be part of the surface. For example consider a ring of constant density. Thecentre of mass is the centre, which is not part of the ring.

51

Mathematically, the three co-ordinates can be defined as

x =

x d

d

,

y =

y d

d

,

z =

z d

d

,

where is the density of the surface. Note that in each case we are dividing by the total mass ofthe surface.

These expressions are called moments and in particular they are the first moments since wehave linear functions of x, y, z in the numerator. In particular the numerators of the three ex-pressions are the moments with respect to yz-, xz- and xy-planes, respectively. These are oftendenoted with Myz,Mxz,Mxy.

Definition 3.20 The moment of inertial measures the rotational inertia of a body. That is the oppositionwe feel when we try and change the rotation of the body about a particular axis. The moments of inertiaabout the x, y, z axes are denoted by Ix, Iy, Iz and are the second moments, defined by,

Ix =

(y2 + z2) d,

Iy =

(x2 + z2) d,

Iz =

(x2 + y2) d.

example 3.21 Find the moment of inertia of a torus about the z axis if the density is constant and equalto one. Recall the torus can be defined as

~g(u, v) = ((R + cos v) cosu, (R + cos v) sinu, sin v), 0 u, v 2pi.where R, are constants.

Solution:Recall that we previously computed ~N = (R + cos v)cos2 v + sin2 v = (R + cos v).Therefore, we can compute the moment of inertia:

Iz =

(x2 + y2) d,

=

2pi0

2pi0

(R + cos v)3 dvdu,

=

2pi0

du

( 2pi0

(R3 + 3R2 cos v +3

2R(1 + cos 2v) + (1 sin2 v) cos v) dv

)= 2pi

[R3v + 3R2 sin v +

3

2R(v +

1

2sin 2v) (sin v 1

3sin3 v)

]2pi0

= 2pi2R[2R2 + 3

].

52


4 Gauss and Stokes Theorems

In this chapter we present the basics of Vector Differential Calculus. But first we review indicialnotation and add one more idea.

Index NotationHere we recall some indicial (tensoral) notation that we mentioned in the first week and definea few more.

We write vectors ~a as ai, where i is a free index. Dot product ~a ~b = aibi. Matrix multiplicaitonA~b = Aijbj . The alternating tensor ink is defined as

ijk =

1 if ijk = 123, 231, 312

0 if 2 or more indices repeat1 if ink = 321, 213, 132

Cross product: (sum over i and j but k is free)

(~a~b)k = ijkaibj

4.1 The vector differential operator4.1.1 Divergence and curl of a vector field

Definition 4.1 Let f : U R3 R be a differential function. The gradient is defined as,

~f = fxex +

f

yey +

f

zez =

f

xi

This is a vector where the coordinate in each direction corresponds to the partial derivative of the function.

Physical Interpretation of Gradient:This we have seen before, in MATH 237, shows us the direction in which the field is increas-

ing most rapidly. Think of it as the steepest part of a hill or valley.

Definition 4.2 Let ~F = (F1, F2, F3) : U R3 R3 be a differential function. The curl is,

~ ~F = ex ey ez

xy

x

F1 F2 F3

= (F3y F2

z,F1z F3

x,F2x F1

y

)= ijk

Fjxi

This is a vector quantity.

53

Physical Interpretation of CurlIf we consider the case where ~F = (u(x, y), v(x, y), 0) is a velocity field, then we see that there

is only one component of the curl that is nonzero, the vertical component is v/xu/y. This,as we have seen before, is the vorticity of a two dimensional flow. What is different is that this isa vector pointing in the z direction, which I point out is orthogonal to the xy-plane.

The other components are similarly defined but in different directions. For a velocity field~F = (u, v, w) then the x component of the curl is the vorticity about the x-axis due to motionin the yz plane. Finally, the y-component of curl is the vorticity about the y axis due to themotion in the xz-plane. That is why, if we assume ~F to be a velocity, then its curl, ~ ~F is thethree-dimensional vorticity, which shows us the local rotation rate about each axis.

Definition 4.3 Let ~F = (F1, F2, F3) : U R3 R be a differential function. The divergence is,

~ ~F =(

x,

y,

z

) (F1, F2, F3),

=

(F1x

+F2y

+F3z

),

=Fixi

.

This is a scalar quantity, which is the sum of partial derivatives.

Physical Interpretation of Divergence:This is something we have not seen before. To see what it does consider a couple of examples.

Suppose we have the two-dimensional case of ~F = (x, y). We saw that the field lines of thisconsists of paths that are moving away from the original at an exponential rate. We can say thepaths are diverging. Note that in this case the divergence is 2. If we change the sign of the RHSthen we will have that the paths will converge to the origin, and the divergence is 2.

Instead consider, ~F = (y, x). The solutions to this vector field are circular paths. If wecompute the divergence we find it is zero.

These are just examples but they should hopefully give you some intuition about what thedivergence of a vector field means. It gives us the idea as to how the solutions are going tospread (if divergence is positive), converge (if divergence is negative) or simply rotate around ifthe divergence is zero.


4.1.2 Identities involving ~G1

~(f + g) = ~f + ~g

G2~(fg) = f ~g + g~f

G3~(~F ~G) = (~F ~)~G+ (~G ~)~F + ~F (~ ~G) + ~G (~ ~F ),

54

D1~ (~F + ~G) = ~ ~F + ~ ~G

D2~ (f ~F ) = f ~ ~F + ~f ~F

D3

~ (~F ~G

)=

xkijkFiGj,

= GjFixk

ijk + FiGjxk

ijk,

= GjFixk

kij + FiGjxk

jki, permute indices

= ~G (~ ~F ) ~F (~ ~G),

C1~ (~F + ~G) = ~ ~F + ~ ~G

C2~ (f ~F ) = f ~ ~F + ~f ~F

C3~ (~F ~G) =

(~G ~

)~F

(~F ~

)~G+ ~F

(~ ~G

) ~G

(~ ~F

)Z1

~ (~f) = ~0Z2

~ (~ ~F ) = 0Note that above we saw terms like (~F ~)~G and (~G~)~F . These can be rewritten in component

form as,

~F ~ = (F1, F2, F3) (

x,

y,

z

),

= F1

x+ F2

y+ F3

z,

= Fk

xi.

It is very important to note that the ~ operator does not commute in general. For example~F ~ 6= ~ ~F .

Definition 4.4 The Laplacian2f of f is defined as

2f ~ (~f) = xi

f

xi=2f

x2i

55

Another important identify is the curl of the curl:

~(~ ~F

)= ~

(~ ~F

)2 ~F .

The Laplacian commutes with ~:L1

~(2f) = 2(~f)

L2~(2 ~F ) = 2(~ ~F )

L3~ (2 ~F ) = 2(~ ~F )

example 4.5 If ~F = ~ and ~ ~F = 0 then2 = 0, or2

x2+2

y2+2

z2= 0.

That is to say that satisfies Laplaces equation.

Theorem 4.6 Helmholtz DecompositionLet ~F be a C2 vector field on a bounded volume V R3. Then ~F can be decomposed into a curl-free anda divergent-free component:

~F = ~ + ~ ~.We define and ~ as the potential function (scalar) and the stream vector.

Note that if compute the divergence and curl we get,

2F = 2,

and~ ~F = ~ ~ ~ = 2~,

where we assume that the stream vector is divergence fee.

Physical Interpretation of Helmholtz decomposition:This theorem, which is not in the AMATH 231 course notes, is a simple and deep physical

interpretation. It says that every vector field (e.g. velocity or force fields) can be thought of as adivergent part and a rotational part. These two are distinct. A vector field could be only diver-gent, only rotational or a combination of the two. If turns out that if we impose the divergenceand curl of a vector field then we can determine the potential function and stream vector.

56


ApplicationsIn physics, electromagnetic fields are composed of electric fields ~E(x, y, z, t) and a magnetic

field ~H(x, y, z, t). By choosing the units appropriately Maxwells equations read,

~E

t= c~ ~H 4pi ~J,

~H

t= c~ ~E,

~ ~E = 4pi,~ ~H = 0.

Where is the charge density, ~J is the current vector and c is the speed of light in a vacuum.These are a famous example of a system of linear partial differential equations (PDEs). Thethree variables that appear are ~E, ~H, ~J . If we suppose that the current is set, then we have twounknowns but four equations. Isnt that over prescribed? In this case no because we need todetermine the rotational and divergent part of each of the electric and magnetic fields.

4.1.3 Expressing ~ in curvilinear coordinatesWe are very used to doing a lot of things in Cartesian coordinates. But, as we have previouslyseen, there is more to life than Cartesian coordinates. For example, there are cylindrical coordi-nates and spherical coordinates. Those are just two. One can define whatever coordinate systemthey want as long as the different variables are orthogonal in some sense.

In Cartesian coordinates our orthonormal basis is {e1, e2, e3} and the del operator takes theform,

~ = e1 x1

+ e2

x2+ e3

x3

A general curvilinear co-ordinate system (v1, v2, v3) inR3 is related to the Cartesian co-ordinatesystem by,

x = f(v1, v2, v3),

y = g(v1, v2, v3),

z = h(v1, v2, v3).

Alternatively, this could be written in vector form as ~x = ~F (~v).If k, l,m are constants then the following are curves (dimension one),

x = f(v1, l,m),

y = g(k, v2,m),

z = h(k, l, v3).

But we know how to find unit tangent vectors of any curves, based on what we learned inChapter 1, and they are,

ei =~xvi ~xvi

57

A convention is to define hi = ~xvi for i = 1, 2, 3.

Note that above we have three unit vectors. If they are linearly independent, then they canbe used to span a basic for R3. If they form a basis then any vector can be written as a sum ofvectors decomposed in terms of that basis.

For example, if we wanted to rewrite the gradient in terms of our new basis we can write,

~u =(~u ei

)ei

Note that above we are using indical (tensoral) notation to sum up over i = 1, 2, 3.We can also find the partial derivative of u with respect to each co-ordinate by using the

Chain Rule:

u

vi=

u

xj

xjvi

,

= ~u ~xvi

,

= hiei ~u.

This can be rewritten as,

ei ~u = 1hi

u

vi.

From this we can generalize to say that the gradient operator can be rewritten in terms of anybasis ei as,

~ = xi

=1

hiei

vi,

~ = ex x1

+ ey

x2+ ez

x3=

1

h1e1

v1+

1

h2e2

v2,+

1

h3e3

v3,

where we recall for completeness that,

hi =

~xvi , and ei = 1hi ~xvi .


Midterm.


Cylindrical CoordinatesDraw a Picture:

x = cos ,

y = sin ,

z = z.

58

This can be written compactly as,

~x(~v) = ( cos , sin , z),

where ~v = (, , z).We derive the unit vecto

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