Chapter 6 Introduction to Vector Calculus

1
Chapter 6 Introduction to Vector Calculus
6.6 Shorter Cut for div and curl 6.7
Integrals 6.8 Line Integrals 6.9 Gausss
Theorem 6.10 Stokes Theorem

• 6.1 Fluid Flow
• 6.2 Vector Derivatives
• 6.3 Computing the divergence
• 6.4 Integral Representation of Curl
• 6.5 The Gradient

2
6.1 Fluid Flow

• Consider a fluid moves through a pipe, what is
  the relationship between the motion of the fluid
  and the total rate of flow through the pipe (take
  a rectangular pipe of sides a and b)?

Now consider the flow through a surface that is
tilted at an angle to the velocity,
the volume that moves past this flat but tilted
surface is
? is the angle between the direction of the
fluid velocity and the normal to the area. This
invites the definition of the area as a vector
3
General Flow, Curved Surfaces

• Now consider the fluid velocity to be some
  function of position, and the surface doesnt
  have to be flat.
• Describe the coordinates on the surface by the
  angle as measured from the midline. Divide the
  surface into pieces that are rectangular strips
  of length a and width b?k/2.

The velocity field is height dependent,
, so the contribution
to the flow rate through this piece of the
surface is
with
4

• Put the pieces together, we then obtain

The total flow is the sum of these over k and
then the limit
Note the result is the same as that for the flat
surface calculation!
5
6.2 Vector Derivatives

• We can describe the flow of a fluid by specifying
  its velocity field (a concept of Vector Field),

One of the uses of ordinary calculus is to
provide information about the local properties of
a function without attacking the whole function
at once. That is what derivatives do. The
geometric concept of derivative is the slope of
the curve at a point. the tangent of the angle
between the x-axis and the straight line that
best approximates the curve at that point.
6

• If a small amount of dye is injected into the
  fluid at some point, it will spread into a volume
  that depends on how much the dye is injected. As
  time goes on this region will move and distort
  and possibly become very complicated, sometimes
  too complicated to grasp in one picture.

Assume that the initial volume of dye forms a
sphere of (small) volume V and let the fluid move
for a short time, you would probably observe
1. In a small time t the center of the sphere
will move. 2. The sphere can expand or contract,
changing its volume. 3. The sphere can rotate. 4.
The sphere can distort.
7
Div, Curl, Strain

• Let us consider the first one, the motion of the
  center, the information will tell you about the
  velocity at the center of the sphere.
• The second one, the volume, gives new
  information. You can simply take the time
  derivative dV/dt to see if the fluid is
  expanding or contracting just check the sign and
  determine if its positive or negative. Thats
  not yet in a useful form because the size of this
  derivative will depend on how much the original
  volume is. If you put in twice as much dye, each
  part of the volume will change and there will be
  twice as much rate of change in the total volume.
  If the time derivative is divided by the volume
  itself this effect will cancel. Therefore, we have

8

• Look at the third way that the sphere can rotate.
  Again, take a very small sphere. The time
  derivative of this rotation is its angular
  velocity, the vector . In the limit as the
  sphere approaches a point, this will tell the
  rotation of the fluid in the immediate
  neighborhood of that point.
• The fourth way that the sphere can change is that
  it can change its shape. In a very small time
  interval, the sphere can slightly distort into an
  ellipsoid. This will lead to the mathematical
  concept of the strain. How much information is
  needed to describe the strain?
• Assume the sphere changes to an ellipsoid,
  what is the longest axis and how much stretch
  occurs along it thats the three components of
  a vector. After that what is the shortest axis
  and how much contraction occurs along it? Thats
  one more vector, but you need only two new
  components to define its direction because its
  perpendicular to the long axis. The total number
  of components needed for this object is
• 32 5.

9
6.3 Computing the divergence

• How to calculate the divergence and compute the
  time derivative of a volume from the velocity
  field?
• Recall the change rate of a volume
• Pick an arbitrary surface to start with and see
  how the volume changes as the fluid moves. In
  time ?t a point on the surface will move by a
  distance v?t and it will carry with it a piece of
  neighboring area A. This area sweeps out a volume
  

10

• If at a particular point on the surface the
  normal is more or less in the direction of
  the velocity then this dot product is positive
  and the change in volume is positive.
• The total change in volume of the whole initial
  volume is the sum over the entire surface of all
  these changes. Divide the surface into a lot of
  pieces with accompanying unit normals
  , then
• The limit of this as all the
• The circle through the integral designates an
  integral over the whole closed surface and the
  direction of is always taken to be outward.
  Finally, divide by and take the limit as
  approaches zero

11

• The is the rate at which the
  area dA sweeps out volume as its carried with
  the fluid. Its a completely general expression
  for
• the rate of flow of fluid through a fixed surface
  as the fluid moves past it.
• Use the standard notation in which the area
  vector combines the unit normal and the area
• If the fluid is on average moving away from a
  point then the divergence there is positive. Its
  diverging!

12
The Divergence as Derivatives

• Express the velocity in rectangular components,
  . For the small
  volume, choose a rectangular box with sides
  parallel to the axes. One corner is at point (x0,
  y0, z0) and the opposite corner has coordinates
  that differ from these by (?x, ?y, ?z). Expand
  everything in a power series about the first
  corner. Instead of writing out (x0, y0, z0) every
  time, abbreviate it by (0).

13

• There are six integrals to do, one for each face
  of the box, and there are three functions, vx,
  vy, and vz to expand in three variables x, y, and
  z.
• If you look at the face on the right in the
  sketch you see that its parallel to the y-z
  plane and has normal . When you
  evaluate only the vx term
  survives flow parallel to the surface (vy, vz)
  contributes nothing to volume change along this
  part of the surface.
• Write the two integrals over the two surfaces
  parallel to the y-z plane, one at x0 and one at
  x0 ?x.

14

• The minus sign comes from the dot product because
  points left on the left side. Evaluate
  these integrals by using their power series
  representations. Take the first integral
• Now put in the second integral, all the terms in
  the above expression that do not have a ?x in
  them will be canceled. The combination of the two
  integrals becomes

15

• The other integrals are the same except that x
  becomes y and y becomes z and z becomes x. The
  integral over the two faces with y constant are
  then
• Add all three of these expressions, divide by
  the volume,
• Take the limit as the volume goes to zero V
  ?x?y?z, then

Note The symbol ? will take other forms in other
coordinate systems.
16
Simplifying the derivation

• The surfaces that have constant values of the
  coordinates are planes in rectangular
  coordinates planes and cylinders in cylindrical
  planes, spheres, and cones in spherical. In every
  one of these cases the constant coordinate
  surfaces intersect each other at right angles.
• The volume elements for these systems come
  straight from the drawings, just as the area
  elements do in plane coordinates. In every case
  you can draw six surfaces, bounded by constant
  coordinates, and surrounding a small box. Because
  these are orthogonal coordinates you can compute
  the volume of the box easily as the product of
  its three edges.

17

• In the spherical case, one side is .
  Another side is . The third side is not
  it is . The
  reason for the factor is that the arc
  of the circle made at constant r and constant
  is not in a plane passing through the origin.
  It is in a plane parallel to the x-y plane, so
  it has a radius .

18
Simplifying the derivation

• Now for the derivation of the divergence, the
  essence is that you find
• on one side of the box in the
  center of the face, and multiply it by the area
  of that side. Do this on the other side,
  remembering that isnt in the same
  direction there, and combine the results.
• Do this for each side and divide by the volume
  of the box, such as

Do this for the other sides, add, and you get the
result. But what if you need to do it in
cylindrical coordinates?
When everything is small, the volume is close to
a rectangular box, so its volume is
19

• The top and bottom present nothing significantly
  different from the rectangular case.

The curved faces of constant r are a bit
different, because the areas of the two opposing
faces arent the same.
Now for the constant sides. Here the areas of the
two faces are the same, so even though they are
not precisely parallel to each other this doesnt
cause any difficulties.
The sum of all these terms is the divergence
expressed in cylindrical coordinates.
20
Divergence in spherical coordinates
The coordinate system is orthogonal if the
surfaces made by setting the value of the
respective coordinates to a constant intersect at
right angles. In the spherical example this means
that a surface of constant r is a sphere. A
surface of constant ? is a half-plane starting
from the z-axis. These intersect perpendicular to
each other. If you set the third coordinate, f ,
to a constant you have a cone that intersects the
other two at right angles.
21
6.4 Integral Representation of Curl

• The calculation of the divergence was facilitated
  by the fact
• could be
  manipulated into the form of an integral
• The similar expression for the curl is

What is that surface integral doing with a
instead of a . ? Just replace the dot product by
a cross product in the definition of the
integral. This time however you have to watch the
order of the factors.
U a vector field that represents pure rigid body
rotation. You are going to take the limit as V?0,
so it may as well be uniform.
22

• Choose a spherical coordinate system with the
  z-axis along ?.

Divide by the volume of the sphere and you have
2? as promised.
23
The Curl in Components

• With the integral representation,
  ,
• available for the curl, the process is much like
  that for computing the divergence.

In the above equation you have
on the right face and on the
left face. This time replace the dot with a cross
(in the right order). On the right
On the left
24

• When you subtract the second from the first and
  divide by the volume, ,
  what is left is (in the limit
  a derivative.

Similar calculations for the other four faces of
the box give results that you can get simply by
changing the labels x ?y?z?x, a cyclic
permutation of the indices. The result can be
expressed in terms of ?.
25
6.5 The Gradient
The gradient is the closest thing to an ordinary
derivative, taking a scalar-valued function into
a vector field. The simplest geometric definition
is the derivative of a function with respect to
distance along the direction in which the
function changes most rapidly, and the direction
of the gradient vector is along that most-rapidly
changing direction.
26
6.6 Shorter path for div and curl
There is another way to compute the divergence
and curl in cylindrical and rectangular
coordinates. The only caution is that you have to
be careful that the unit vectors are inside the
derivative, so you have to differentiate them too.
For example is the divergence of
, and in cylindrical coordinates
27

• But for ,
  will change with . This can be noticed by
  first showing that

and differentiating with respect to ?. This gives
Put all these together, from we obtain
This agrees with
28
Note

• Similarly, we can find the derivatives of the
  corresponding vectors in
• spherical coordinates. The non-zero values are

The result is for spherical coordinates
The expressions for the curl are (this is left
for homework)
29
6.7 Integrals
Consider the integral
The basic idea to do this is that you first
divide a complicated thing into little pieces to
get an approximate answer. You then refine the
pieces into still smaller ones to improve the
answer and finally take the limit as the
approximation becomes perfect.
Divide the curve into a lot of small pieces, then
if the pieces are small enough you can use the
Pythagorean Theorem to estimate the length of
each piece.
By using a parametric representation of the curve
30
The integral for the length becomes
where v is the speed.
Think of this as
By using
If the curve is expressed in polar coordinates,
the integral for the length of a curve is then
By using ? as a parameter
31
Weighted Integrals
Let us examine one loop of a logarithmic spiral
The length of the arc from ? 0 to ? 2p is
The time for a particle to travel along a short
segment of a path is dt ds/v where v is the
speed. The total time along a path is of course
the integral of dt.
How much time does it take a particle to slide
down a curve under the influence of gravity?
32
1. Take the straight-line path from (0, 0) to
(x0, y0). The path is y x y0/x0
33
2. Take another path for which its easy to
compute the total time. Drop straight down in
order to pick up speed, then turn a sharp corner
and coast horizontally. Compute the time along
this path and it is the sum of two pieces.
34
6.8 Line Integrals
Work, done on a point mass in one dimension is an
integral
If the system is moving in three dimensions, but
the force happens to be a constant, then work is
simply a dot product
The general case for work on a particle moving
along a trajectory in space is a line integral
along an arbitrary path for an arbitrary force.
Divide the specified curve into a number of
pieces, at the points . Between points k
-1 and k you had the estimate of the arc length
as , but here you
need the whole vector from to in
order to evaluate the work done as the mass moves
from one point to the next. Let
, then
35
How do you evaluate these integrals? Start with
the simplest method and assume that you have a
parametric representation of the curve
, then and the integral
is
This an ordinary integral over t
Start with , take the dot product
with and manipulate the expression.
The integral of this from an initial point of the
motion to a final point is
This is the work-energy theorem. Note in most
cases you have to specify the whole path, not
just the endpoints.
36

• For example, consider that
  , what is the work
  done going from point (0, 0) to (L, L) along the
  three different paths indicated?

37
Gradient
38
6.9 Gausss Theorem
Recall the original definition of the divergence
of a vector field
39
The reason for this is that each interior face of
volume is matched with the face of an
adjoining volume . The latter face will
have pointing in the opposite direction,
so when you add all the interior surface
integrals they cancel. All thats left is the
surface on the outside and the sum over all those
faces is the original surface integral.
In the above equation multiply and divide every
term in the sum by the volume
Now taking the limit as all the
approach zero. The quantity inside the brackets
becomes the definition of the divergence of
and you then get
Gausss Theorem
40
Example
Verify Gausss Theorem for the solid hemisphere,
Use the vector field
Doing the surface integral on the hemisphere,
and on the bottom
flat disk,
. The surface integral is then in two pieces,
41
Now do the volume integral of the divergence
The two sides of the Gausss theorem agree.
42
6.10 Stokes Theorem
The expression for the curl in terms of integrals
is
Use exactly the same reasoning as that was used
in the case of the Gausss theorem, this leads to
43
Here we use the subscript 1 for the top surface
and 2 for the surface around the edge.
Look at around the thin edge. The
element of area has height and length
along the arc. Call the unit normal out
of the edge.
The product
, using the
property of the triple scalar product. The
product is in the direction along
the arc of the edge, so
44
Note
Put all these pieces together and you have
45
C
46
Multiply and divide each term in the sum by
and you have
Now increase the number of subdivisions of the
surface, finally taking the limit as all the
, and the quantity inside the
brackets becomes the normal component of the curl
of in

. The limit of the sum is
the definition of an integral, so
Stokes Theorem
47
Example
Verify Stokes theorem for that part of a
spherical surface
Use for this example the vector field
First we calculate the curl of the field
Only the component of the curl because the
surface integral uses only the normal ( )
component. The surface integral of this has the
area element .
48
The other side of Stokes theorem is the line
integral around the circle at angle 0.
The two sides of the theorem agree.
49
Conservative Fields
An immediate corollary of Stokes theorem is that
if the curl of a vector field is zero throughout
a region, then line integrals are independent of
path in that region. To state it a bit more
precisely, in a volume for which any closed path
can be shrunk to a point without leaving the
region, if the curl of equals zero, then
depends on the endpoints of
the path, and not on how you get there. To see
why this follows, take two integrals from point a
to point b.
50
The difference of these two integrals is
This equations happens because the minus sign is
the same thing that you get by integrating in the
reverse direction. For a field with
, Stokes theorem says that this closed path
integral is zero, and the statement is proved.
The converse of this theorem is also true. If
every closed-path line integral of is
zero, and if the derivatives of are
continuous, then its curl is zero. Stokes
theorem tells you that every surface integral of
is zero, so you can pick a point and a
small at this point. For small enough
area whatever the curl is, it wont change much.
The integral over this small area is then
, and by assumption this is zero. Its
zero for all values of the area vector. The only
vector whose dot product with all vectors is zero
is itself the zero vector.
51
Potentials
The relation between the vanishing curl and the
fact that the line integral is independent of
path leads to the existence of potential
functions.
If in a simply-connected domain
(thats one for which any closed loop can be
shrunk to a point), then can be expressed
as a gradient, -grad . The minus sign is
conventional. That line integrals are independent
of path in such a domain means that the integral
is a function of the two endpoints alone. Fix
and treat this as a function of the upper limit
. Call it . The defining
equation for the gradient is
52
Compare the last two equations and because
is arbitrary, you can immediately get
53
Vector Potentials
When a vector field has zero curl then its a
gradient. When a vector field has zero divergence
then its a curl. In both cases the converse is
simple, and its what you see first
In both cases, there are extra conditions needed
for the statements to be completely true. To
conclude that a conservative field (
) is a gradient requires that the domain be
simply-connected, allowing the line integral to
be completely independent of path. To conclude
that a field satisfying can
be written as requires
something similar that all closed surfaces can
be shrunk to a point.
54
Note Neither the scalar potential nor the vector
potential are unique. You can always add a
constant to a scalar potential because
the gradient of a scalar is zero and it doesnt
change the result. For the vector potential you
can add the gradient of an arbitrary function
because that doesnt change the curl.
Example Verify that
is a vector potential for the uniform field
.