0. Chapter 7 Introduction to Vector Calculus
7.6 Shorter Cut for div and curl
7.7 Integrals
7.8 Line Integrals
7.9 Gauss’s Theorem
7.10 Stokes’ Theorem
7.1 Fluid Flow
7.2 Vector Derivatives
7.3 Computing the divergence
7.4 Integral Representation of Curl
7.5 The Gradient

1. 7.1 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:

2. General Flow, Curved Surfaces
Now consider the fluid velocity to be some function of position, and the surface doesn’t 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

3. 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!

4. 7.2 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.

5. 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.

6. Div, Curl, Strain
1. 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.
2. The second one, the volume, gives new information. You can simply take the time derivative dV/dt to see if the fluid is expanding (dV/dt >0) or contracting (dV/dt <0. That’s 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

7. 3+2= 5 (a symmetric matrix will do the job).
3. 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.
4. 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 — that’s the three components of a vector. After that what is the shortest axis and how much contraction occurs along it? That’s one more vector, but you need only two new components to define its direction because it’s perpendicular to the long axis. The total number of components needed for this object is
3+2= 5 (a symmetric matrix will do the job).

8. 7.3 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

9. 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

10. The is the rate at which the area dA sweeps out volume as it’s carried with the fluid. It’s 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. It’s diverging!

11. 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).

12. 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 it’s 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.

13. 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

14. Add all three of these expressions, divide by the volume,
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.

15. 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.

16. In the spherical case, one side is. Another side is
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
volume
area

17. 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 is not 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

18. 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 aren’t the same.
Now for the faces of constant . Here the areas of the two faces are the same, so even though they are not precisely parallel to each other this doesn’t cause any difficulties.
The sum of all these terms is the divergence expressed in cylindrical coordinates.

19. Divergence in spherical coordinates
The spherical coordinate system is also 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 φ is a half-plane starting from the z-axis. These intersect perpendicular to each other. If you set the third coordinate, θ , to a constant you have a cone that intersects the other two at right angles.
volume
area

20. 7.4 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. However you have to watch the order of the factors.
Let v be 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.

21. Choose a spherical coordinate system with the z-axis along ω.
Divide by the volume of the sphere and you have 2ω as promised.

22. The Curl in Components On the left 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

23. When you add the second and 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 ▽.

24. 7.5 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.

25. 7.6 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
Note: The unit vectors , , and don’t change as you alter r or z. They do change as you alter θ (except for )
i.e.,

26. and differentiating with respect to θ. This gives
But for , will change with This can be noticed by first showing that
and differentiating with respect to θ. This gives
Put all these together, from
we obtain
This agrees with

27. Note: The result is for spherical coordinates
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):

28. 7.7 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

29. where v is the speed. The integral for the length becomes
Think of this as
where v is the speed.
By using
If the curve is expressed in polar coordinates, the integral for the length of a curve is then
By using θ as a parameter

30. Weighted Integrals Let us examine one loop of a logarithmic spiral:
The length of the arc from  = 0 to  = 2π 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?

31. 1. Take the straight-line path from (0, 0) to (x0, y0)
1. Take the straight-line path from (0, 0) to (x0, y0). The path is y =x ·y0/x0

32. 2. Take another path for which it’s easy to compute the total time
2. Take another path for which it’s 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. Note

33. 7.8 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

34. Start with , take the dot product with and manipulate the expression.
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.

35. Line Integration with Mathematica
For example, consider that , what is the work done going from point (0, 0) to (L, L) along the three different paths indicated?
Line Integration with Mathematica

36. Gradient
What is the line integral of a gradient? Consider that
. The integral of the gradient is then
where the indices represent the initial and final points. That is when you integrate a gradient, you need the function only at its endpoints. The path doesn’t matter.
Is this statement sound familiar with the work-energy theorem.

37. 7.9 Gauss’s Theorem
Recall the original definition of the divergence of a vector field:
Fix a surface and evaluate the surface integral of over the surface:
Now divide this volume into a lot of little volumes, with individual bounding surfaces If you do the surface integrals of
over each of these pieces and add all of them, the result is the original surface integral.

38. 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 that’s 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
Gauss’s Theorem

39. Example Verify Gauss’s 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,

40. Note: Hint: Now do the volume integral of the divergence and consider
The two sides of the Gauss’s theorem agree.
Integration of Vector Field over a Surface with Mathematics
Applying Divergence Theorem with Mathematica

41. 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 Gauss’s theorem, this leads to
Let us first apply it to a particular volume, one that is very thin and small. Take a tiny disk of height , with top and bottom area Let be the unit normal vector out of the top area. For small enough values of these dimensions, is simply the value of the vector inside the volume times the volume itself.

42. Take the dot product of both sides with , and the parts of the surface integral from the top and the bottom faces disappear. That’s just the statement that on the top and the bottom, is in the direction of , so the cross product makes perpendicular to
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

43. Note: Put all these pieces together and you have
Divide by and take the limit as Recall that all the manipulations above work only under the assumption that you take this limit.
This form is easier to interpret than was the starting point with a volume integral. The line integral of is called the circulation of around the loop. Divide this by the area of the loop and take the limit as the area goes to zero and you then have the circulation density of the vector field.

44. The component of the curl along some direction is then the circulation density around that direction. Notice that dictates the right-hand rule that the direction of integration around the loop is related to the direction of the normal
Pick a surface A with a boundary C. The surface doesn’t have to be flat, but you have to be able to tell one side from the other. Divide the surface into a lot of little pieces , and do the line integral of
around each piece. Add all these pieces and the result is the whole line integral around the outside curve. C
As before, on each interior boundary between area and the adjoining , the parts of the line integrals on the common boundary cancel because the directions of integration are opposite to each other. All that’s left is the curve on the outside of the whole loop.

45. If is a conservative field, then
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 .
The limit of the sum is the definition of an integral, so
Stokes’ Theorem
Note: if , then
If is a conservative field, then

46. 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

47. The other side of Stokes’ theorem is the line integral around the circle at angle 0.
The two sides of the theorem agree.
Applying Stokes’ Theorem with Mathematica

48. 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 (simply-connected domain), 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.

49. The converse of this theorem is also true
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 won’t change much. The integral over this small area is then , and by assumption this is zero. It’s zero for all values of the area vector. The only vector whose dot product with all vectors is zero is itself the zero vector.
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 difference of these two integrals is

50. 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 (that’s 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

51. Note: How does the integral change when you change a bit?
Compare the last two equations and because is arbitrary, you can immediately get

52. Vector Potentials
When a vector field has zero curl then it’s a gradient. When a vector field has zero divergence then it’s a curl. In both cases the converse is simple, and it’s what you see first:
We can construct the function because It is also possible, if , to construct the function such that
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.

53. Example: 1. Verify that is a vector potential for the uniform field .
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 doesn’t change the result.
For the vector potential you can add the gradient of an arbitrary function because that doesn’t change the curl. .
Example: 1. Verify that is a vector potential for the uniform field
2. Prove that
is conservative, and find a scalar potential such that