A tiny bit of review ...
Remember that if we start with a function, f, and if
∇=(∂/∂x)i+(∂/∂y)j+(∂/∂z)k,
then grad f is ∇f, a vector field:
(∂f/∂x)i+(∂f/∂y)j+(∂f/∂z)k.
If F=Pi+Qj+Rk is a vector field then
curl F is ∇xF. Note that (with a bit of effort!) this is
(Q_{z}R_{y})i(P_{z}R_{x})j+(P_{y}Q_{x})k.
This is horrible, and difficult to remember but the notation
(especially the cross product and ∇) is designed to help.
Suppose we ∂/∂x the i component, ∂/∂y the
j component, and ∂/∂z the k component. Here in
this 21^{st} century (!) math course, I seem to just be doing
this for fun. The people who actually invented these results had the
computations forced on them (really!) because they wanted
descriptions of certain aspects of reality involving fluid flow and
electromagnetism. These are the three results we get for the
differentiations:
(Q_{zx}R_{yx}) (P_{zy}R_{xy}) (P_{yz}Q_{xz}).
I wrote these on the board, in a vertical way, and asked people to
look at them. A number of students observed that if we added them, the
result would be 0! This is weird and wonderful (or weird and
remarkable, a phrase which might
be used either positively or negatively about many parts of this
course). Our conclusion is that a vector field G for which
∇·G is not 0 cannot be the curl of another
vector field. This is a "compatibility condition" for being a curl.
Example (from a textbook problem attempted last night)
The vector field <0,0,z^{2}> is not the curl of
another vector field, because 0+0+2z is not always 0.
Divergence
Suppse F is
F=Pi+Qj+Rk, a vector field. Then
the divergence of F is a function,
∇·F=∂P/∂x+∂Q/∂y+∂R/∂z.
It is also called div F.
In fluid dynamics, this quantity sort of represents the
source rate of the fluid (more fluid if positive, less fluid if
negative) at a point.
Statement of the Divergence Theorem
Suppose E is a solid bounded region in space (R^{3}) and S is
the boundary of E, with N the outward pointing normal on
S. Suppose also that F is a vector field with differentiable
coefficients. Then:
∫∫_{S}F·N dS=∫∫∫_{E}div F dV.
The ingredients
Here S divides up space, R^{3}, into two pieces (examples
follow). One of the pieces is a bounded region, E. The surface S is
always oriented in this "scenario" to have its N pointing
outward, which means away from the bounded region E.
The vector field, F(x,y,z), should be
P(x,y,z)i+Q(x,y,z)j+R(x,y,z)k and P, Q, and R
should be differentiable functions. The divergence of F is
∇·F:
(∂P/∂x)+(∂Q/∂y)+(∂R/∂z).
Simple examples of regions and surfaces Most "concrete" computations with the Divergence Theorem will likely involve fairly simple shapes.


 

Fluids and the Divergence Theorem equation
The equation ∫∫_{S}F·N dS=∫∫∫_{E}div F dV
itself has meaning in fluid dynamics. The righthand side is the net
flow(flux) in/out of region E (in fact, if the region is a cube, it is
commonly called a flow box in that field). The lefthand side
is the triple integral of the local source rate of the fluid flow.
Proving the Divergence Theorem for the unit cube
I wanted to "demystify" the Divergence Theorem by explaining why it is
true for the unit cube in R^{3}.
The unit cube is a parallelopiped whose vertices (corners) have entries 0 or 1. There are 8 vertices: (0,0,0), (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1), and (1,1,1). There are 12 edges. Edges join at two vertices whose coordinates differ by one entry. There are 6 faces, each obtained by holding one coordinate equal either to 0 or to 1. By the way, the unit cube and its generalizations in higher dimensions turn out to be very interesting. One reason is the existence of Gray codes.
Now the triple integral side of the Divergence Theorem is ∫∫∫_{The cube}(∂P/∂x)+(∂Q/∂y)+(∂R/∂z) dV. I will split this into three separate integrals, and analyze each part.
Following the suggestion of Mr. Forrest, we looked at
∫∫∫_{The cube}(∂Q/∂y) dV.
We decided to write dV here as dy dA_{x,z}. The reason
for this is that the integration with respect to y will "undo" the
∂/∂y. Several students saw this, and I was happy that they
did.
The innermost integral is then ∫_{y=0}^{y=1}(∂Q/∂y) dy. The Fundamental Theorem of Calculus immediately applies and we get Q(x,y,z)]_{y=0}^{y=1}=Q(x,1,z)Q(x,0,z). We then integrate both of these: ∫∫_{x,z between 0&1}Q(x,1,z) dA_{x,z}∫∫_{x,z between 0&1}Q(x,0,z) dA_{x,z} Now look at the surface. The (x,1,z) part of the surface integral has j as normal, and the (x,0,z) part of the surface has j as normal. The surface integral of F·n will be Q(x,0,z) and will be +Q(x,1,z). In both cases, x and z will range from 0 to 1. The Fundamental Theorem of Calculus yields a minus sign when "stuff" is at the lower end of the integral. Geometrically, we get a minus sign on part of the boundary because the normals are directed outward. 
Let's look at ∫∫∫_{The cube}(∂R/∂z) dV. I would write dV as dz dA_{x,y}. The
Fundamental Theorem of Calculus would apply to the innermost
integral: ∫_{z=0}^{z=1}(∂R/∂z) dz=Q(x,y,z)]_{z=0}^{z=1}=Q(x,y,1)Q(x,y,0). Again integrate both of these: ∫∫_{x,y between 0&1}Q(x,y,1) dA_{x,y}∫∫_{x,y between 0&1}Q(x,y,0) dA_{x,z} The minus sign comes from the Fundamental Theorem of Calculus and it comes from the +/orientation of the normals. 
Finally the last term is ∫∫∫_{The cube}(∂P/∂x) dV. I hope that you see dV here should be written as
dx dA_{y,z}. Then the Fundamental Theorem of Calculus
applies and we've got this: ∫_{x=0}^{x=1}(∂P/∂x) dx=P(x,y,z)]_{x=0}^{x=1}=P(1,y,z)Q(1,y,z). Each of these terms needs to be integrated with respect to y and z from 0 to 1. The + part (that is, at (1,y,z) in the cube) has a normal vector of i and the  part (that is, at (0,y,z) in the cube) has a normal vector of i. So this part of the triple integral, after using the Fundamental Theorem of Calculus, gives the flux over the two indicated pieces of the boundary of the cube. 
If now we add up the three pieces of the triple integral we will get the surface integral of F·N over the boundary of the cube with the "correct" (outward) orientation. I wanted to tell you that the Divergence Theorem is a version of the Fundamental Theorem of Calculus, and that the signs checked. Algebraically, the signs occur because of FTC and ]. Geometrically, they come from the outward choices.
Two old computations redone
We introducted flux computations on December 3. The first example I
gave was:
Suppose our vector field is F=<xy,z^{2},3>. What is the total outward flux of F through the surface of the unit cube, 0≤x≤1, 0≤y≤1, and 0≤z≤1.Our answer was: "... the total flux is 1/2." The computation was not difficult but it was a bit tedious. Now let's do it using the Divergence Theorem. Well, div F=y+0+0, not too difficult. And then the triple integral were supposed to compute is ∫∫∫_{The cube}y dV. If we order the d's as, say, dy dx dz, then dy gives us y^{2}/2 and the limits give us 1/2. This "integrates" to 1/2 dx from 0 to 1, and then 1/2 again dz. The answer is indeed 1/2, and with a tiny effort you could even do the computation "in your head".
Here is the other example we considered then.
If F(x,y,z)= x^{2}i+yzj4zk, and the surface is the sphere of radius 5 centered at the origin, what is the total flux of F through this sphere (directed outwards).Please look at what we did on December 3. The computation was a good deal of work. We can also use the Divergence Theorem on this problem.
A textbook problem
Here is a standard textbook problem in the Divergence Theorem section
of a U.S. calculus book. The reasoning needed for this problem
resembles some of the problems we handled with Green's Theorem. Here
the vector field is
F(x,y,z)=z arctan(y^{2})i+z^{3}ln(x^{2}+1)j+zk.
We need to find the flux of F across the part of the paraboloid
x^{2}+y^{2}+z=2 which lies above the plane z=1 and is
oriented upward.
Discussion and solution
The divergence of F is 0+0+1: we've gotten rid of a great deal
of mess! In fact, it is the presence of the ludicrous (?) functions
arctan(y^{2}) and ln(x^{2}+1) which sort of signals
me, declaring that I'd better try to compute the desired quantity
indirectly. Of course, it doesn't also hurt (!) that the problem
occurs at the end of the section discussing the Divergence Theorem!
The paraboloid is z=2x^{2}y^{2}: it "opens" down. The vertex or top is at (0,0,2). The normals to the paraboloid vary a great deal. While it might be possible to compute the flux directly, the Divergence Theorem states that the integral of 1 (that's div F) over the solid region above z=1 and below z=2x^{2}y^{2} will equal the flux through the parabolic cap plus the flux through the disc on the plane z=1. That disc has radius 1, centered at the origin, since the boundary is 1=2x^{2}y^{2} or x^{2}+y^{2}=1. Also the outward normal on the disc is constant because the disc is flat, and the outward normal is k.
Let's compute the triple integral: ∫∫∫_{The cup}1 dV. Probably this is
simplest to compute with cylindrical coordinates. θ will go from 0
to 2Π, and r will go from 0 to 1. That's a polar description of the
base of the solid. What's the height? The bottom is at z=1, and the
top is at z=2x^{2}y^{2}, or (in "polar")
z=2r^{2}. So we compute
∫_{θ=0}^{θ=2Π}∫_{r=0}^{r=1}∫_{z=1}^{z=2r2}1 dz r dr dθ=∫_{θ=0}^{θ=2Π}∫_{r=0}^{r=1}(2r^{2}1)r dr dθ=
∫_{θ=0}^{θ=2Π}∫_{r=0}^{r=1}(rr^{3})dr dθ=
∫_{θ=0}^{θ=2Π}(r^{2}/2r^{4}/4)]_{r=0}^{r=1}dθ=∫_{θ=0}^{θ=2Π}(1/4)dθ=Π/2.
Now the surface integral over the "bottom" disc. F·N is (z arctan(y^{2})i+z^{3}ln(x^{2}+1)j+zk)·(k) which is z. But z=1 on this disc, so we need to integrate 1 over a disc bounded by a circle of radius 1: the answer is Π, 1 multiplied by the area of the area.
We now have: Π/2 (the divergence integral) is equal to the flux over the paraboloid plus Π (the flux over the disc). Therefore the flux over the paraboloid must be (3Π)/2.
Other uses
While textbook problems are (sometimes) nice, more interesting uses of
the Divergence Theorem include a discussion of heat transfer and
Gauss's Law for electric charges. We have no time, but I just want to
remark that the results are remarkable and really interesting.
Gauss's Law is discussed in many physics books. Also you can look at
pages 1037 and 1038 of the textbook. The heat equation, which tries to
describe heat transfer, is discussed in lots of engineering courses
and in several math courses, including Math 421 which many of the
students in this class likely will take.
FTC through the ages ...
We considered this at the beginning of the lecture. The three
semesters of calculus are a tour of results originating around 1630 or
so to around 1870 or so. We go through, therefore, 250 years of
mathematical development, and certainly this semester, multivariable
calculus, has had its share of really clever ideas. The Fundamental
Theorem of Calculus appeared in the first semester. This semester had
a 2dimensional version (Green's Theorem), a 2.5dimensional version
(Stokes' Theorem), and even today a 3dimensional version, the
Divergence Theorem. Wow! And now we are all the way up to the late
Nineteenth Century! Imagine if you took a series of
chemistry or physics or bio courses which would have left you at that
time of the discipline. Just think ...
Thank you for discussing all of this really interesting stuff with me this semester.
Some notation which many people think is helpful
The conditions P_{y}=Q_{x} and
P_{z}=R_{x} and Q_{z}=R_{y} may be
difficult to remember. Some new notation which is supposed to help
follows. Here is a quote from a web page entitled Earliest Uses of
Symbols of Calculus:
The vector differential operator, now written as an upsidedown delta, ∇, and called nabla or del, was introduced by William Rowan Hamilton (18051865). 
So "del" is given by: ∇=(∂/∂x)i+(∂/∂y)j+(∂/∂z)k. Here are some uses:
( i j k ) det ( ∂/∂x ∂/∂y ∂/∂z ) ( P(x,y,z) Q(x,y,z) R(x,y,z) )It turns out this is equal to (if I don't foul up the signs!) the following vector field:
The ingredients for Stokes' Theorem
Stokes' Theorem was developed in response to ideas of electromagnetism
and fluid dynamics. Just like Green, Stokes was interested in both
mathematics and physics, and he attempted to construct mathematical
models for rapidly evolving fields of physics. I will attempt to list
the ingredients for a (relatively!) straightforward version of Stokes'
Theorem.
So this is a curve in space (R^{3}) with START=END and which has no other selfintersections. 
• How the surface and curve interact (by their orientations)
The word "orientation" here means how to select t, the unit
tangent vector on the boundary curve, and N, the unit normal on
the surface. The boundary curve will be a parameterized curve. It has
a unit tangent vector, t, pointing in the direction of
increasing parameter value. If we "walk" along the boundary curve in
this direction, the surface should be to our left. Now we have
t and a direction to the left. Complete this to a righthanded
coordinate system. The selection of N, the unit normal vector
to the surface, is made so N points in the direction of the
last entry of a righthanded coordinate system which begins with
t and the inside surface direction. I think in the accompanying
picture to the right, the N would point "out" of the page, and
towards the "inside" of the cupshaped surface.
Under these conditions, then the Stokes Theorem Equation is true:
∫_{The boundary curve}F·t ds=∫∫_{The surface}curl F·N dS
The textbook writes this in a slightly different way as
∫_{The boundary curve}F·ds=∫∫_{The surface}curl F·dS
So the work or circulation of F around the boundary is equal to the flux through the surface of the curl F. This is a wellknown complicated theorem. If the curve is in R^{2} and the surface is the inside of the curve, then the result is "just" Green's Theorem, which is already quite complicated. I'd like to spend most of the time in this lecture just checking both sides of the Stokes' Theorem equation, and getting some familiarity with it that way.
A textbook problem
Here is a problem from a calculus textbook:
Verify that Stokes' Theorem is true for the vector field
F(x,y,z)=y^{2}i+xj+z^{2}k
and the surface is the part of the paraboloid
z=x^{2}+y^{2} that lies below the plane z=1, oriented
upward.
Some discussion
The plane z=1 intersects the paraboloid in a circle. This is a circle
of radius 1 centered at (0,0,1). The paraboloid "overlays" a region
inside a circle of radius 1 centered at the origin in the xyplane. We
will compute both integrals in Stokes' Theorem and (I hope!) get the
same answers. If the paraboloid is "oriented upward" then I presume
that the N points up. Going around the blue circle in the
standard (counterclockwise/positive) direction will orient the
boundary curve "compatibly": the t, the leftish piece of
surface next to the boundary curve, and the up N form a
righthanded triple. This took some time to see in class.
The work integral
So I need to compute ∫_{The curve}y^{2}dx+x dy+z^{2}dz.
The curve is a circle, and can be parameterized as:
x=1cos(t) dx=sin(t)dt y=1sin(t) dy=cos(t)dt z=1 dz=0and the parameterization interval for the whole circle is [0,2Π]. Then ∫_{The curve}y^{2}dx+x dy+z^{2}dz becomes
The surface integral
Now we need to compute
∫∫_{The paraboloid}curl F·N dS.
The curl
This is ∇xF, so:
( i j k ) det( ∂/∂x ∂/∂y ∂/∂z )=0i0j+(12y)k ( y^{2} x z^{2} )Parameterizing the surface, etc.
( i j k ) det( 1 0 2u )=2ui2vk+1k ( 0 1 2v )We discussed the magical cancellation in the last lecture. V·N dS became V·(r_{u}xr_{v}) dA_{u,v}. curl F here is (12y)k=(12v)k so that curl F·N dS=(12v)k·(2ui2vk+1k)dA_{u,v}=(12v)dA_{u,v}.
Computation of the surface integral
We need to identify the domain in the uvplane which parameterizes our
little cup. The uvplane is the xyplane in different clothing, but
the cup is the graph over the region inside the unit circle:
u^{2}+v^{2}≤1. So we need
∫∫_{Inside the unit circle}(12v)dA_{u,v}
But the 2v integrates to 0, since the region is symmetric in v and 2v is "odd" (the + and  cancels totally). The 1 in the integrand just gives the area, and the area inside the unit circle is Π(1^{2}), and this is Π.
This instantiation (?) of Stokes' Theorem is verified: Π=Π.
Another textbook problem
Here is a slightly more vicious (viscous?) problem from the Stokes'
Theorem section of a calculus text by Robert A. Adams:
Find ∫∫_{The surface}curl F·N dS where the surface is that part of
the sphere x^{2}+y^{2}+(z2)^{2}=8 which lies
above the xyplane, and N is the outward unit normal on the
surface, and F is
y^{2}cos(xz)i+x^{3}e^{yz}je^{xyz}k.
Since the problem occurs in the Stokes' Theorem section of the text we
should probably use Stokes' Theorem. The region of the sphere is shown
to the right. The sphere is centered at (0,0,2) and its radius is
sqrt(8)=2sqrt(2). So a portion of the sphere extends below the
xyplane.
The boundary of the top portion occurs if z=0 in the
equation
x^{2}+y^{2}+(z2)^{2}=8. Then
x^{2}+y^{2}+(2)^{2}=8 and
x^{2}+y^{2}=4. This is a circle of radius 2 centered
at the origin in the xyplane. We should establish the orientation of
this circle. If we look closely at a small piece of the surface near
the boundary curve, the outward unit normal points slightly down. We
must "walk" along the curve so that the surface is to the left. The
t direction is the standard counterclockwise direction on the
boundary circle. I hope the local picture to the right helps to
convince you of that. Again, the problem of deciding the resulting
orientation of one chunk (surface, boundary curve) from the other
(boundary curve, surface) seemed in class to be the most complicated
qualitative aspect of this problem.
Now Stokes' Theorem applies:
∫∫_{The spherical surface}curl F·N dS=∫_{The boundary circle}F·t ds.
But notice:
this circle is also the correctly oriented boundary of the disc of
radius 2 centered at the origin in the xyplane. So I can use Stokes'
Theorem a second time to change the line integral to a much
simpler surface integral:
∫_{The boundary circle}F·t ds=∫∫_{The disc}curl F·N dS
This is simpler for several reasons. The region over which we're
integrating is flat, a disc in the xyplane. The correctly oriented
normal, N, is just k. I hope the picture convinces you
of that.
We should compute curl F. Wait, we just need to compute
the k part of curl F:
( i j k ) det ( ∂/∂x ∂/∂y ∂/∂z )=Blah!iBlah, blah!j+[3x^{2}e^{yz}2ycos(xz)]k ( y^{2}cos(xz) x^{3}e^{yz} e^{xyz})A further simplification occurs. We're on the xyplane, where z=0. So the k component, 3x^{2}e^{yz}2ycos(xz), becomes 3x^{2}2y because cos(0)=1 and e^{0}=1.
Comment
I did this problem because using the same boundary
curve to switch surfaces is a very common "trick" done in
electromagnetism and fluid flow. If two surfaces have the same
boundary and if the vector field is nice, then the flux of the curls
of the vector fields through the two surfaces must be the same. This
is weird and wonderful, and people use it. See the discussions on page
1024 and 10261027.
Green's Theorem If the boundary curve is in R^{2} and the "surface" is a region in R^{2} then Stokes' Theorem is Green's Theorem. Why is this true? If the simple closed curve is oriented counterclockwise as usual, then the normal will be +k. So if the vector field is Pi+Qj+Rk, the normal N is k and the k component of the curl of the vector field is Q_{x}P_{y}. The Stokes' Theorem equation declares that the integral of Pdx+Qdy over the boundary curve (with the usual orientation) equals the double integral of Q_{x}P_{y} over the interior. 
Surface integrals
If we believe a piece of surface in R^{3} could be a
mathematical model of a thin curved plate, then the plate could have
varying density. We could add up the density multiplied by the area to
get mass. This is a surface integral:
∫∫_{Region in (u,v)}(Density) dS.
Textbook example
Let me try an artificial (textbook!) example. Here is
problem 22 of section 16.4. It asks us to calculate
∫∫_{Rect}f(x,y,z) dS if the surface is given by
r(u,v)=<u cos(v),u sin(v),v>, Rect is the
rectangle in u and v where 0≤u≤1 and 0≤v≤2π, and
f(x,y,z)=sqrt(x^{2}+y^{2}).
This whole example is arranged so that things will work out "well".
The surface is called a helicoid.
If we hold v constant, then
the curves with u varying are straight line segments of length
1, at height v (a number between 0 and 2π), radiating out from the
zaxis. The curve drawn in green shown in the
picture to the right is what happens when v=2.5. So it is the curve
(cos(2.5)u,sin(2.5)u,2.5) for u between 0 and 1.
If we hold u constant, then the curve is actually a helix wrapping
around the zaxis. The curve drawn in blue shown in
the picture to the right is what happens when u=.3. So it is the curve
(.3cos(v),.3sin(v),v) for v between 0 and 2π.
Computation
Since
r(u,v)=<u cos(v),u sin(v),v>, we have
r_{u}=<cos(v),sin(v),0>
r_{v}=<u sin(v),u cos(v),1>
The cross product:
( i j k ) r_{u}xr_{v}=det( cos(v) sin(v) 0 ) ( u sin(v) u cos(v) 1 )So this is sin(v)i–(cos(v))j+(u(cos(v))+u(sin(v))^{2})k=sin(v)icos(v)j+uk. The magnitude of this vector is sqrt(1+u^{2}). So here dS=sqrt(1+u^{2}) du dv.
The function or "density" is sqrt(x^{2}+y^{2}) but in the helicoid parameterization, x=u cos(v) and y=u sin(v). So sqrt(x^{2}+y^{2}) accidentally (!) turns out to be u because again cos^{2}+sin^{2}=1.
The integral ∫∫_{Rect}f(x,y,z) dS is also accidentally ∫_{v=0}^{v=2Π}∫_{u=0}^{u=1}u sqrt(1+u^{2}) du dv. By total coincidence, we can compute this exactly. It is 2Π(sqrt(2)/31/3). Wow. Wowie. I guess this is terrific (and all totally arranged, of course).
Nontextbook example
What's the surface integral of x^{3}+y^{2}+z of that
part of the graph z=3x^{2}+5y^{4} which is over the
unit square, 0≤x≤1 and 0≤y≤1? The parameterization is the
straightforward
r(u,v)=ui+vj+(3u^{2}+5v^{4})k
so that r_{u}=1i+6uk and
r_{u}=1j+20v^{3}k and the
crossproduct r_{u}xr_{v} is
6ui20v^{3}j+1k. (I did this in
class. If you weren't there [$40} please check the result!). Therefore
dS is sqrt(1+36u^{2}+400v^{6}) du dv (not so
agreeable, huh?).
And x^{3}+y^{2}+z changes to
u^{3}+v^{2}+3u^{2}+5v^{4}. So we just need to compute
∫_{v=0}^{v=1}_{u=0}^{u=1}(u^{3}+v^{2}+3u^{2}+5v^{4})sqrt(1+36u^{2}+400v^{6}) dv du
Maple knows a lot more about antidifferentiation than I do, and it can't "do" this, although on my home PC it takes about a second to admit this. With evalf, we get 26.484 as an approximate numerical value in less than a tenth of a second.
Most surface integrals can't be evaluated exactly.
This is a pity, since flux, an important physical quantity, is defined to be a surface integral. So we need to discuss flux, and, just as before, "Strangely enough , it all turns out well."
Flux
The horrible factor r_{u}xr_{v} makes a "random" surface integral almost
impossible to compute in terms of antidifferentiations involving
familiar functions. It is very nice that the surface integrals of most
interest in physical and engineering problems are not "random" but
result from computations of flux, and it turns out that the horrible
factor disappears for such computations.
Suppose we have a vector field, F in R^{3}. We could
imagine a surface in R^{3}, and then try to see how the flow
of the vector field interacts with the surface. The picture to the
right is quite imaginary. I've never seen the arrows of a vector
field, and I want the surface, sort of like a net, not to give any
resistance to the imaginary arrows. It is, of course, an imaginary
surface.
The flux is the net flow through the surface.
What do these words mean?
net
The direction the fluid flows means something. It is possible that at
some points the fluid crosses the surface in different directions. We
should have some way of giving a sign to the flow, left to right/right
to left, inward/outward, and then totaling the different
contributions, with signs, to see whether the net flow is
positive or negative.
through
The flow through the surface is important. The same piece of surface
("dS") can have different flux, even if the vector field is constant
 always the same direction and magnitude. What can then change is
the angle of the dS piece relative to the flow. If it is perpendicular
to the flow, there will be the most flux. If the dS is parallel to the
flow, there will be no flux. In between, there will be some "in
between" amount. In fact, if you think about this, the amount of flux
will depend on the cosine of the angle the surface makes with the
vector field. We can compute this with F·N where N is a unit vector normal or
perpendicular to the surface. Notice that the choice of N is
important. N is also a unit normal, and using that choice will
change the sign of the flux.
The pictures above are supposed to be crosssectional so they cut the
surface perpendicularly. The flux is the normal component, and
the work is the tangential component. So in this lecture and
also in the last two, we will be studying the normal component.
The whole
surface
If we want to compute the net flux through the whole surface, then we
will need to assign unit normal vectors at every point of the
surface. There are some surfaces which can't accept such
assignments. The simplest example is the Mobius strip
(take a long rectangle, make a halftwist in the long direction, and
attach the short edges together). If you give an N at any one
point, and then follow around the assignment continuously, when you
get back to the point, you'll discover that you have reversed the
normal! So there will be no nice way to define and compute flux
through surfaces which don't permit nice "assignments" of normals. I
will assume such a problem will not occur in the remainder of this
course (hey, it doesn't for planes and spheres and toruses and
... almost anything you will encounter in applications).
The specific technical words used are at the beginning of section
16.5: our surfaces will be oriented surfaces, where it is
possible to specify a choice of normal continuously at every point of
the surface. So the Mobius strip is not an oriented surface.
An example on a cube
Suppose our vector field is F=<xy,z^{2},3> and
the surface we're interested in is the surface of the cube defined by
0≤x≤1, 0≤y≤1, and 0≤z≤1. It is not an accident that
this surface is an example of a closed surface which divides
all of space, R^{3}, into two pieces, points inside and points
outside the surface. This is similar to a closed curve in
R^{2}. Such surfaces are important in physical
applications. We will choose the normal to be the outwardpointing
unit normal, so flows out are positive. I would like to compute the
total flux of F through the surface of this cube. There are a
total of 6 faces. I'll do one face carefully, and then the others more
rapidly.
Magical cancellation!
If our surface is parameterized there is a natural way to get a
unit normal. Just take r_{u} and r_{v}:
these are velocity vectors for curves on the surface, and are tangent
to the surface. Their crossproduct will be perpendicular to the
surface. If we then normalize (divide by the magnitude) we'll get an
acceptable N. So we can take N to be
r_{u}xr_{v}
divided by the scalar r_{u}xr_{v}. If Flux=∫∫_{The surface}F·N dS this will be the same as
∫∫_{The surface}[F·(r_{u}xr_{v})/r_{u}xr_{v}] r_{u}xr_{v} dA_{uv}
Look at the marvelous cancellation (much the same as what occurred in
the line integral case).
Therefore the flux integral is
∫∫_{The surface}F·(r_{u}xr_{v}) dA_{uv}.
While a "plain" surface integral needs to be very carefully prepared
to be "computable" (as the two examples considered earlier show), the
cancellation here means no horrible square root terms, and many flux
integrals should be computable. Let me now compute another flux over a
closed surface.
An example on the sphere
Suppose F is x^{2}i+yzj4zk and I
would like to know the flux through the sphere of radius 5 centered at
the origin. We considered this surface before and
there we learned
r_{u}xr_{v}=25cos(u)[sin(v)]^{2}i25sin(u)[sin(v)]^{2}j25sin(v)cos(v)k.
Is this the inward or outward pointing normal? Look at v=Π/2 and
u=0. Then (substitute  we did this in class!) the point on the
surface is (5,0,0) and the vector we just computed is
<25,0,0>. This points inward and we are therefore
considering flow in to be positive. This is different from the
previous cube example, and that's o.k. as long as we realize what we
are doing!
Also, the parameterization itself was x(u,v)=5cos(u)sin(v); y(u,v)=5sin(u)sin(v); z(u,v)=5cos(v). This means that F described in (u,v) terms becomes 25sin(u)^{2}sin(v)^{2}i+25sin(u)sin(v)cos(v)j20cos(v)k
We need to integrate F·N dS,
but this, because of the cancellation of the horrible factor r_{u}xr_{v} becomes just F·(r_{u}xr_{v}) dA_{u,v}. Let
me match up the components and get the integrand:
625sin(u)^{2}sin(v)^{4}cos(u)625
sin(u)^{2}sin(v)^{3}cos(v)+500cos(v)^{2}sin(v).
This must be integrated from 0 to 2Π in u and from 0 to Π in v.
I emphasized in class that this integration is not impossible, even by
hand. You should at least vaguely remember integrating powers
of sines and cosines: they really weren't too hard. Maple used about a tenth of a second of CPU
time to tell me the value of this double integral: (2000/3)Pi. Next
week, I'll show you how to get the value of this flux integral using
the Divergence Theorem with practically no effort!
Problem 18 in section 16.5 This is a more qualitative problem. There's a picture given, resembling what is to the right: a half of a circular cylinder whose axis of symmetry is the zaxis, and the halfcylinder is the "forward" half, where x≥0. An n, a unit normal, is given, and this determines the selection of the unit normal at all other points (it will be the normal pointing away from the zaxis). The problem asks to determine whether the flux integral ∫∫_{halfcylinder}F·dS representing the total flux is positive, negative, or zero. The statement adds, "Explain your reasoning." Several F's are given. I didn't have time to discuss this in class. I believe (hope?) that the answers below are correct, and they are done using simple reasoning about dot products and symmetry.

Line integrals and work are to
Green's Theorem
as
Surface integrals and flux are
to Stokes' Theorem and the Divergence
Theorem
How to describe a plane
Suppose I give you a point, p=(3,1,2), and two vectors,
A=<4,1,3> and B=<1,5,2>. I would like to
describe algebraically the plane which contains the point p in the
direction of the vectors A and B. Here is how we did
this problem earlier in the course.
An implicit description of the plane
Compute the crossproduct of A and B:
( i j k ) det( 4 1 3 )=17i5j+21k. ( 1 5 2 )It is easy to check (I just did with some dot products) that the resulting vector is perpendicular to both A and B and is a normal vector to the plane we'd like to describe. A point with coordinates (x,y,z) is on this plane if
An explicit description
Change the point p=(3,1,2) to a position vector,
P=<3,1,2>. Then every point on this plane has a unique
description as P plus some multiple of A plus some
(possibly other) multiple of B. Your text calls these
multiplies u and v, so I will also. So the plane is anything which can
be written as P+uA+vB. We can write this with
more details:
<3,1,2>+u<4,1,3>+v<1,5,2>=<3+4u+1v,1u+5v,2+3u+2v>. The
textbook writes this as a vector position function:
r(u,v)=(3+4u+1v)i+(1u+5v)j+(2+3u+2v)j
Notation comment
The textbook uses Φ(u,v) for the vectorvalued position
function.
I am going to write r(u,v) instead, since I have trouble with
capital Greek letters.
The i component is called x(u,v), the j component is
called y(u,v), and the k component is called z(u,v), Each point
on the plane has a unique "address" in terms of the pair of numbers
(u,v). This is like the central Manhattan street grid: streets and
avenues. So 3^{rd} Avenue and 47^{th} Street is a
unique point, and every intersection has a unique address.
A sphere
Creating a "mapping" from a piece of the (u,v) plane to a surface in
R^{3} is frequently called a coordinate chart (although
not in your book, where it is called a
parameterization). Getting a coordinate chart for a plane is
not difficult. Getting (useful!) coordinate charts for curved surfaces
can be much harder. Here's an example which is only moderately
difficult because we have studied spherical coordinates: a sphere of
radius 5 centered at the origin in R^{3}.
Now let's use spherical coordinates, with ρ fixed at 5. We see that
a point on the sphere will be described by (I use u for θ and v for φ):
x(u,v)=5cos(u)sin(v); y(u,v)=5sin(u)sin(v); z(u,v)=5cos(v).
I want points on the sphere to have unique (u,v) addresses, and the conventional choice for restriction of the (u,v) domain is 0<=u<=2π and 0<=v<=π.
Horizontal lines in the (u,v) domain become circles parallel to the
xyplane on the sphere (latitudes?). Vertical lines in the (u,v)
domain, where v varies and u is fixed, become half great circles on
the sphere, going from the "North Pole" to the "South Pole". The
region in the (u,v) domain where 0<=u<=π and π/2<=v<=π
becomes the lower (z<=0) and "forward" (y>=0) quarter sphere.
A torus
I'll try now to give a parametric description of a torus. So the torus
(the surface of a doughnut) will be a surface which is gotten when a
circle perpendicular to the xyplane, and radially oriented, is
revolved around the zaxis. One view of what I'm trying to describe is
to the right. The center of the circle to be revolved around the
zaxis is moved so that it describes a circle in the xyplane centered
at the origin. There are two more views below of the surface. One is
from "above", looked down the zaxis. The other is from the "side",
looking along the yaxis.
I can give points on the torus a unique "address" in terms of two
numbers. These two numbers will represent angles. One angle will be
θ, but I'll call it u to agree with the text.
I'll make this specific torus more definite: the length of the vector
from the zaxis to the center of the circle will be 4, and the radius
of the circle being revolved around the zaxis will be 2. The left
picture below shows the angle u. The right picture shows a slice
perpendicular to the xyplane, through the vector of length 4. I will
specify a point on the torus by letting v be the angle made by a line
from the point on the torus to the center of the circle compared to
the xyplane itself.
The z coordinate of the point only involves v. The z coordinate is the
"opposite" side of a right triangle with acute angle v and hypoteneuse
2. So z=2sin(v). The vector from the origin to the center of the
circle has length 4. The angle v adds on another 2cos(v) to that
length. But then we use this total length, 4=2cos(v), along with the
angle u (secretly, θ) to get the x and y coordinates. So
x=(4+2cos(v))cos(u) and y=(4+2cos(v))sin(u).
Therefore the torus will be given by the following vectorvalued function:
r(u,v)=x(u,v)i+y(u,v)j+z(u,v)k with
x(u,v)=(4+2cos(v))cos(u) and
y(u,v)=(4+2cos(v))sin(u) and z(u,v)=2sin(v).
The domain of the coordinate chart which will give each point on the
torus surface a unique (u,v) address is 0<=u<=2π and
0<=u<=2π.
In the 21^{st} century I can check using a picture what I've just written. Here is a Maple command and its output. I should remark, if you've never been in my office watching me try to use Maple, that typing this command and looking at its output took 5 attempts before I was successful. Such wonderful effort may explain why I wouldn't trust myself to use Maple in class while people watched and giggled, and we all wasted time.
plot3d([4+2*cos(v))*cos(u),4+2*cos(v))*sin(u),2*sin(v)], u=0..2*Pi,v=0..2*Pi, scaling=constrained,axes=normal);
Finally, we can consider horizontal lines across the domain of the
coordinate chart, r(u,v), and ask what kinds of curves these
create on the torus. These lines, where u changes and v is unchanged,
become circles around the zaxis. Similarly, the vertical lines in the
domain of r become lines "around" the torus, for fixed u=θ.
If we restrict to region where 0<=u<=π and π<=v<=2π,
then the part of the torus which results is "forward" of the xzplane
(where y>=0) and below the xyplane (where z<=0). This is a
quarter of the torus.
A graph
Here I considered a rather simple surface given by a graph:
z=3x^{2}+5y^{4}. Certainly all I expected people to
see and say was this this surface is (vaguely) cupshaped, and its
lowest point is at (0,0,0). There's a really simple way to
parameterize such surfaces: just use x and y. So
r(u,v)=ui+vj+(3u^{2}+5v^{4})k. The
parameterization geometrically consists of pushing up a grid from the
xyplane until it hits the graph.
Surface area
Here once again was used the wonderful integral calculus mantra:
Cut up, approximate, sum, limit
Small rectangles in the uv domain, of dimension Δu by Δv
were changed to approximate parellelograms. The curve with u varying
and v fixed has a tangent vector, and that tangent vector is
r_{u}, that is, ∂r/∂u (take the
partial derivative of the component functions with respect to u). The
text calls this T_{u} but I think we have lots of T's
in this course. The tangent vector along the curve with v varying and
u fixed is r_{v}. The resulting area of the approximate
parellelogram is obtained by taking a cross product (you need to
remember a much earlier result, which is that the area of a
parellelogram is the magnitude of the cross product of vectors forming
adjacent sides), and we get r_{u}xr_{v}ΔuΔv. Then add these up and take
the limit. The surface area for a part of the surface which is
parameterized by a domain, D, in the uvplane turns out to be
∫∫_{D}r_{u}xr_{v} du dv. The
textbook calls r_{u}xr_{v} du dv by the
name, dS, for a piece of surface area. It will also be useful to
realize that r_{u}xr_{v}, a crossproduct of two
vectors tangent to the u and v coordinate curves on the surface, is
a vector normal to the surface.
The surface area of the sphere
Here
r(u,v)=5cos(u)sin(v)i+5sin(u)sin(v)j+5cos(v)k
Therefore
r_{u}=5sin(u)sin(v)i+5cos(u)sin(v)j+0k
and
r_{v}=5cos(u)cos(v)i+5sin(u)cos(v)j5sin(v)k
( i j k ) r_{u}xr_{v}=det(5sin(u)sin(v) 5cos(u)sin(v) 0 ) ( 5cos(u)cos(v) 5sin(u)cos(v) 5sin(v) )The i component is just 25cos(u)[sin(v)]^{2} and the j component is 25sin(u)[sin(v)]^{2} and the k component is 25[sin(u)]^{2}sin(v)cos(v)25[cos(u)]^{2}sin(v)cos(v). The last component simplifies (sin(u)^{2}+cos(u)^{2}=1) and the result is
The surface area of the torus You could try this yourself. The answer is 32π^{2}, and a detailed explanation is below.
Let's try this:
r(u,v)=(4+2cos(v))cos(u)i+(4+2cos(v))sin(u)j+2sin(v)k. ( i j k ) r_{u}xr_{v}=det((4+2cos(v))sin(u) (4+2cos(v))cos(u) 0 ) ( 2sin(v)cos(u) 2sin(v)sin(u) 2cos(v) )The i component is 8cos(v)cos(u)+4cos(v)^{2}cos(u)=4cos(v)(2cos(u)+cos(v)cos(u)). This is 4cos(v)cos(u)(2+cos(v)). The j component is 8cos(v)sin(u)+4[cos(v)]^{2}sin(u)=4cos(v)(2sin(u)+cos(v)sin(u)). This is 4cos(v)sin(u)(2+cos(v)). The k component is 8[sin(u)]^{2}sin(v)+4[sin(u)]^{2}cos(v)sin(v)+8sin(v)[cos(u)]^{2}+4[cos(u)]^{2}sin(v)cos(v). This simplifies (again with sin^{2}+cos^{2}) to 8sin(v)+4sin(v)cos(v)=4sin(v)(2+cos(v)). I have never done this computation before, but I hope it will work out. Now let us square and add the components: 4^{2}cos(v)^{2}cos(u)^{2}(2+cos(v))^{2}+ 4^{2}cos(v)^{2}sin(u)^{2}(2+cos(v))^{2}+ 4^{2}sin(v)^{2}(2+cos(v))^{2} Again sin^{2}+cos^{2} is used, and we get: 4^{2}cos(v)^{2}(2+cos(v))^{2}+ 4^{2}sin(v)^{2}(2+cos(v))^{2} One last time, sin^{2}+cos^{2}, and this becomes: 4^{2}(2+cos(v))^{2} This is r_{u}xr_{v}^{2}, so we take the square root and get 4(2+cos(v)) (notice that 2+cos(v) is always nonnegative so the square of its square is itself). Now to get the area of the whole torus we need to integrate this over the uvsquare which is [0,2π] by [0,2π]. So the area is ∫_{0}^{2π}∫_{0}^{2π}4(2+cos(v)) du dv. The answer is 32π^{2}. This answer is correct. A torus is such a symmetric figure that its surface area can be determined using a (sort of) calculusfree method called the Theorem of Pappus. 
Evaluating line integrals
I basically know three ways to evaluate line integrals:
George Green, whose name is attached to this result (and to others in mathematical physics) was essentially selftaught and his biography is fascinating. See here and here.
Rectangular Green's Theorem
I stated a result called Green's Theorem for a rectangle with
corners at (0,0), (a,0), (a,b), and (0,b).
The result was the following:
∫_{Rectangular bdry}P(x,y)dx+Q(x,y)dy=∫∫_{Inside of rect}∂Q(x,y)/∂x∂P(x,y)/∂y dA.
Silly example
I first tried a rather silly example which was something like the
following. I think I took a=3 and b=2, and specified that
P(x,y)=5y+Junk_{1}(x) and
Q(x,y)=x^{2}+Junk_{1}(y). Here the
Junk functions were some absurd formulas:
arctan(y^{3}) or e^{cos(x)}. I wanted to compute the
line integral of P(x,y)dx+Q(x,y)dy around the boundary of the
rectangle. Of course I used the version of Green's Theorem stated
above, and the two Junk terms disappeared after the partial
differentiations, and the double integral had integrand 2x5, which is
rather simple to compute over a rectangle.
Since this is a math course, I decided I should give some
deductive "reason" that Green's Theorem is true. I tried to verify
∫_{rectangle bdry}P(x,y)dx=∫∫_{rectangle}–(∂P/∂y) dA.
The verification amounted to some juggling with the
Fundamental Theorem of Calculus. I only discussed the Gree's Theorem
equation with P(x,y). The verification for Q(x,y) is quite similar. I
started with ∫∫_{rectangle}–(∂P/∂y) dA. Mr. Laflotte suggested dy dx, so the
double integral became the iterated integral
∫_{x=0}^{x=a}∫_{y=0}^{y=b}–(∂P/∂y) dy dx.
Let's look at the inside. The dy and ∂/∂y "cancel". I
really mean that FTC (the Fundamental Theorem of Calculus applied to
the y variable) can be applied. So
∫_{y=0}^{y=b}–(∂P/∂y) dy=P(x,y)^{y=b}_{y=0}=P(x,b)+P(x,0).
Now we are supposed to integrate this result from 0 to a with respect to x. We worked hard and realized that P(x,0) integrated dx from 0 to a was the lower horizontal side of the line integral, and –P(x,b) integrated from 0 to a dx was the upper horizontal side of the line integral  the minus sign corresponds to integrating on that segment from right to left. What about the up and down parts of that line integral? Well, P(x,y)dx on the up and down parts is 0 because x does not change on those parts of the line integral, so the result is 0. We therefore have verified Green's Theorem for P(x,y)dx. The Q part works in the same way, and Green's Theorem is really mathematically true for a rectangle.
Computation on a semicircle
This is another example created for the classroom, but it is more
complex, and more resembles some real uses of Green's Theorem. Here I
considered the line integral ∫_{S}y dx+x^{2}dy
where S represents the upper half of the
circle of radius 3 centered at the origin. I think this integral
certainly can be computed straightforwardly by a parameterization but
I'd like to show another technique. First, what is the value of ∫_{L}y dx+x^{2}dy where L is the line segment along the xaxis stretching
from (3,0) to (3,0)? Since y doesn't change on L, dy=0. Also notice that y=0 on all of L. Therefore y dx+x^{2}dy is 0 on all
of L.
So ∫_{S}y dx+x^{2}dy=∫_{S}y dx+x^{2}dy+∫_{L}y dx+x^{2}dy=∫_{S+L}y dx+x^{2}dy, and the curve S+L, which is S followed by L, encloses the upper half of the area inside a circle of radius 3 centered at the origin: a semidisc (great word!). I'll call it Q.
Then a version of Green's Theorem allows us to replace ∫_{S+L}y dx+x^{2}dy by ∫∫_{Q}∂(x^{2})/∂x∂y/∂y dA. The
integrand is 2x1, and we need to integrate it over Q.
The integral ∫∫_{Q}2x dA is actually 0, because
the region Q is symmetric with
respect to the yaxis, and 2x takes equally positive and negative
values on this region. So the values cancel. And the integral
∫∫_{Q}1 dA is
minus onehalf the area of a circle of radius 3. Therefore the
original line integral, ∫_{S}y dx+x^{2}dy, has the
value 9π/2.
I would agree that this is an awkward, even ludicrous way of evaluating the line integral. (But it is slick!) I went through it because in a week we'll be trying computations like this in three dimensions and maybe things there won't be as clear.
The text's version of Green's Theorem
Suppose C is a positively (counterclockwise) oriented piecewise smooth
simple closed curve, and D is the region bounded by C. Suppose that
P(x,y) and Q(x,y) are functions with continuous first partial
derivatives in D. Then:
∫_{C}P(x,y)dx+Q(x,y)dy=∫∫_{D}∂Q(x,y)/∂x∂P(x,y)/∂y dA.
Definitions of terms
The statement above is complicated, and I think many of the words need
some explanation.
Information transported from the boundary to the inside
Suppose P(x,y)=(1/2)y and Q(x,y)=(1/2)x. Then
∂Q(x,y)/∂x∂P(x,y)/∂y just "happens" to be 1 (the
P and Q were selected carefully!). And the double integral of 1 over
the region is equal to its area. So apparently
∫_{C}(1/2)y dx+(1/2)x dy is the area of D. Let
me help you understand this result, which I consider rather
remarkable. The curve C should be parameterized. So x=x(t) and y=y(t)
for t between a and b. The line integral gets translated into the
following Riemann integral:
∫_{t=a}^{t=b}(1/2)y(t)x´(t)+(1/2)x(t)y´(t) dt.
So suppose you are in a race car, driving around a "closed
course". Your position could be measured with a GPS device, so you
would know <x(t),y(t)> (the position vector). You could also
imagine that you have some idea of your velocity:
<x´(t),y´(t)>. You could approximate this, after all,
but doing arithmetic on successive position vectors over a small time
interval. So from this "boundary data" you could
compute (at least approximately) the integral
∫_{t=a}^{t=b}(1/2)y(t)x´(t)+(1/2)x(t)y´(t) dt
and this must be the area of enclosed by the race track. This is
astonishing to me.
People designed interesting mechanical linkages to compute areas based on such results. These instruments are called planimeters. People can be very clever.
Analogous results in higher dimensions (R^{3}) would allow some knowledge of the heart based upon electrical readings on the skin (EKG's?). Or the results can be used to try to deduce the presence of cracks inside structures (steel beams?) as a result of measurements on the outside. Detect conditions inside an object using surface measurements, so there is no need for invasive procedures.
Another example, somewhat paradoxical
Here I will tell again you what P(x,y) and Q(x,y) should be. These are
certainly not randomly chosen but are specific functions which are
used to model important physical phenomena.
P(x,y)=y/(x^{2}+y^{2}) and
Q(x,y)=x/(x^{2}+y^{2})
It just happens (!!) that:
∂P/∂y=[(x^{2}+y^{2})+y(2y)]/(x^{2}+y^{2})^{2}=[y^{2}x^{2}]/(x^{2}+y^{2})^{2}
and
∂Q/∂x=[(x^{2}+y^{2})2x(x)]/(x^{2}+y^{2})^{2}=[y^{2}x^{2}]/(x^{2}+y^{2})^{2}.
Therefore in this case, P_{y}=Q_{x} so that
Q_{x}P_{y} is 0.
Now let me compute the line integral ∫_{C}P(x,y)dx+Q(x,y)dy, where C is, say, the circle of
radius 3 centered at (0,0). The first parameterization you should
consider is this: x=3cos(t) and y=3sin(t) so that dx=3sin(t) dt
and dy=3cos(t) dt. Then x^{2}+y^{2}=3^{2}
because sin^{2}+cos^{2}=1. Thus
∫_{C}y/(x^{2}+y^{2})dx+x/(x^{2}+y^{2})dy
becomes ∫_{t=0}^{t=2π}([3sin(t)/3^{2}](3sin(t))+[3cos(t)/3^{2}]3cos(t))dt. If the arithmetic is done correctly we need to
integrate 1 from 0 to 2π (almost everything cancels!) and the result
is 2π.
But but but ... the integrand on the "other side" of Green's Theorem is Q_{x}P_{y} and we computed that to be 0. A circle certainly is a simple closed curve, positively oriented, etc. We haven't made any errors. Maybe it is true that 2π=0?
Technicalities matter.
I've ignored the requirements on P(x,y) and Q(x,y). Here the
requirements definitely are relevant. Both P(x,y) and Q(x,y), and
their first partial derivatives, are quotients of polynomials, and
have x^{2}+y^{2} in the denominator (the bottom of the
fraction). That expression is 0 at the origin, (0,0). In fact, P(x,y)
and Q(x,y) and their first partial derivatives are not
continuous at (0,0), which I will call the "bad point". This bad point
is inside the circle. The example shows that even if the hypothesis
"P(x,y) and Q(x,y) are functions with continuous first partial
derivatives" fails at just one point of D, the equality predicted by
Green's Theorem can fail very very emphatically (I think "zero equals
something nonzero" is a fairly emphatic failure).
Information transported from one curve to another, using what's in
between (!?) The example considered is, however, very important in practice. Let me show why. Let's consider another path, W, which is a quadrilateral. W is a simple closed curve made of four straight line segments. It starts at (8,0), goes to (2,6), then to (7,1), then to (0,5), and finally back to (8,0). I would like to compute ∫_{W}y/(x^{2}+y^{2})dx+x/(x^{2}+y^{2})dy. Here I am not so sure I could "easily" compute the integrals which would come out of parameterizing the four line segments, and changing the x,y language to t language. At least the task would be lengthy. So let me show you another way. Again, I can't just apply Green's Theorem to W and the region inside W, since W encloses the same bad point. But I can do something very clever using Green's Theorem.
Crosscut
Now let me split up the crosscut into two distinct line segments, one
directed from W to C and one directed backwards. The points A and B
become doubled, and each pair is slightly separated. Now the closed
curve is a simple closed curve. The region this simple closed
curve bounds does not include the origin, the bad point. Inside
the region, P_{y}=Q_{x} and therefore the double
integral side of the Green's Theorem equation is 0. The line integral
side is also 0. This will work for any separation, no matter how
small. Now the line integral will vary continuously as the
separation →0. And therefore, when the separation=0, the line
integral will still be 0. And so, clearly the integral along W
minus the integral along C is 0 (because the crosscuts cancel!) so the
integral along W is the same as the integral along C: it must be 2π!
This crosscutting technique is so familiar to people who use it that it is rarely discussed in detail. In fact, any curve which goes around the origin exactly once will have the line integral of our P(x,y)dx+Q(x,y)dy equal to 2π for similar reasons.
A physical model?
Another application (important!) of Green's Theorem 
Exams returned
I returned exams to students who were in class today. There's some
effort involved in prompt and accurate grading of exams, but I believe
that such grading is important so I try to do it. 78 students took the
exam, and 21 students did not pick up their exams. The mean grade of
the exams not picked up is 43.33, about 14 points less than the class
mean, and the median (which more accurately reflects the grade
distribution here) is 39, about 18 points less than the class
median. Most of the exams not picked up were F's, and, indeed, most of
the F's (and D's!) in the course are reported for students who do not
regularly attend class. Or, I believe, do much else connected with the
course. What are these people thinking? What are they doing with their
lives? I hope these students' actions reflect their reasoned decisions
which I respect but which I strongly regret.
It is my privilege and task to try to help people learn the intricate and (to me) beautiful material of this course. Most of the students in the course indicate majors for which the course material is likely very relevant. I also feel sadness and frustration.
An example
The examples I've been giving in class (and here) have almost all
concerned 2dimensional vector fields. I'll discuss one more such
example, and then the remaining part of the discussion in this lecture
will have 3dimensional examples, even though 2dimensional ones could
be give. 2 is good, but so is 3. Please get used to both of them. And
4 ... and 5 ... and ...
Computing some work
Let's consider integrating
xy dx+y^{2}dy from (1,2) to (3,10). If you wish you could
think that this involves computing the work against the force field
xyi+y^{2}j.
One work computation
Let's move from (1,2) to (3,2) along a straight line segment (y=2),
and then from (3,2) to (3,11) along another straight line segment
(x=3). I'll use a parameterization for each of the two pieces.
Another work computation
We could move from (1,2) to (3,11) along y=x^{2}+1. For this
path, we can choose x=t, so y=t^{2}+1, 1≤t≤3, dx=dt, and
dy=2t dt. Then xy dx+y^{2}dy integrated on this
parabola becomes
∫_{1}^{3}t(t^{2}+1)(1)+(t^{2}+1)^{2}(2t) dt=
∫_{1}^{3}t^{3}+t+(t^{2}+1)^{2}(2t) dt=
(1/4)t^{4}+(1/2)t^{2}+(1/3)(t^{2}+1)^{3}]_{t=1}^{t=3}=
(1/4)3^{4}+(1/2)3^{2}+(1/3)(3^{2}+1)^{3}
(1/4)1^{4}(1/2)1^{2}+(1/3)(1^{2}+1)^{3}=1064/3.
You may be amused to know when I did this by myself,
unwatched, and then checked the computation with
Maple I had, of course, made a
mistake. Oh well.
Different?
So the integral of xy dx+y^{2}dy along two paths joining
(1,3) and (2,11) is different. Why should we expect the vlaues
to be the same, anyway? Maybe the math models a situation where we are
moving a pebble along a sort of flat area, and there is more friction
or the path is bumpier in part of the area. Sometimes, though, the
values are always the same. It turns out this is true for some models
of physical reality and some forces that people have measured. Such a
integral, or rather, a force, is called path dependent.
What if the integrals are the same?
This is called path independence or, a physics word,
conservative. It turns out that gravity is a conservative
vector field, and so is a point charge's electric field, and so are
magnetic fields. (Not totally obvious  I will check it later).
Path independence is a very strong property. Let me be more precise about it. Take any two points p and q. Take any path connecting p to q. Then the work done will depend only on p and q, and not on the choice of path.
Which vector fields are conservative?
Now let me consider what happens in R^{3} and try to
understand conservative vector fields. This is such an absurdly strong
property that you shouldn't be surprised if remarkable things
happen. Suppose
F=<F_{1},F_{2},F_{3}> is
conservative in R^{3}. What can we do then? Let me show you
some very clever things.
We could pick a point at random, say (2,6,3). Such a "random" point was picked by a student in the class. This is called the ground state physically. Then we could consider the function g(x,y,z), the total work, which results from taking any path from (2,6,3) to (x,y,z). Since all paths result in the same work, this does define a function. What can we say about g(x,y,z)? Well, select an order from x and y and z. An order of the variables was picked by various students in the class, I think the order was y to x to z. Then I will show you that something neat happens. If we want to move from (2,6,3) to (x,y,z), then we could move in pieces, each by a line segment, and only changing one variable at a time. We've done this repeatedly in 251. Changing lots of variables at once is difficult, so try to change one at a time  maybe things will be easier.
So I want to write the line integral of F_{1}(x,y,z) dx+F_{2}(x,y,z) dy+F_{3}(x,y,z) dz over these three line segments. Here we go, but watch carefully, because the details are quite important.
The total integral which is g(x,y,z) is therefore the sum of all three of these.
g(x,y,z)=∫_{6}^{y}F_{2}(2,t,3) dt+∫_{2}^{x}F_{1}(t,y,3) dt+∫_{3}^{z}F_{3}(x,y,t) dt.
Look at this mess. Actually, it isn't a mess, but it is somewhat complicated. How many y's are there in the formula? If you count carefully, there are 3. And there are two x's. But there is only one z. So let me try to understand how g(x,y,z) depends on z. What if I ∂/∂z this formula for g(x,y,z)? The first two integrals have no z at all in them. So from the z point of view they are constants. Therefore the result of ∂/∂z of those integrals is 0. The last integral has z in the upper bound of the integral. This is a first semester calculus exercise. The Fundamental Theorem of Calculus says that the way to differentiate a function defined by a definite integral with a variable upper bound is to plug that variable into the integrand. Here the integrand is F_{3}(x,y,t) and we are integrating with respect to t. Therefore ∂g/∂z=F_{3}(x,y,z).
We could have picked other orders of the variables from the list of x and y and z, and written similar straight line segment paths going from (2,6,3) to (x,y,z). The value of g(x,y,z) would be the same because of path independence. If, say, y were the last variable picked, then I would ∂/∂y the result. The y path would have as integrand F_{2}. The result would be ∂g/∂y=F_{2}(x,y,z).
And if we picked an order where the last named variable was x, and did a similar computation, the result would be ∂g/∂x=F_{1}(x,y,z).
A conclusion
Suppose F=<F_{1},F_{2},F_{3}> is
path independent. If g(x,y,z) is the work function resulting from
picking a ground state and integrating from there to (x,y,z), then
∂g/∂x=F_{1}(x,y,z); ∂g/∂y=F_{2}(x,y,z); ∂g/∂z=F_{3}(x,y,z).
That is, ∇g=F and F is a gradient vector field, with g as one potential. The effect, by the way, of picking another ground state, is just to alter g by an additive constant, and getting a different potential. So the same would result in the other two variables. Indeed, ∇g=F. Therefore, if a vector field is conservative, then it is a gradient vector field. Differentiate the upper value of the integral.
If F is a conservative vector field, then F is a gradient vector field.
The other way!
It happens that the converse of
the preceding statement is true. That is, any gradient vector field
must be conservative. Let me show you why, because the why here
is not an irrelevant verification, but shows a remarkably useful
computational shortcut.
So let me now assume that F=<F_{1},F_{2},F_{3}> and that I happen to know a function g so that ∂g/∂x=F_{1}(x,y,z);, ∂g/∂y=F_{2}(x,y,z), and g/∂z=F_{3}(x,y,z). I would like to compute the line integral of F_{1}(x,y,z) dx+F_{2}(x,y,z) dy+F_{3}(x,y,z) dz along a curve going from p where t=Start to q where t=End. So we have a parameterization, (x(t),y(t),z(t)) of the curve.
So dx=(dx/dt) dt, dy=(dy/dt) dt, and dz=(dz/dt) dt. We
need to "plug in" the parameterization to the components of
F. So the line integral, ∫_{C}F·Tds, becomes
∫_{Start}^{End}(F_{1}(x(t),y(t),z(t))(dx/dt)+F_{2}(x(t),y(t),z(t))(dy/dt)+F_{3}(x(t),y(t),z(t))(dz/dt)) dt.
"Strangely enough , it all turns out well." Well, it is not too
strange if you realize that 150 years of effort have gone into what
I'm showing you today. Several students recognized that the integrand,
F_{1}(x(t),y(t),z(t))(dx/dt)+F_{2}(x(t),y(t),z(t))(dy/dt)+F_{3}(x(t),y(t),z(t))(dz/dt)
the whole thing, is the derivative with respect to t of
g(x(t),y(t),z(t)). This is exactly because F_{1} is
∂g/∂x, F_{2} is ∂g/∂y, and F_{3}
is ∂g/∂z and we are multiplying each of these by the
appropriate dx/dt, dy/dt, and dz/dt. This is a consequence of the
several variable Chain Rule (actually, this is the most important use
of the Chain Rule in the course). So what we have is
∫_{Start}^{End}d/dt(g(x(t),y(t),z(t))dt. Another use
of FTC is that the integral of a derivative can be evaluated by just
forgetting the derivative, and taking the value of the original
function at the top limit minus its value at the bottom limit. Well,
if you do that, the result is
The work done is the value of the potential at the End minus its value at the Start..
This is such a marvelous fact that I want to use it immediately.
Example
Suppose g(x,y,z)=x^{3}y+xz^{2}+5. Then ∇g is
(3x^{2}+z^{2})i+x^{3}j+2xzk. I'll
call this F. Now consider the helix with position vector
C(t)=<5cos(2t),4t,5sin(2t)>. Let's compute the work done
against F from t=π/2 to t=3π on the helix. A Maple picture of this helix is shown to the
right. Its axis of symmetry is the yaxis, and it spins around that
axis in the x and z directions as y increases.
One way to compute the work is to use the parameterization, substitute things. etc. But since I know that F=∇g, I can take advantage of the result we just discovered, and find the values of g at the End and Start and subtract them.
Start
This is when t=π/2. So
we plug this into the curve and get the starting position. So we need
<5cos(2[π/2]),4[π/2],5sin(2[π/2])>=<5,2π,0)>. Now
we need to compute g at this point: g(5,2π,0)=
(5)^{3}(2π)+(5)0^{2}+5=250&pi+5.
End
This is when t=3π. So
we plug this into the curve and get the ending position. So we need
<5cos(2[3π]),4[3π],5sin(2[3π])>=<5,12π,0)>. Now
we need to compute g at this point: g(5,12π,0)=
(5)^{3}(12π)+(5)0^{2}+5=1,500&pi+5.
The value of the line integral is g(End)g(Start)=(1,500&pi+5)(250&pi+5)=1,750π.
This seems much easier to me than working the with parameterization, differentiating, substituting, integrating, and evaluating. Your opinion may be different but I think you'd be wrong!
Continue with this vector field and compute another example
This one might seem extremely weird to you. The "curve" will need some
description. I want to begin at (0,0,0), the origin. The first piece
will be a semicircle in the yz plane whose other end will be
(0,0,6). I want to follow that with a line segment from (0,0,6) to
(5,0,6). Then a line segment from (5,0,6) down to (5,0,0) on the
xaxis. And then a line segment from (5,0,0) to (5,5,0). And finally,
we will finish up with a line segment from (5,5,0) to (0,0,0). A
attempt at drawing this is shown to the right. This is all C. I would
like to compute
∫_{C}(3x^{2}+z^{2})dx+x^{3}dy+2xzdz.
And, of course, after all this, a number of students saw the answer immediately. Of course I wanted all of the drama to myself, but I was just as happy that they did see it. The important fact here is that Start=End, so that g(End)–g(Start)=0. I don't need to do any computation at all!
Another definition
A closed curve is one where Start=End.
Big result, allowing easy computation of many line integrals
These three statements are logically equivalent:
An important physical example
We looked at inverse square vector fields in R^{2}. A three
dimensional version is not too difficult to exhibit:
[x/(x^{2}+y^{2}+z^{2})^{3/2}]i+[y/(x^{2}+y^{2}+z^{2})^{3/2}]j+[z/(x^{2}+y^{2}+z^{2})^{3/2}]k
This vector field points from (x,y,z) to (0,0,0) because
it is a positive multiple of xiyjzk. The
reason for the (initially) strange scaling factor of
(x^{2}+y^{2}+z^{2})^{3/2} is that
when you compute the length, you will see it is exactly (distance from (x,y,z) to (0,0,0))^{2}: inverse square.
Now "notice" that ∂/∂x of (x^{2}+y^{2}+z^{2})^{1/2} is x/(x^{2}+y^{2}+z^{2})^{3/2}], the i component of the inverse square vector field. Also ∂/∂y and ∂/∂z applied to the same function give the j and k components. So the inverse square vector field is a gradient vector field! Therefore work computations are path independent and the results will be the difference of the gravitational potential. Just verifying the derivatives, just these very basic computations, allows us now to conclude many things about work for this vector field. I think this is remarkable.
One physical result
What does this mean? Well, if you could imagine a rocket making two
trips from a Start somewhere on
the surface of the Earth to an End at the same point in space, then
(ideally, ignoring all sorts of complicating factors like friction and
nonspherical Earths, etc.) the same amount of work has been done. And
the work done could be computed as the difference of the gravitational
potential at the two points. This intellectually represents a huge
amount of freedom to do the computation of the physics involved.
How can you tell if a vector field is a gradient vector field?
Well, we discussed this a bit last time. We could integrate each of
the components with respect to the appropriate variable. I know that
integrating, or, better, antidifferentiating, is
difficult. Differentiation is usually much easier. So if I want a
quick way to check if
<F_{1},F_{2},F_{3}> is eligible to be a
gradient vector field, then I could compute first derivatives, and
consider (I know that second "cross partials" should be equal) the
results.
Example of the checking procedure
Suppose you want to consider the vector field
xy^{2}i+(x^{2}y+z)j+5z^{2}k. Is
it a gradient vector field? I will check some derivatives.
What can we say if the components of the vector field "pass" these tests? (By the way, these equations are called compatibility conditions because they show that the components of the vector field get along (!) with each other.) Can we then conclude that there is a potential? Well, there's some complications, and this will be covered later in the course.
Derivative 0 always means the function must be constant (1
variable)
In calc 1, the following implication is noted and used almost
everywhere: if f´ is always 0, then f is constant. This is used
most prominently when people find antiderivatives. So, for
example, when I know that (1/3)x^{3} is an antiderivative of
x^{2}, I can then assert that all antiderivatives of
x^{2} are (1/3)x^{3}+Constant, for any value of
"Constant". And that uses the previous fact about derivatives being 0.
Gradient 0 always means the function must be constant (2
variables)
In two variables, I wanted an analogous statement. Here what we have
as hypotheses is gradient of f (a function of both x and y) is 0. But
∇f=0 means (first component) ∂f/∂x=0 and (second
component) ∂f/∂y=0. So I would like to conclude that such an
f must be constant. This was my first observation. I had
included this assertion in the previous diary entry. There I had
verified that two potentials of the same vector field differ by a
constant and I needed as part of that to look at the
difference. Please look here.
How to get from the gradient back to a
potential?
Suppose we suspect that f(x,y)i+g(x,y)j is ∇φ?
How can we guess φ? Ms. Krupnik suggested a method on Tuesday
night. She suggested that we "integrate" or, rather, compute
antiderivatives of each component. Let's try this with an example.
Suppose that we are told the vector field
<2xycos(x^{2}y)+x^{7},x^{2}cos(x^{2}y)y^{5}>
is a gradient vector field. So it is
<∂φ/∂x,∂φ/∂y>. What can we do now?
Combining the descriptions of φ
This process gives us two descriptions of φ and we need to compare
them to discover what we can do. So:
sin(x^{2}y)+(1/8)x^{8}+Any function of y
sin(x^{2}y)(1/6)y^{6}+Some function of x
We need to think and compare definitions. Any part of φ involving
both x and y must appear in both descriptions, and that here is just
sin(x^{2}y). Any y alone function must appear in the y
description and any x alone function must appear in the x
description. Anything else is truly just a constant  it can't in any
way depend on either x or y. So combining the descriptions we get this
as our answer, a complete description of all potentials of the
given vector field:
φ(x,y)=sin(x^{2}y)+(1/8)x^{8}(1/6)y^{6}+Constant.
Try again ...
The vortex flow we looked at last time was yi+xj. If
∂φ/∂x=y then integrating with respect to x we get
φ(x,y)=yx+Const involving y. If ∂φ/∂y=x,
then integrating with respect to y we get
xy+Const involving x.
Compare these descriptions!
Can we have yx+(Const involving y)=xy+(Const involving x)?
Well, the xy and yx cannot be buried in each other's constants,
because they involve both x and y. And yx is not the same as xy. So,
darn it, there is no possible φ here. This vector field
cannot be a gradient vector field. This agrees with what we learned
last time, where a different method gave the same conclusion. People
usually like the method used last time which involved differentiation,
because in practice differentiation is much easier than
antidifferentiation.
How about 3 dimensions not just 2?
Similar statements are true for 3 dimensional vector fields. Look at
the textbook.
Segue
I've mostly heard the word "segue" used with music or movies or plays,
and it means "Smooth transition from one thought/idea to another."
These comments on integration are intended to help you realize why we
need the next technical tool of the course.
Line integrals
We need an additional technical tool to investigate vector fields:
line integrals. Line integrals will allow the computation of work for
force fields, and flux for fluid flow, and something for gradient
vector fields. Work and flux and other things are important physical
quantities and turn out to have nice mathematical properties. So here
we go.
Mass of a wire
I'll introduce line integrals using the metaphor of the mass of a wire.
The integral calculus mantra is:
chop up, approximate, sum, limit.
Here I'll assume that there is a wire sitting in R^{2}. Its
geometry is simple, with a constant crosssectional area (assumed to
be 1 in the measurement system used). The density of the wire will
vary according to the position on the wire: at some points the density
will be high, and the wire heavy, and at other points, the density
will be low, and the wire rather light. The task is to create some
technical tool which will represent the total mass of the wire. By the
way, the accompanying picture looks like a particularly ugly
worm or snake. I am sorry.
Suppose I cut the wire up into lots of little chunks. How little?
Well, the length of each chunk will be ds, a tiny piece of arc length
(we discussed this in several lectures given late January, long, long
ago). If I assume that the density D(x,y) varies only a small amount
because the chunk of the wire is very small, then dm, the mass of this
piece, is nearly D(x,y) ds (remember I made the crosssectional
area equal to 1). Further, I can add up these pieces of mass to get
the total mass. I can take the limit as the number of pieces gets
large and the length of the pieces gets small. The result is that the mass
of the wire should be given by
∫_{The wire}D(x,y) ds.
This integral is called a line integral, where the word
line doesn't mean "straight line" but is used in the sense "A
thin continuous mark, as that made by a pen, pencil, or brush applied
to a surface." Now I need to show you what this means. So I will
compute a very artificial example.
Silly example
My example was the following: the wire follows the part of the circle
of radius 3 which is in the first quadrant. The density of the wire at
the point (x,y) is 7y+5. Find the mass of the wire. Well, I will take
the integral ∫_{The wire}D(x,y) ds and parameterize
everything in sight. To me the natural parameterization of an arc of a
circle uses essentially the angle from the center. The text likes to
use t here, so I will parameterize this quarter circle with x=3cos(t)
and y=3sin(t). The interval of this parameterization is [0,Pi/2]. Now
ds is sqrt([dx/dt]^{2}+[dy/dt]^{2}) dt. The
square root stuff is the magnitude of the velocity vector, the
speed. And ds=SPEED·dt is a translation of
distance=rate·time of course. In this case, dx/dt=3sin(t) and
dy/dt=3cos(t) so that
sqrt([dx/dt]^{2}+[dy/dt]^{2}) just happens (!!!) to
simplify to 3. And ds=3 dt. What about the density? Since
D(x,y)=7y+5, we know that the density is 7(3sin(t))+5. And the
integral should go from 0 to Pi/2. Therefore
The mass=∫_{The wire}D(x,y) ds=∫_{t=0}^{t=Pi/2}{21sin(t)+5}3 dt=63cos(t)+15t]_{t=0}^{t=Pi/2}=63+(15/2)Pi.
This is a totally insignificant, physically unrealistic (to me) computation. The only thing I had fun with is drawing the picture, which with its varying colors is supposed to suggest the increasing density of the wire as y increases. There is one very important fact which should be mentioned now:
Independence of parameterization
In this line integral and in all line integrals, the result will be
the same no matter which parameterization is used. Verification of
this uses the one variable chain rule and is not too interesting right
now. So let me go on.
Random example
I then featured a really random example. I asked students for some
random (positive) integers which I then used to construct an example
very much like the one presented in what follows. The previous example
was arranged so that ds was nice. I now tried to compute something
like the mass of a wire where the wire was sitting on the curve
y=x^{4} from, say, (0,0) to (2, 16), and the density is
D(x,y)=x^{4}y^{7}. I not being totally idiotic
here. Any computational strategy in calculus should be able to handle
lowdegree polynomials fairly efficiently, even by hand. So here's
what we did.
A simple parameterization of a graph of a function is just to use the
independent variable as the parameter. So I used x=t and
y=t^{4} and then the density x^{4}y^{7} became
t^{4}(t^{4})^{7}=t^{28}. This isn't
too bad, but I am saving the worst for last, and in this computation
the worst is ds. So dx/dt=1 and dy/dt=4t^{3},
and therefore ds/dt=sqrt(1+16t^{6}). The line integral for
this "mass" translates in tland as follows:
∫_{The wire}D(x,y) ds=∫_{t=0}^{t=2}t^{28}sqrt(1+16t^{6}) dt.
This integral cannot be computed exactly in terms of standard
"familiar" functions. I found to my amazement that Maple does
have a storehouse of weird functions which it can use to "evaluate"
this integral. But then I have no feeling for the functions it used
and certainly not for the values of the functions.
ds is the obstacle
What is horrible about most line integrals is ds. Almost no ds except
for those included in textbook problems lead to familiar
antiderivatives. At the same time, important quantities such as work
and flux are defined as mathematical objects in terms of line
integrals, and the general expectation is that these quantities
can be computed exactly in terms of familiar functions (or
should be computable in terms of familiar functions).
To me this is an excellent example of the psychological phenomenon
known as cognitive
dissonance:
Cognitive dissonance arises from conflicting cognitions. Cognitive dissonance is the perception of incompatibility between two cognitions, which for the purpose of cognitive dissonance theory can be defined as any element of knowledge, attitude, emotion, belief or value, as well as a goal, plan, or an interest. In brief, the theory of cognitive dissonance holds that contradicting cognitions serve as a driving force that compels the mind to acquire or invent new thoughts or beliefs, or to modify existing beliefs, so as to minimize the amount of dissonance (conflict) between cognitions.
Work
Let's consider the physical concept called work. This is
"force" times "distance", loosely, but we're going to need to be a bit
more specific. For example, consider a mass being pushed up a
frictionless triangle. If the triangle is steeper, then more force is
needed. In the picture, the force of gravity is directed down. The
blue arrows on each inclined plane (triangle) show the force component
needed to move the box up that triangle.
So instead of dieting, avoid steep triangles and your weight
will decrease ... no no no
... this is just more bad information from the lecturer!
Anyway, what really matters is the component of the force in the
direction of motion (we could take the dot product of the force and
the displacement also).
Work with a varying vector field along a curve
Again, chop up, approximate, sum,
limit.
I would like to describe how to compute the work done against a
varying force field F as we travel from p to q (in
R^{2} but the same ideas will work in R^{3}, also)
along a curve, C. We chop up C into small pieces (the red lines are
the chopping places). One of them is displayed under a "magnifying
glass" (sort of) in the picture. The piece is so small that it is
almost a straight line, and its length is ds. The piece is also so
small that the vector field, F, is almost constant near the
piece. Remember that a unit tangent vector, pointing in the direction
of the piece, is called T (go back and think of January!). Then
the part of the force field along the curve is F·T and
the piece of the work will be approximately F·Tds. All
of the work will be the integral along C of this quantity. Therefore
the work is ∫_{C}F·Tds.
A computation
Let us "test" this definition with a random computation: well,
sort of "random". C will be the curve y=x^{4} and p will be
the point (0,0) and q will be the point (2,16). I will have the force
field be
x^{2}y^{3}i+xy^{5}j. Now let
us compute. We need to change everything
into tland, where I will choose a most routine parameter: x=t so that
y=t^{4}.
ds
As before, the speed becomes
sqrt([dx/dt]^{2}+[dy/dt]^{2})=sqrt([1]^{2}+[4t^{3}]^{2})=sqrt(1+16t^{6})
which is horrible enough. Thus ds=sqrt(1+16t^{6}) dt.
F
At the point (x,y), F is
x^{2}y^{3}i+xy^{5}j so that at
(t,t^{4}) on the curve, F is
t^{2}(t^{4})^{3}i+t(t^{4})^{5}j=t^{14}i+t^{21}j.
T
Now comes almost the miracle. In the movie Shakespeare in
Love, one
character states, "The natural condition is one of insurmountable
obstacles on the road to imminent disaster." He then says almost
immediately, "Strangely enough , it all turns out well." When asked
"How", the character replies, "I don't know. It's a mystery." So, if
not a miracle, let me show you the little mystery here.
The unit tangent vector, T, is a vector in the direction of the curve. The position vector of the curve is <t,t^{4}> so that the velocity vector is <1,4t^{3}>. But we need a unit vector to get the projection of F in the direction of the curve. Divide by the magnitude of <1,4t^{3}>. Therefore, T=<1,4t^{3}>/sqrt(1+16t^{6}).
Assembling the work integral
∫_{C}F·T ds=∫_{t=0}^{t=2}(t^{14}i+t^{21}j)·
(<1,4t^{3}>/sqrt(1+16t^{6}))sqrt(1+16t^{6}) dt. This is
∫_{t=0}^{t=2}t^{14}+4t^{24} dt.
which with totally routine polynomial calculations can be evaluated:
it is 2^{15}/15+(4/25)2^{25}.
What happens?
The speed comes in to squeeze down the velocity vector to get the unit
tangent vector. The speed comes in as the factor which multiplies dt
to get ds. The two appearances of the speed cancel. They will always cancel!.
Using the notation to help
Here is how people use notation to guide their way through the
computation. No one computes the speed (the square root stuff) when
computing work because it will cancel. So:
Problem statement
Compute the work if
F(x,y)=x^{2}y^{3}i+xy^{5}j
and the curve is y=x^{4} going from (0,0) to (2,16).
A solution
So I initially write ∫_{C}F·T ds but then I immediately change
to ∫_{C}x^{2}y^{3}dx+xy^{5}dy. I
evaluate this line integral again by changing everything to tland. So
x=t and y=t^{4} and dx=1 dt and dy=4t^{3} dt, Also,
x^{2}y^{3}=t^{2}(t^{4})^{3}=t^{14}
and
xy^{5}=t(t^{4})^{5}=t^{21}. Therefore
x^{2}y^{3}dx+xy^{5}dy=t^{14} dt+t^{21}4t^{3} dt.
Therefore ∫_{C}x^{2}y^{3}dx+xy^{5}dy=∫_{t=0 [START]}^{t=2 [END]}
t^{14}+4t^{24} dt. And the result will be the
same. Almost everyone uses this notation, and never bothers with
computing T and ds and then canceling, etc. Of course, the
ideas are important: the physical quantity we are computing has
certain properties which make this computation extremely interesting.
Irrelevance awards
I asked in the previous diary entry for the identity of the individual
pierced by arrows and for the painter. The answers are, respectively,
St. Sebastian and Andreas Mantegna
(c. 14311506). St. Sebastian was martyred c.288 at Rome. Of course,
there is a
Wikipedia article about him. I found the
following information which I did not know before:
As a protector from the plague, Sebastian was formerly one of the Fourteen Holy Helpers (until the suppression of that cult in 1969) . The connection of the martyr shot with arrows with the plague is not an intuitive one. In GrecoRoman myth, Apollo, the archergod is the deliverer of pestilence; the figure of Sebastian Christianizes this familiar literary trope.Mr. Forrest discovered these facts. He will get something, as soon as I figure out what the something should be!
For example, consider a big "chunk" of "stuff". Look at the heat in the chunk. Sometimes some of the heat might originate inside the object, and sometimes there might be cooling. And there are also boundary effects. If there's something really hot near part of the boundary, heat will flow in around that part, and, similarly, there might be cool things (?) near some other part of the boundary, and heat will flow out that part of the boundary. This is all very complicated. Somehow, there should be some way of balancing the heat sources and sinks inside the region, and the boundary flow of heat, and expressing this in a neat way. The tools we develop in this last portion of the course will allow us to create a model of this situation.
The Fundamental Theorem of Calculus, one of the wonderful results of
calc 1, is something like this:
If F´=f, then
∫_{x=a}^{x=b}f(x) dx=F(b)F(a)
This
is a wonderful result. Think about it. It states that some sort
of stuff inside the interval [a,b] (we could think about height times
width, f(x)dx, or some accumulation of "stuff": f(x) inside [a,b] and
added up over the whole interval, [a,b]) is equal to an edge
quantity. There isn't much "edge" to an interval. Here the edge
quantity is F(b)F(a). You can, with some effort, see things as some
sort of balance between interior accumulation and stuff in and out the
edges. The remainder of the course is an effort to generalize this to
dimensions 2 (Green's Theorem) and 3 (Divergence Theorem) and even
dimension (sort of) 2.5: Stokes' Theorem and Gauss's Theorem. These
results provide a language for considering edge effects compared to
inside effects for things like heat flow and fluid flow and ... lots
of other phenomena. So please listen and look carefully. We still need
to learn a few more vocabulary notions.
Vector fields
I'm going to begin with two dimensional vector fields, since the
pictures are easier to draw. Those of you who have seen Euler's method
in calculus will remember bunches of arrows in the plane. So in the
most elementary sense, a vector field is such a "bunch of arrows". In
fact, it is an arrow, a vector, sitting at every point, (x,y).
To the right is a "picture" of part of a vector field. The picture was
produced by Maple using a command I
don't know very well (there are thousands of commands, and I believe I
know less than a tenth of one percent of them). This command is in the
plots package. The picture was produced by
fieldplot([x^2+10*y,3*xy^2], x=1..5,y=1..5,
arrows=slim, grid=[10,10], fieldstrength=maximal(.9),
thickness=2);
This vector field is the vector
(x^{2}+10y)i+(3xy^{2})j at the point
(x,y). The Maple command sketches a few
of the arrows ([10,10] gives 100 of them). Sometimes such a picture of
a vector field can be very helpful.
Models?
Vector fields are mathematical models of some basic physical
phenomena. Two that are immediate are force fields and fluid flows.
The force field (gravity, electromagnetism, etc.) might be the simplest. The "arrow" at every point denotes the presence of a "force" which can act on the proper kind of object (gravitation: an object with mass, magnetism, an object with magnetic "stuff", etc.). In Newton's law of gravitation, we could think of a powerful mass, M, (the sun?) at the center of the universe, and other very small masses, m's, scattered around. Ignore interactions between the m's. Each m will be attracted to the M with a force of magnitude GmM/(dist)^{2} where "dist" is the distance from m to M. The direction of the attraction is from m to M. To get a vector field, just sketch an arrow in the direction of the force, with a length proportional to the magnitude of the force. Then to complete the creation (?) of the force field, just remove all the little m's and leave the arrows. This is quite an idea, really a big leap of imagination. I have never seen any of these arrows, but imagining them is sometimes quite helpful. A quote from Albert Einstein describing radio may be appropriate here:
You see, wire telegraph is a kind of a very, very long cat. You pull his tail in New York and his head is meowing in Los Angeles. Do you understand this? And radio operates exactly the same way: you send signals here, they receive them there. The only difference is that there is no cat.
More detail about the inverse square law Suppose we stand at (x,y) with a mass m, and the origin has a mass, M. Then the gravitational attraction is a force of magnitude GmM/(x^{2}+y^{2}). What's on the bottom is the square of the distance from (x,y) to (0,0). I'll forget the m now (erasing the mass m and leaving the arrow). So we need a force of magnitude GM/(x^{2}+y^{2}). The direction should be from (x,y) to (0,0): the direction of xiyj. But that vector has length sqrt(x^{2}+y^{2}). So to get a vector of the correct magnitude and direction we need to divide by sqrt(x^{2}+y^{2}) (that creates a unit vector pointing in the correct direction) and then multiply by GM/(x^{2}+y^{2}). So an inverse square law pointing at the origin (dropping the constant multipliers) is [x/(x^{2}+y^{2})^{3/2}]i[y/(x^{2}+y^{2})^{3/2}]j. Is this complicated enough? It can be difficult to see the inverse square law under all this algebra. To the right is a picture of this inverse square vector field in the first quadrant. The magnitude of the vector field decreases rapidly away from the origin.  
Fluid flow: vortex flow  
I tried to discuss one of the standard examples of a fluid flow: a vortex. This is a counterclockwise velocity field of a fluid spinning with uniform angular velocity. Suppose the vector field was Ai+Bj at the point (x,y). This "vortex" would need to be perpendicular to the circle centered at (0,0) going through (x,y). That means the vector field would be perpendicular to the radius vector, xi+yj. We can check perpendicularity with the dot product: Ax+By must be 0. But the magnitude of the vector, sqrt(A^{2}+B^{2}), also should increase directly in proportion to the distance of (x,y) to the origin, so the fluid flow will have constant angular velocity. After some computation we saw that this would have to be Kyi+Kxj, with K positive to keep this counterclockwise.  
Other flows: 3i+0j (uniform flow to the right) Please note that adjustments were made to the fieldstrength option in this fieldplot command and others to avoid having the arrows overlap each other and to display better some of the other vector fields.  
3xi+0j, a flow which doesn't move on the yaxis, but flows away from the yaxis otherwise, and faster, the farther the fluid was away from the axis.  
x^{2}i+y^{2}j. This flow wasn't too easy to describe in words. A Maple of it is shown to the right. I don't think we drew enough "arrows" to show the flow. We did decide that the flow followed y=x along that line, and was also perpendicular to the line y=x. But ... otherwise this seems difficult to understand. The flow seems to always go up and to the right. I'll come back to this later.  
Associated to a fluid flow are streamlines, which would describe how a fluid actually flows (the velocity vector field only would give what's called the "instantaneous" flow). Such streamlines are investigated and can sometimes be described exactly by solving differential equations. 

Vector fields
The simplest ways students discover vector fields in their studies
is:
Force fields A typical example is the inverse square
"law" analyzed last time.
Fluid flow A vortex is one example. Explicitly, it was
yi+xj. This sort of describes a swirling flow around
the origin.
Gradient vector fields We haven't discussed this so far, and
these are an important type of vector fields.
Gradient vector fields
This could be in either two or three variables. Let's suppose for
simplicity we have a function φ(x,y) of two variables (I'm
following your textbook's notation here). Then the gradient of φ
is
∇φ=[∂φ/∂x]i+[∂φ/∂y]j
and this is a vector field: it is a function multiplied by i
plus a function multiplied by j.
An example
I looked at φ(x,y)=x^{2}+y^{2}. Here I recalled
that the contour lines were
x^{2}+y^{2}=Constant. These are all circles centered
at the origin. If we draw a collection of contour lines for equally
spaced constants (for example, for 1 and 2 and 3 and 4 and ...) then
it turns out that the lines (level curves: they are curvy!) get
closer and closer together as the equally spaced constants
increase. This is because the function is increasing more rapidly as x
and y get larger in absolute value. The associated gradient vector
field is 2xi+2yj. At (x,y), these vectors are
perpendicular to the circles as theory declares and point away from
the origin in the direction of increase of φ. As (x,y) "moves"
farther from (0,0), the magnitude of the vectors increases.
To the right is a picture of this gradient vector field together with
some contour lines. Previously I had defined g:=x^2+y^2;. The
picture displays the output of these Maple
commands:
> contourplot(g,x=2.5..2.5,y=2.5..2.5, thickness=2,
contours=[1,2,3,4,5], grid=[40,40]);
>
fieldplot(2*x,2*y],x=2.15..2.15,y=2.15..2.15,arrows=slim,grid=[10,10],
fieldstrength=maximal(.9)color=blue,thickness=2);
Another example This example uses a function which is less symmetric and less simple. I don't think I'd like to try to understand this "by hand" unless you gave me a heck of a lot more time than I could take during a class meeting. The function is x^{2}y2x+y^{2}+3y. The level curves are shown for the values 12, 10, 8, 6, 4, 2, 0, 2, 4, 6, 8, and 10: the even integers from 12 to 12. Notice that the arrows seem to be perpendicular to the level curves as they are supposed to be, and that the arrows are longer when the spacing between the contour lines (corresponding to equal differences in the constants) gets closer.
Comment 
In general ...
What was written above is generally true about a gradient vector field
and the contour lines or curves (contour surfaces in
R^{3}). That is, the gradient vector field "arrows" are
perpendicular to the level curves. They "point" towards increasing
values of the function. Their magnitude (the length of the arrows) is
a measure of the density (?) of the contour lines: if the function is
rapidly changing (more dense contour lines) then the length of the
gradient will be longer.
By the way, if ∇φ=F,
then φ is called a potential of the vector field,
F. The word and concepts connected with it are important in
physics. I'll discuss more about this during the next class.
How many potentials?
If φ=x^{2}+y^{2}, then
∇φ=2xi+2yj. I asked if the vector field could
have other potentials? After some thought, we came up with
x^{2}+y^{2}+1 and x^{2}+y^{2}17 as
other potentials, because the ∇ "process" would make the
constants go away.
Then I asked a harder question: could we list all possible potentials of the vector field 2xi+2yj? Well, there was a bit of discussion, and then I wrote x^{2}+y^{2}+Constant, where "Constant" means any possible constant. Why is this true? Well, if two functions have the same gradient, then their difference will have gradient equal to 0. Now the question is: why would a function, let's call it g(x,y), with ∂g/∂x=0 always and ∂g/∂y=0 always have to be constant?

Is the vortex flow a gradient vector field?
Look at yi+xj. Is this vector field a gradient vector
field? If it is, can we find a potential for it? Some attempts were
made to guess a φ for this vector field. I then remarked that
maybe we should turn the question around, and consider: if there is
no potential for this vector field, can we If it is not,
explain why? This turns out to be a serious and interesting question,
because of things we will learn very soon. Also it has some
interesting physical consequences. Well, here is one way to decide the
answer.
Suppose φ(x,y) is a potential for yi+xj. That
means
∂/∂y φ_{x}=y > φ_{xy}=1 ∂/∂x φ_{y}=x > φ_{yx}=1But "mixed" partial derivatives are supposed to be equal (that is, if some hypotheses are satisfied, but these are all very reasonable functions and nothing weird happens). Since 1 is not equal to 1 we know that the vortex flow is not a gradient vector field. I don't think this is an obvious fact.
Return to x^{2}i+y^{2}j Is x^{2}i+y^{2}j a gradient vector field? Let's see: if there is a potential φ then x^{2} should be ∂&phi/∂x which makes me think of x^{3}/3. And y^{2} should be ∂&phi/∂y which makes me think of y^{3}/3. So I decide to guess that φ(x,y)=(1/3)x^{3}+(1/3)y^{3} is a potential for this vector field. By the way, so is (1/3)x^{3}+(1/3)y^{3}205.66783 but I will stick with plain (1/3)x^{3}+(1/3)y^{3}. According to general theory, the vector field will be perpendicular to the level curves of φ. I will have Maple draw some of these curves and display them along with the arrows of the vector field that we saw before. The result is to the right. I hope but I certainly can't guarantee that the contour lines help you understand the fluid flow. The streamlines of this flow are perpendicular to the contour lines. But maybe you can see how the existence of the potential function makes the picture more interesting. Potentials also have other consequences, not just artistic. Comment These level curves are not equally spaced. I played with the contourplot command a bit to get nicelooking curves. The values of the function that I used were [35,20,10,5,1.5,0,1.5,5,10,20,35]. 
Please look at the textbook, which also discusses and shows some 3dimensional vector fields.
Maintained by greenfie@math.rutgers.edu and last modified 11/11/2008.