Math 251 diary, fall 2008
In reverse order: the most recent material is first.

This is the third part of the diary.
A link to earlier material
A link to much earlier material

Wednesday, December 10

The last differentiation operation in the course is something called divergence, or, more briefly, div.

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 (Qz-Ry)i-(Pz-Rx)j+(Py-Qx)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 21st 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:
     (Qzx-Ryx)     (Pzy-Rxy)     (Pyz-Qxz).
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,z2> is not the curl of another vector field, because 0+0+2z is not always 0.

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 (R3) 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:
     ∫∫SF·N dS=∫∫∫Ediv F dV.

The ingredients
Here S divides up space, R3, 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.
  • The sphere
    Here the spatial region is the inside of a sphere (a ball). The surface is the sphere, and the normals, a few of which are shown, point outward from the center of the sphere.
  • A parallelopiped
    This is supposed to be an object with six flat sides, with the opposite sides parallel in pairs. The surface has exactly 6 outward normal vectors, one for each side.
  • A torus
    The region in space is the region inside a torus. The surface has normals pointing out, but now the surface is more complicated. Indeed there are even some normals which point at each other!

Fluids and the Divergence Theorem equation
The equation ∫∫SF·N dS=∫∫∫Ediv F dV itself has meaning in fluid dynamics. The right-hand 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 left-hand 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 R3.

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 dAx,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=0y=1(∂Q/∂y) dy. The Fundamental Theorem of Calculus immediately applies and we get Q(x,y,z)]y=0y=1=Q(x,1,z)-Q(x,0,z). We then integrate both of these:
∫∫x,z between 0&1Q(x,1,z) dAx,z-∫∫x,z between 0&1Q(x,0,z) dAx,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 dAx,y. The Fundamental Theorem of Calculus would apply to the innermost integral:
z=0z=1(∂R/∂z) dz=Q(x,y,z)]z=0z=1=Q(x,y,1)-Q(x,y,0). Again integrate both of these:
∫∫x,y between 0&1Q(x,y,1) dAx,y-∫∫x,y between 0&1Q(x,y,0) dAx,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 dAy,z. Then the Fundamental Theorem of Calculus applies and we've got this:
x=0x=1(∂P/∂x) dx=P(x,y,z)]x=0x=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,z2,3>. 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 cubey dV. If we order the d's as, say, dy dx dz, then dy gives us y2/2 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)= x2i+yzj-4zk, 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.
The flux would be the (triple) integral over the whole sphere (its inside!) of div F=(∂(x2)/∂x)+(∂(yz)/∂y)+(∂(-4z)/∂z)=2x+z-4.
The integral of 2x over the whole sphere will be 0: there are positive and negative chunks of the sphere which will cancel. Similarly, the integral of z over the whole sphere will be 0 (the sphere is terrific, and has "balance" with respect to all of its variables). by the way, the total integral of x33 and y47 and z2003 over this sphere will be 0, for the same reasons. I don't know what the integral of even powers would be. They could be computed but it would be work.
Therefore the flux is equal to the integral of -4 over the sphere. And that's -4 multiplied by the volume of the sphere: (4/3)Π·53. The result is -(2000/3)Π, that same answer as we got from a direct computation. But I could use the Divergence Theorem and symmetry/assymetry rather rapidly. So this is good.

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(y2)i+z3ln(x2+1)j+zk. We need to find the flux of F across the part of the paraboloid x2+y2+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(y2) and ln(x2+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=2-x2-y2: 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=2-x2-y2 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=2-x2-y2 or x2+y2=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 cup1 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=2-x2-y2, or (in "polar") z=2-r2. So we compute
θ=0θ=2Πr=0r=1z=1z=2-r21 dz r dr dθ=∫θ=0θ=2Πr=0r=1(2-r2-1)r dr dθ= ∫θ=0θ=2Πr=0r=1(r-r3)dr dθ= ∫θ=0θ=2Π(r2/2-r4/4)]r=0r=1dθ=∫θ=0θ=2Π(1/4)dθ=Π/2.

Now the surface integral over the "bottom" disc. F·N is (z arctan(y2)i+z3ln(x2+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 2-dimensional version (Green's Theorem), a 2.5-dimensional version (Stokes' Theorem), and even today a 3-dimensional 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.

Monday, December 8

Several weeks ago we asked when a vector field could be a gradient vector field. That is, given F=Pi+Qj+Rk, when is there φ so that ∇φ=F? Although we can integrate and compare the various descriptions, integration is frequently tedious and difficult. I mentioned that a quick check can be gotten by looking at the "cross second derivatives". The resulting equations are also callled "compatibility conditions". Here is a way of encoding this idea.

Some notation which many people think is helpful
The conditions Py=Qx and Pz=Rx and Qz=Ry 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 upside-down delta, ∇, and called nabla or del, was introduced by William Rowan Hamilton (1805-1865).

So "del" is given by: ∇=(∂/∂x)i+(∂/∂y)j+(∂/∂z)k. Here are some uses:

There will be one further use of ∇ defined on Wednesday.

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.

  • A simple closed curve
    So this is a curve in space (R3) with START=END and which has no other self-intersections.
  • A piece of surface
    This should be a piece of a surface, all of whose boundary is the curve mentioned above. As several students remarked, specifying the boundary curve does not mean there's only one surface. In fact, there are many really neat and clever computations which depend on changing the surfaces involved. I'll show you one of these in a few minutes.

  • Work and flux
    Stokes' Theorem connects the work of a vector field around a closed curve with the flux of a related vector field over a surface. So this means that we need to have a direction on the curve (how we push things around) and we also need to make a selection of normal vector on the surface. These choices need to be made together.
    The textbook interprets the work in the fluid dynamics sense, as a circulation. We didn't have enough time in class to consider circulation. There is discussion of this in sections 17.1 (on page 1010) and 17.2 (see page 1025).

     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 right-handed 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 right-handed 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 cup-shaped surface.

    Under these conditions, then the Stokes Theorem Equation is true:

    The boundary curveF·t ds=∫∫The surfacecurl F·N dS

    The textbook writes this in a slightly different way as

    The boundary curveF·ds=∫∫The surfacecurl 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 well-known complicated theorem. If the curve is in R2 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)=y2i+xj+z2k and the surface is the part of the paraboloid z=x2+y2 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 xy-plane. 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 right-handed triple. This took some time to see in class.

    The work integral
    So I need to compute ∫The curvey2dx+x dy+z2dz. 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=0
    and the parameterization interval for the whole circle is [0,2Π]. Then ∫The curvey2dx+x dy+z2dz becomes
    I can "compute" this integral with tricks. It can also be computed using the things done in Calc 2. But we're near the end of the term, and tricks make the computations flow faster.
    First, look at sine on the interval [0,Π/2], and then look at [sin(x)]3. Both the curves go up from 0 to 1. The appearance is flipped left/right on [Π/2,Π], and then the appearance on [0,Π] is flipped down|up on [Π,2Π]. The total integral from 0 to 2Π must be 0 because of the cancellation. The first picture below shows my drawings of sine and the cube of sine. The red/green picture with two curves shows a Maple graph of the two curves. Consequence: ∫t=0t=2Π[sin(t)]3dt=0.

    How about the integral of [cos(t)]2 on [0,2Π]? The value should certainly be the same as the integral of [sin(t)]2 on the same interval since the shapes are the same, just one quarter period out of phase. The sum of these curves is 1 (sin2+cos2) which on [0,2Π] has integral 2Π. So ∫t=0t=2Π[cos(t)]2dt must be half of that and it equals Π.

    The line integral side of Stokes' Theorem is Π.

    The surface integral
    Now we need to compute ∫∫The paraboloidcurl F·N dS.

    The curl
    This is ∇xF, so:

       (  i     j   k   )
    det( ∂/∂x ∂/∂y ∂/∂z )=0i-0j+(1-2y)k
       (  y2    x   z2  )
    Parameterizing the surface, etc.
    Since the surface is presented as a graph, try the graph function itself as a parameterization:
    r(u,v)=ui+vj+(u2+v2)k so ru(u,v)=1i+2uk and rv(u,v)=1j+2vk. Then ruxrv=
       ( i  j  k )
    det( 1  0 2u )=-2ui-2vk+1k
       ( 0  1 2v )
    We discussed the magical cancellation in the last lecture. V·N dS became V·(ruxrv) dAu,v. curl F here is (1-2y)k=(1-2v)k so that curl F·N dS=(1-2v)k·(-2ui-2vk+1k)dAu,v=(1-2v)dAu,v.

    Computation of the surface integral
    We need to identify the domain in the uv-plane which parameterizes our little cup. The uv-plane is the xy-plane in different clothing, but the cup is the graph over the region inside the unit circle: u2+v2≤1. So we need
    ∫∫Inside the unit circle(1-2v)dAu,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(12), 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 surfacecurl F·N dS where the surface is that part of the sphere x2+y2+(z-2)2=8 which lies above the xy-plane, and N is the outward unit normal on the surface, and F is y2cos(xz)i+x3eyzj-e-xyzk.
    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 xy-plane. The boundary of the top portion occurs if z=0 in the equation x2+y2+(z-2)2=8. Then x2+y2+(-2)2=8 and x2+y2=4. This is a circle of radius 2 centered at the origin in the xy-plane. 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 surfacecurl F·N dS=∫The boundary circleF·t ds.
    But notice: this circle is also the correctly oriented boundary of the disc of radius 2 centered at the origin in the xy-plane. So I can use Stokes' Theorem a second time to change the line integral to a much simpler surface integral:
    The boundary circleF·t ds=∫∫The disccurl F·N dS
    This is simpler for several reasons. The region over which we're integrating is flat, a disc in the xy-plane. 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!i-Blah, blah!j+[3x2eyz-2ycos(xz)]k
        ( y2cos(xz) x3eyz  -e-xyz)
    A further simplification occurs. We're on the xy-plane, where z=0. So the k component, 3x2eyz-2ycos(xz), becomes 3x2-2y because cos(0)=1 and e0=1.
    So we need ∫∫The disc3x2-2y dAu,v. Just as in the previous problem, the -2y integral over the disc is 0, because there is cancellation of the positive and negative contributions of y. I see no clever way to compute the 3x2 integral and will do this using polar coordinates (with x=rcos(θ)):
    ∫∫The disc3x2dAu,v=3∫θ=0θ=2Πr=0r=2r2[cos(θ)]2r dr dθ= 3∫θ=0θ=2Π[cos(θ)]2dθ∫r=0r=2r3dr. The θ integral is Π (a trick used before) and the r integral is 16/4. So the flux is 12Π.

    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 1026-1027.

    Green's Theorem
    If the boundary curve is in R2 and the "surface" is a region in R2 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 Qx-Py. The Stokes' Theorem equation declares that the integral of Pdx+Qdy over the boundary curve (with the usual orientation) equals the double integral of Qx-Py over the interior.

    Wednesday, December 3

    The initial 20 minutes were devoted to finishing up 16.4. I fell behind last time, perhaps because I didn't feel well. I'm sorry.

    Surface integrals
    If we believe a piece of surface in R3 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 ∫∫Rectf(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(x2+y2).

    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 z-axis. 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 z-axis. 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.

    Since r(u,v)=<u cos(v),u sin(v),v>, we have
          ru=<cos(v),sin(v),0>       rv=<-u sin(v),u cos(v),1>
    The cross product:

             (    i          j      k )
    ruxrv=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)i-cos(v)j+uk. The magnitude of this vector is sqrt(1+u2). So here dS=sqrt(1+u2) du dv.

    The function or "density" is sqrt(x2+y2) but in the helicoid parameterization, x=u cos(v) and y=u sin(v). So sqrt(x2+y2) accidentally (!) turns out to be u because again cos2+sin2=1.

    The integral ∫∫Rectf(x,y,z) dS is also accidentally ∫v=0v=2Πu=0u=1u sqrt(1+u2) du dv. By total coincidence, we can compute this exactly. It is 2Π(sqrt(2)/3-1/3). Wow. Wowie. I guess this is terrific (and all totally arranged, of course).

    Non-textbook example
    What's the surface integral of x3+y2+z of that part of the graph z=3x2+5y4 which is over the unit square, 0≤x≤1 and 0≤y≤1? The parameterization is the straightforward r(u,v)=ui+vj+(3u2+5v4)k so that ru=1i+6uk and ru=1j+20v3k and the cross-product ruxrv is -6ui-20v3j+1k. (I did this in class. If you weren't there [$40} please check the result!). Therefore dS is sqrt(1+36u2+400v6) du dv (not so agreeable, huh?). And x3+y2+z changes to u3+v2+3u2+5v4. So we just need to compute
          ∫v=0v=1u=0u=1(u3+v2+3u2+5v4)sqrt(1+36u2+400v6) 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."

    The horrible factor ||ruxrv|| 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 R3. We could imagine a surface in R3, 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?
    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.

    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 cross-sectional 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 half-twist 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,z2,3> 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, R3, into two pieces, points inside and points outside the surface. This is similar to a closed curve in R2. Such surfaces are important in physical applications. We will choose the normal to be the outward-pointing 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.

    So the total flux is 1/2. I know a secret way (called the Divergence Theorem) to check this result (I just did, in my head!). I think the answer is correct.

    Magical cancellation!
    If our surface is parameterized there is a natural way to get a unit normal. Just take ru and rv: these are velocity vectors for curves on the surface, and are tangent to the surface. Their cross-product 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 ruxrv divided by the scalar ||ruxrv||. If Flux=∫∫The surfaceF·N dS this will be the same as
    ∫∫The surface[F·(ruxrv)/||ruxrv||] ||ruxrv|| dAuv
    Look at the marvelous cancellation (much the same as what occurred in the line integral case).
    Therefore the flux integral is ∫∫The surfaceF·(ruxrv) dAuv.
    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 x2i+yzj-4zk 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 ruxrv=-25cos(u)[sin(v)]2i-25sin(u)[sin(v)]2j-25sin(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)2sin(v)2i+25sin(u)sin(v)cos(v)j-20cos(v)k

    We need to integrate F·N dS, but this, because of the cancellation of the horrible factor ||ruxrv|| becomes just F·(ruxrv) dAu,v. Let me match up the components and get the integrand:
    -625sin(u)2sin(v)4cos(u)-625 sin(u)2sin(v)3cos(v)+500cos(v)2sin(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 z-axis, and the half-cylinder 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 z-axis). The problem asks to determine whether the flux integral ∫∫half-cylinderF·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.
    1. F=i; the total flux is positive.
    2. F=j; the total flux is zero.
    3. F=k; the total flux is zero.
    4. F=yi; the total flux is positive.
    5. F=-yj; the total flux is negative.
    6. F=xj; the total flux is zero.

    Monday, December 1

    Now we will discuss parameterized surfaces and surface integrals (material in section 16.4). Let's see: the "ratio" I wrote on the board actually guides the last 4 meetings of course:

    Line integrals and work are to Green's Theorem
    Surface integrals and flux are to Stokes' Theorem and the Divergence Theorem

  • Surface integrals
    We will begin a discussion of parametric surfaces today, and then will go on to considerations of surface area. We will meet surface integrals themselves at the next lecture, Tuesday.

    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 cross-product of A and B:

       ( i   j  k )
    det( 4  -1  3 )=-17i-5j+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
    This implicit description, using a normal vector, was very convenient for parts of the course since the gradient always gives a vector normal to a level surface, and therefore normal to a plane which is tangent to that level surface. But there's another way to describe the plane.

    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,1-u+5v,2+3u+2v>. The textbook writes this as a vector position function:

    Notation comment
    The textbook uses Φ(u,v) for the vector-valued 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 3rd Avenue and 47th 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 R3 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 R3.

    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 xy-plane 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 xy-plane, and radially oriented, is revolved around the z-axis. One view of what I'm trying to describe is to the right. The center of the circle to be revolved around the z-axis is moved so that it describes a circle in the xy-plane centered at the origin. There are two more views below of the surface. One is from "above", looked down the z-axis. The other is from the "side", looking along the y-axis.
    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 z-axis to the center of the circle will be 4, and the radius of the circle being revolved around the z-axis will be 2. The left picture below shows the angle u. The right picture shows a slice perpendicular to the xy-plane, 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 xy-plane 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 vector-valued 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 21st 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 z-axis. 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 xz-plane (where y>=0) and below the xy-plane (where z<=0). This is a quarter of the torus.

    A graph
    Here I considered a rather simple surface given by a graph: z=3x2+5y4. Certainly all I expected people to see and say was this this surface is (vaguely) cup-shaped, 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+(3u2+5v4)k. The parameterization geometrically consists of pushing up a grid from the xy-plane 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 ru, that is, ∂r/∂u (take the partial derivative of the component functions with respect to u). The text calls this Tu 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 rv. 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 ||ruxrv||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 uv-plane turns out to be ∫∫D||ruxrv|| du dv. The textbook calls ||ruxrv|| du dv by the name, dS, for a piece of surface area. It will also be useful to realize that ruxrv, a cross-product 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 ru=-5sin(u)sin(v)i+5cos(u)sin(v)j+0k
    and rv=5cos(u)cos(v)i+5sin(u)cos(v)j-5sin(v)k

             (      i               j           k    )
    ruxrv=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)]2sin(v)cos(v)-25[cos(u)]2sin(v)cos(v). The last component simplifies (sin(u)2+cos(u)2=1) and the result is
    Now for the magnitude of this vector: the square root of the sum of the squares of the components. There's a common factor of 252[sin(v)]2 when everything is squared, so I'll take that out of the square root (sin(v) is positive when v is between 0 and π as it is here). So we must multiply 25sin(v) by {cos(u)sin(v)]2+[sin(u)sin(v)]2+[cos(v)]2. Now look at this: one use of sin(u)2+cos(u)2=1 turns this into sin(v)2+cos(v)2 and that's 1 also! So therefore we square root things and see that |ruxrv|=25sin(v). We get the whole sphere when the coordinate patch is 0<=u<=2π and 0<=v<=pi, so the whole surface area is ∫0π025sin(v) du dv. This integral can be computed, and its value is 100π. The surface area of a sphere of radius r is supposed to be 4π r2 (four "great circles") so when r=5 this is the textbook answer.

    The surface area of the torus
    You could try this yourself. The answer is 32π2, 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.
    Then ru=-(4+2cos(v))sin(u)i+(4+2cos(v))cos(u)j+0k
    and rv=(-2sin(v))cos(u)i+(-2sin(v))sin(u)j+2cos(v)k

             (          i               j               k   )
    ruxrv=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)2cos(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)]2sin(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)]2sin(v)+4[sin(u)]2cos(v)sin(v)+8sin(v)[cos(u)]2+4[cos(u)]2sin(v)cos(v). This simplifies (again with sin2+cos2) 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:
    42cos(v)2cos(u)2(2+cos(v))2+ 42cos(v)2sin(u)2(2+cos(v))2+ 42sin(v)2(2+cos(v))2
    Again sin2+cos2 is used, and we get: 42cos(v)2(2+cos(v))2+ 42sin(v)2(2+cos(v))2
    One last time, sin2+cos2, and this becomes: 42(2+cos(v))2
    This is ||ruxrv||2, so we take the square root and get 4(2+cos(v)) (notice that 2+cos(v) is always non-negative so the square of its square is itself). Now to get the area of the whole torus we need to integrate this over the uv-square which is [0,2π] by [0,2π]. So the area is ∫004(2+cos(v)) du dv. The answer is 32π2.
    This answer is correct. A torus is such a symmetric figure that its surface area can be determined using a (sort of) calculus-free method called the Theorem of Pappus.

    Monday, November 24

    Today I'll deviate significantly from the published syllabus by altering the order of the topics. The reason is that I'd like to discuss a result called Green's Theorem before we develop yet another kind of integral in the course. This is section 17.1. I will return with the concluding sections of chapter 16 after Thanksgiving.

    Evaluating line integrals
    I basically know three ways to evaluate line integrals:

    1. Parameterize them, and hope that the Fundamental Theorem of Calculus can handle the resulting definite integral.
    2. Hope that the integrand results from a gradient vector field, and try to create a potential. Then the value of the line integral is the difference of the values of the potential at the End minus its value at the Start.
    3. Green's Theorem, which tells how to compute line integrals on closed curves.
    Green's Theorem will look at first rather special but the uses are really remarkable. I should emphasize that although this is not a course on physics or on engineering applications, Green's Theorem was actually developed in the 19th century to handle certain problems about electricity and magnetism, and then also about fluid flow. The result was not "invented" in some abstract situation, but really did follow from attempts to understand how to describe real world problems.

    George Green, whose name is attached to this result (and to others in mathematical physics) was essentially self-taught 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 bdryP(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+Junk1(x) and Q(x,y)=x2+Junk1(y). Here the Junk functions were some absurd formulas: arctan(y3) or ecos(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 2x-5, 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 bdryP(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=0x=ay=0y=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=0y=b–(∂P/∂y) dy=P(x,y)|y=by=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 ∫Sy dx+x2dy 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 ∫Ly dx+x2dy where L is the line segment along the x-axis 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+x2dy is 0 on all of L.

    So ∫Sy dx+x2dy=∫Sy dx+x2dy+∫Ly dx+x2dy=∫S+Ly dx+x2dy, 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+Ly dx+x2dy by ∫∫Q∂(x2)/∂x-∂y/∂y dA. The integrand is 2x-1, and we need to integrate it over Q.
    The integral ∫∫Q2x dA is actually 0, because the region Q is symmetric with respect to the y-axis, and 2x takes equally positive and negative values on this region. So the values cancel. And the integral ∫∫Q-1 dA is minus one-half the area of a circle of radius 3. Therefore the original line integral, ∫Sy dx+x2dy, 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:
    CP(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=at=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=at=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 (R3) 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/(x2+y2) and Q(x,y)=x/(x2+y2)
    It just happens (!!) that:
    Therefore in this case, Py=Qx so that Qx-Py is 0.

    Now let me compute the line integral ∫CP(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 x2+y2=32 because sin2+cos2=1. Thus
    C-y/(x2+y2)dx+x/(x2+y2)dy becomes ∫t=0t=2π([-3sin(t)/32](-3sin(t))+[3cos(t)/32]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 Qx-Py 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?

    No! That is not true.

    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 x2+y2 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 non-zero" 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/(x2+y2)dx+x/(x2+y2)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.

    Let me select a point A on W and a point B on C and join them by a line segment. Now I want to describe a closed curve in the plane, and this is very very clever. I'll go first from (8,0) on W to A. Then I will go from A to B on the crosscut. Then around C in reverse (note that the picture now has the arrowhead on C reversed and the label states "-C"). Then I go from B to A, and finally, I follow the remainder of W around to (8,0). What about the integral of P(x,y)dx+Q(x,y)dy along this path? The integrals on the crosscut line segments cancel, and the integral on -C is minus the integral along C, and the remainder is the integral on W. So the result is the integral along W minus the integral along C.

    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, Py=Qx 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!
    That clearly may be the worst clearly in the course. I don't think an accusation of excessive rigor (too much proving: "the quality of being logically valid") would be correct in Math 251, and the statement I am asserting as clear needs proof. What I am asserting is that if one path is close to another, and if P(x,y) and Q(x,y) are, say, at least continuous, then the work along the two paths will be close. To me this is physically reasonable. I hope it is to you.

    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?
    I think the computations I've just done are difficult to understand. Here is another physical model. Line integrals can also be thought of as measuring flux, the net flow through the curve. We will investigate this situation in R3 soon. But consider the following physical setup: two parallel planes of plastic, and a hose inserted through one of the planes, pumping water in at a steady rate (the blue things in the picture are not minnows, they are water drops!). If we surround this source with some sort of circular detection "fence" then we can look at how the water goes through the fence, and from that deduce the rate at which water is being pumped in. But we could also install another detection fence. If we do the same computations for that fence, the source rate should be the same, because the hose is pumping water in steadily: its rate is not varying. The mathematical computation with the crosscuts above is a version of this physical situation. When Py-Qx=0, away from the hose (at the origin) the source rate is 0. Any comprehensive measurement of flow surrounding the origin will get the same answer.
    Another physical setup which this P and Q model is an isolated point charge. There the work done in moving around the charge will always be the same. We will encounter this in three dimensions soon, and maybe there it will be easier to understand.

    Another application (important!) of Green's Theorem
    I did not get a chance to discuss this in class. When is a vector field P(x,y)i+Q(x,y)j conservative? Certainly when it is a gradient vector field: so there is a potential function, f(x,y), with ∂f/∂x=P(x,y) and ∂f/∂y=Q(x,y). We can try to find f(x,y) by integrating and matching the descriptions. But this can be difficult and lengthy. Differentiation is easier, and certainly if there is a potential, f(x,y), then ∂P/∂y=∂Q/∂x by Clairaut's Theorem. But the example above (the isolated electric charge or the fluid flow point source) shows that we can have ∂P/∂y=∂Q/∂x without having a potential. Why? If we had a potential then the integral around closed curves would be 0, but we saw that such an integral for our example may not be 0. If we had no bad points, then the integral around closed curves would be 0 using Green's Theorem. So any region where there are no bad points has the converse true:
    If ∂P/∂y=∂Q/∂x and there are no "bad points" for P or Q, then there is a potential.
    This can sometimes be very useful because differentiation is ... easier! So people have given a name to regions which have no holes. A region is simply connected if it has no holes. In such a region, if the condition ∂P/∂y=∂Q/∂x is correct at all points, then there is a potential, so the vector field is conservative. A disc (the inside of a circle) is simply connected. But if you remove a point, the result is not simply connected.

    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.

    Monday, November 17

    Today finally we will begin to get what I think of as significant payoffs from 251. There are some remarkable results which I'll begin telling you about today. In particular, I really know only a few ways to evaluate line integrals. So far, we've seen "Parameterize and use the definition." Today we'll see a distinctly different method, one which has great physical significance.

    An example
    The examples I've been giving in class (and here) have almost all concerned 2-dimensional vector fields. I'll discuss one more such example, and then the remaining part of the discussion in this lecture will have 3-dimensional examples, even though 2-dimensional 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+y2dy from (1,2) to (3,10). If you wish you could think that this involves computing the work against the force field xyi+y2j.

    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.

  • From (1,2) to (3,2): x=t, y=2, 1≤t≤3, dx=dt, and dy=0dt. Then xy dx+y2dy integrated on this horizontal line segment becomes ∫13t(2) dt=t2]t=1t=3=9-1=8.
  • From (3,2) to (3,11): x=3, y=t, 2≤t≤11, dx=0dt, and dy=dt. Then xy dx+y2dy integrated on this vertical line segment becomes ∫211t2dt=(1/3)t3]t=2t=11=(1,331-8)/3=1,223/3.

    Another work computation
    We could move from (1,2) to (3,11) along y=x2+1. For this path, we can choose x=t, so y=t2+1, 1≤t≤3, dx=dt, and dy=2t dt. Then xy dx+y2dy integrated on this parabola becomes ∫13t(t2+1)(1)+(t2+1)2(2t) dt= ∫13t3+t+(t2+1)2(2t) dt= (1/4)t4+(1/2)t2+(1/3)(t2+1)3]t=1t=3= (1/4)34+(1/2)32+(1/3)(32+1)3 -(1/4)14-(1/2)12+(1/3)(12+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.

    So the integral of xy dx+y2dy 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 R3 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=<F1,F2,F3> is conservative in R3. 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 F1(x,y,z) dx+F2(x,y,z) dy+F3(x,y,z) dz over these three line segments. Here we go, but watch carefully, because the details are quite important.

  • First segment
    Here y is changing and x and z do not change. So a simple parameterization is x=-2,y=t,z=3, and then dx=0 dt, dy=1 dt, and dz=0 dt. The integral becomes ∫6yF2(-2,t,3) dt.
  • Second segment
    Here x is changing and y and z do not change. So a simple parameterization is x=t,y=y (a constant!),z=3, and then dx=1 dt, dy=0 dt, and dz=0 dt. The integral becomes ∫-2xF1(t,y,3) dt.
  • Third segment
    Here z is changing and x and y do not change. So a simple parameterization is x=x (a constant),y=y (a constant),z=t, and then dx=0 dt, dy=0 dt, and dz=1 dt. The integral becomes ∫3zF3(x,y,t) dt.

    The total integral which is g(x,y,z) is therefore the sum of all three of these.
          g(x,y,z)=∫6yF2(-2,t,3) dt+∫-2xF1(t,y,3) dt+∫3zF3(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 F3(x,y,t) and we are integrating with respect to t. Therefore ∂g/∂z=F3(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 F2. The result would be ∂g/∂y=F2(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=F1(x,y,z).

    A conclusion
    Suppose F=<F1,F2,F3> 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=F1(x,y,z);     ∂g/∂y=F2(x,y,z);     ∂g/∂z=F3(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=<F1,F2,F3> and that I happen to know a function g so that ∂g/∂x=F1(x,y,z);, ∂g/∂y=F2(x,y,z), and g/∂z=F3(x,y,z). I would like to compute the line integral of F1(x,y,z) dx+F2(x,y,z) dy+F3(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, ∫CF·Tds, becomes
    StartEnd(F1(x(t),y(t),z(t))(dx/dt)+F2(x(t),y(t),z(t))(dy/dt)+F3(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,
    the whole thing, is the derivative with respect to t of g(x(t),y(t),z(t)). This is exactly because F1 is ∂g/∂x, F2 is ∂g/∂y, and F3 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 ∫StartEndd/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.

    Suppose g(x,y,z)=x3y+xz2+5. Then ∇g is (3x2+z2)i+x3j+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 y-axis, 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.

    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)02+5=-250&pi+5.

    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)02+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 x-axis. 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(3x2+z2)dx+x3dy+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:

    1. F=<F1,F2,F3> is path independent (or, if you prefer, conservative).
    2. If F=<F1,F2,F3>=∇g, and C is any curve, then ∫CF·Tds=∫CF1(x,y,z)dx+F2(x,y,z)dy+F3(x,y,z)dz=g(End)-g(Start).
    3. If C is any closed curve, then ∫CF1(x,y,z)dx+F2(x,y,z)dy+F3(x,y,z)dz.

    An important physical example
    We looked at inverse square vector fields in R2. A three dimensional version is not too difficult to exhibit:
    This vector field points from (x,y,z) to (0,0,0) because it is a positive multiple of -xi-yj-zk. The reason for the (initially) strange scaling factor of (x2+y2+z2)-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 -(x2+y2+z2)-1/2 is -x/(x2+y2+z2)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 non-spherical 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 <F1,F2,F3> 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 xy2i+(x2y+z)j+5z2k. Is it a gradient vector field? I will check some derivatives.

  • Then ∂/∂y(xy2)=2xy should be the same as ∂/∂x(x2y+z)=2xy. (should because ∂2/∂x∂y=∂2/∂y∂x.) That's true, but there's more to check.
  • Also ∂/∂z(xy2)=0 should be the same as ∂/∂x(5z2)=0. (should because ∂2/∂x∂z=∂2/∂z∂x.) That's true, but there's more to check.
  • And finally, ∂/∂z(x2y+z)=1 should be the same as ∂/∂y(5z2)=0. (should because ∂2/∂y∂z=∂2/∂z∂y.) That is not true, so this vector field is not a gradient vector field, and therefore it can't be conservative.

    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.

    Wednesday, November 12

    I began with might have seemed some irrelevant observations, but these "observations" were an effort to get students to expect that certain kinds of integrals would be useful in investigating vector fields.

    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)x3 is an antiderivative of x2, I can then assert that all antiderivatives of x2 are (1/3)x3+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(x2y)+x7,x2cos(x2y)-y5> is a gradient vector field. So it is <∂φ/∂x,∂φ/∂y>. What can we do now?

  • ∂φ/∂x=2xycos(x2y)+x7, and therefore to recover φ we should compute ∫(∂φ/∂x)dx=∫2xycos(x2y)+x7dx. We can do this but we need only remember that y is a constant in this computation since we are integrating with respect to x. We can do the cosine part of this with a simple substitution, and the result will be sin(x2y)+(1/8)x8+Some stuff. O.k.: the trick now is to think about what Some stuff could represent. Indeed, it is the stuff that when we do ∂/∂x to it, we get 0. Certainly constants like 12π+sqrt(117) qualify, but also (we are in two variable calculus!) any function of y disappears. Look: ∂/∂x of sin(e5y2)) is 0. So, in fact, Some stuff is Any function of y.
  • We also know that ∂φ/∂y=x2cos(x2y)-y5. We can use this to recover information about φ directly by computing another integral: ∫(∂φ/∂y)dy=∫x2cos(x2y)-y5dy. This we can do also. Again, remember that anything involving x is a constant for this integration. The first piece is a simple substitution and the second is easy. In any case, the result is sin(x2y)-(1/6)y6+Some other stuff. This is the result of a dy integral, so Some other stuff is any function of x (hey, it could be x/arctan(log(x)) or just about anything!) involving y.

    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(x2y)+(1/8)x8+Any function of y
              sin(x2y)-(1/6)y6+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(x2y). 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:

    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.

    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 R2. Its geometry is simple, with a constant cross-sectional 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 cross-sectional 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 wireD(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 wireD(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 wireD(x,y) ds=∫t=0t=Pi/2{21sin(t)+5}3 dt=-63cos(t)+15t]t=0t=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=x4 from, say, (0,0) to (2, 16), and the density is D(x,y)=x4y7. I not being totally idiotic here. Any computational strategy in calculus should be able to handle low-degree 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=t4 and then the density x4y7 became t4(t4)7=t28. 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=4t3, and therefore ds/dt=sqrt(1+16t6). The line integral for this "mass" translates in t-land as follows:
    The wireD(x,y) ds=∫t=0t=2t28sqrt(1+16t6) 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.

    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 R2 but the same ideas will work in R3, 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 ∫CF·Tds.

    A computation
    Let us "test" this definition with a random computation: well, sort of "random". C will be the curve y=x4 and p will be the point (0,0) and q will be the point (2,16). I will have the force field be x2y3i+xy5j. Now let us compute. We need to change everything into t-land, where I will choose a most routine parameter: x=t so that y=t4.

    As before, the speed becomes
    sqrt([dx/dt]2+[dy/dt]2)=sqrt([1]2+[4t3]2)=sqrt(1+16t6) which is horrible enough. Thus ds=sqrt(1+16t6) dt.

    At the point (x,y), F is x2y3i+xy5j so that at (t,t4) on the curve, F is t2(t4)3i+t(t4)5j=t14i+t21j.

    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,t4> so that the velocity vector is <1,4t3>. But we need a unit vector to get the projection of F in the direction of the curve. Divide by the magnitude of <1,4t3>. Therefore, T=<1,4t3>/sqrt(1+16t6).

    Assembling the work integral
    CF·T ds=∫t=0t=2(t14i+t21j)· (<1,4t3>/sqrt(1+16t6))sqrt(1+16t6) dt. This is ∫t=0t=2t14+4t24 dt. which with totally routine polynomial calculations can be evaluated: it is 215/15+(4/25)225.

    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)=x2y3i+xy5j and the curve is y=x4 going from (0,0) to (2,16).

    A solution
    So I initially write ∫CF·T ds but then I immediately change to ∫Cx2y3dx+xy5dy. I evaluate this line integral again by changing everything to t-land. So x=t and y=t4 and dx=1 dt and dy=4t3 dt, Also, x2y3=t2(t4)3=t14 and xy5=t(t4)5=t21. Therefore x2y3dx+xy5dy=t14 dt+t214t3 dt.
    Therefore ∫Cx2y3dx+xy5dy=∫t=0 [START]t=2 [END] t14+4t24 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. 1431-1506). 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 Greco-Roman myth, Apollo, the archer-god 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!

    Monday, November 10

    Where are we? Where are we going?
    About 75% of the course has gone by. Much of what we've done will be very useful for many of the students, but, perhaps, some of what we've done will be neeeded less often (I am trying to be "diplomatic"). The last portion of the course, about what is frequently called Vector Calculus, can help students understand and model many complicated phenomena they will likely encounter.

    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=ax=bf(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*x-y^2], x=1..5,y=1..5, arrows=slim, grid=[10,10], fieldstrength=maximal(.9), thickness=2);
    This vector field is the vector (x2+10y)i+(3x-y2)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.

    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/(x2+y2). 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/(x2+y2). The direction should be from (x,y) to (0,0): the direction of -xi-yj. But that vector has length sqrt(x2+y2). So to get a vector of the correct magnitude and direction we need to divide by sqrt(x2+y2) (that creates a unit vector pointing in the correct direction) and then multiply by GM/(x2+y2). So an inverse square law pointing at the origin (dropping the constant multipliers) is -[x/(x2+y2)3/2]i-[y/(x2+y2)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
    The velocity field of a fluid flow may be less familiar. Here we could think of water flowing in between two parallel planes, close together. At a point we could insert a drop of ink, and look a very short time later. The ink might be stretched into a short segment. This would be the velocity vector of the fluid at that point. It could change from one point to another, both in length and in direction. The length of the segment will (hopefully!) depend on the speed of the fluid, and the direction of the line segment will be the direction of the fluid 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(A2+B2), 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 y-axis, but flows away from the y-axis otherwise, and faster, the farther the fluid was away from the axis.
    x2i+y2j. 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.
    Not this kind of arrow!
    Identify the historical figure and the painter and win BIG!

    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)=x2+y2. Here I recalled that the contour lines were x2+y2=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 x2y-2x+y2+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.

    I looked at this picture and wondered what the heck happened where the arrows got really small. What does happen? Well, we can look for a critical point: φx=2xy-2=0 and φy=x2+3y+3=0. The first equation tells me that y=1/x and then the second equation becomes x3+3x+3=0, and I can't "solve" that. So I asked fsolve and was told there was one critical point, at (approximately) x=-.596 and y=-1.678: this looks correct. What kind of critical point? (Max/min/saddle) If you look at the picture and the level curves you can probably guess. I actually computed the Hessian, and it is a saddle. Ain't math fun?

    In general ...
    What was written above is generally true about a gradient vector field and the contour lines or curves (contour surfaces in R3). 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=x2+y2, then ∇φ=2xi+2yj. I asked if the vector field could have other potentials? After some thought, we came up with x2+y2+1 and x2+y2-17 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 x2+y2+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?

    I didn't have time for this in class (it is in section 16.1, though). But here is the information we know:
         ∂g/∂x=0 always
         ∂g/∂y=0 always
    and I would like to convince you that g's values don't change.
    The green path
    Take the point p in the plane with coordinates (x0,y0) and another point q, on the same horizontal line. So q's coordinates are (x1,y0). I would like to "compare" g's values at p and q. So I would like to somehow "compute" g(x1,y0)-g(x0,y0). Please notice that nothing is happening in the second variable. But the Fundamental Theorem of Calculus applies, and this is the same as ∫t=x0t=x1(∂g/∂x)(t,y0) dt. The difference in the x variable is the same as the definite integral of the partial derivative with respect to x from x0 to x1. But we are assuming that this partial derivative is always equal to 0. Therefore g(x1,y0)-g(x0,y0)=0 (the definite integral of 0 is 0) so that g(x1,y0)=g(x0,y0).
    The blue path
    Take the point q in the plane with coordinates (x1,y0) and another point r, on the same vertical line. So r's coordinates are (x1,y1). I would like to "compare" g's values at q and r. So I would like to somehow "compute" g(x1,y1)-g(x1,y0). Please notice that nothing is happening in the first variable. But the Fundamental Theorem of Calculus applies, and this is the same as ∫s=y0s=y1(∂g/∂y)(x1,s) ds. The difference in the y variable is the same as the definite integral of the partial derivative with respect to y from y0 to y1. But we are assuming that this partial derivative is always equal to 0. Therefore g(x1,y1)-g(x1,y0)=0 (the definite integral of 0 is 0) so that g(x1,y1)=g(x1,y0).
    This is very clever. As long as we can get from one point to another by a succession of vertical and horizontal line segments, the values of the potential won't change. In this course, we will always have functions defined in such regions (regions which are connected -- they have just one "piece"). So indeed we can conclude that if ∇φ=0, then φ is constant. And further, if two functions have the same gradient, they must differ by a constant. And even further, if we can "guess" one potential, then we know all potentials -- just add on an arbitrary constant. In physics terms, the choice of the value of that constant is called "selecting a ground state". Talk to the physics people about these words if you like.

    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

    φx=-y -----> φxy=-1
    φy=x  -----> φyx=1
    But "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 x2i+y2j
    Is x2i+y2j a gradient vector field? Let's see: if there is a potential φ then x2 should be ∂&phi/∂x which makes me think of x3/3. And y2 should be ∂&phi/∂y which makes me think of y3/3. So I decide to guess that φ(x,y)=(1/3)x3+(1/3)y3 is a potential for this vector field. By the way, so is (1/3)x3+(1/3)y3-205.66783 but I will stick with plain (1/3)x3+(1/3)y3. 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 nice-looking 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 3-dimensional vector fields.

    Maintained by and last modified 11/11/2008.