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.

The curl
If F(x,y,z)=P(x,y,z)i+Q(x,y,z)j+R(x,y,z)k, then the curl of F is another vector field, ∇xF. It is a cross-product, and can be evaluated by taking the determinant:

    (    i        j        k     )
det (   ∂/∂x     ∂/∂y     ∂/∂z   )
    ( P(x,y,z) Q(x,y,z) R(x,y,z) )
It turns out this is equal to (if I don't foul up the signs!) the following vector field:
(Ry–Qz)i–(Rx–Pz)j+(Qx–Py)k.
F satisfies the compatibility conditions above if curl F=∇xF is 0. This isn't an accident. Hamilton invented the del so that it would be useful!

If a vector field is a gradient, its curl is 0
A way of testing if a vector field is a gradient vector field: take the curl: the result should be 0. Hamilton would want you to remember this by writing ∇x∇f=0 since a vector crossed with itself is 0.

The formula for curl is horrible, and difficult to remember but the notation (especially the cross product and ∇) is supposed to help. Suppose we ∂/∂x the i component of curl F, ∂/∂y the j component, and ∂/∂z the k component. Here in this 21st century (!) math course, I may 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).
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).

If a vector field is a curl, its divergence is 0
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. Hamilton would want you to remember this by writing ∇·∇xF=0 since a vector dotted with something which is a cross product with itself must be zero!

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

    Pretty darn embarrassing that I forgot the "curl" the first time I wrote the statement of the theorem. I kept telling myself on the way to Hill Center this morning, "Don't screw up the statement of Stokes' Theorem. Don't screw up the statement of ..." so of course what did I do? I am so darn sorry.

    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
    ∫t=0t=2Π–[sin(t)]3+[cos(t)]2dt.
    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 a few lectures ago. 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.

    I think I also made a mistake about this is class. I think what's written here is correct and what I've written here is correct.

    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=2Π∫r=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.

    Comment
    I did this problem because using the same boundary curve to switch surfaces is a very common "trick" done in electromagnetism and fluid flow. If two surfaces have the same boundary and if the vector field is nice, then the flux of the curls of the vector fields through the two surfaces must be the same. This is weird and wonderful, and people use it. See the discussions on page 1024 and 1026-1027.

    Also since the divergence of a curl is 0, the flux of a curl vector field on a closed surface must be 0 (the Divergence Theorem) so again the previous result is verified.

     
    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.

    FTC through the ages ...
    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 3-dimensional version, the Divergence Theorem, and even today a 2.5-dimensional version (Stokes' 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 ...