We will deal with a homogeneous, isotropic material. Here "homogeneous" means the material is the same from one point to the next, and "isotropic" means that the directions around a point all have similar properties. Bean soup is not homogeneous because of the beans and the soup. A piece of lumber is not isotropic because behavior with/against the grain is different.
Assumption #1
The heat inside a small piece of material is directly proportional to
the mass of the material and the temperature. The constant of
proportionality is called specific heat. Using the language of
Math 251, we can write that the heat inside a small piece of
material=_{The piece}K(Density)u(x,y,z,t) dV.
Here K is the specific heat, and Density is the density of the
material (so Density dV is a small amount of mass). Density and K
are constants in our circumstances. u(x,y,z,t) is the temperature at
the point (x,y,z) of the material at time t. In class, and in the reference cited
above some specific heats are discussed (water has high K and
metals, lower).
Assumption #2
Heat flows from hot to cold. The heat flow is directly proportional to
both the difference in temperature and to the crosssectional area
through which the heat flows. If there is a big surface through which
the heat can flow, than there will be much more heat flow. And if the
temperature difference is high, there will be more heat flow. The
constant of proportionality here is called the conductivity,
c. Again, we discussed these assumptions in class and the reference
has further information. I can stir a hot stew safely with a wooden
spoon (relatively low conductivity) but I'd better be careful if I use
a spoon made of metal (higher conductivity).
Hot inside But I need to be careful. Consider a small part of the surface of our small piece. Suppose, for example, that the inside of the piece is warmer than the outside. The gradient of the temperature, u, points in the direction of increasing temperature. So it will point towards the inside of the piece. If we compare an outwardpointing unit normal, then u·n will be negative because the angle between the two vectors is between Pi/2 and Pi and cosine is negative in that range. So if the inside is hotter, u·n is negative.
Hot outside
If the temperature is hotter outside, then the gradient vector of the
temperatur will point outwards, towards the higher temperature. Then
if we compute u·n the result will positive
because the angle between the two vectors is between 0 and Pi/2. So
the "contribution" of the dot product is positive if the outside is
hotter than the inside.
So the dot product u·n is positive when the heat flows
towards the piece and negative if the heat should flow away.
Art! By the way, the pictures are supposed to show hot with flames and cold with ice cubes. This is tremendously imaginative and wonderful. maybe
Therefore the amount of heat coming in/going out is _{The surface of the piece}cu·ndS.
But the change in the heat over time is also (/t)_{The piece}K(Density)u(x,y,z,t) dV.
This is one place where our simplifying assumptions help a great
deal: K and the Density are constant. Only u varies with time. So
this is equal to
_{The piece}K(Density)(u/t) dV.
Now comes the magic! We apply the Divergence
Theorem to the surface integral giving the change in heat and get
_{The piece}divergence of cu dV.
Well, cu is
cu_{x}i+cu_{y}j+cu_{z}k
and the divergence of this (c is a constant) is
c(u_{xx}+u_{yy}+u_{zz}). By the way, the
collection of derivatives u_{xx}+u_{yy}+u_{zz}
also has a specific name: it is called the Laplacian. Notation
for the Laplacian of u varies. In some places, ^{2}u is used, and, otherwise,
u is used.
And so:
We now know that for any little piece of the material, the following
triple integrals are equal because they both give the change in heat
of the little piece:
_{The piece}K(Density)(u/t) dV=_{The piece}c(u_{xx}+u_{yy}+u_{zz}) dV.
If we take a really really tiny piece, then the functions inside each
integral are approximately constant (because they are continuous!) so
we can approximate the integrals by the function values and the
volume:
K(Density)(u/t)The Volume=c(u_{xx}+u_{yy}+u_{zz})The Volume
Divide both sides by The Volume and get The Heat Equation:
K(Density)(u/t)=c(u_{xx}+u_{yy}+u_{zz})
Then we can "study" this equation: try to find approximate solutions
or to study qualitative aspects of solutions, or ... well, I will
admit it: only very rarely can exact solutions which are physically
interesting be found. It turns out that the heat equation and its
"solutions" are good models of many physical situations.
Another way to evaluate double integrals
I began with the following example: compute
_{R}(xy)^{40}(x+y)^{50}dA where R is
the rectangular region with corners (1,1), (2,0), (0,2), and (1,1).
This is an irritating integral. But there is some not well concealed
symmetry. The boundaries of rectangle can be written as x+y=2,
x+y=0, xy=2, and xy=2.
It almost seems as if the integrand and the region are begging
us to rewrite everything in terms of u and v where u=xy and
v=x+y. Then the region of integration can be described 2<=u<=2
and 0<=v<=2. The integrand becomes
u^{40}v^{50}. Notice that if we add the equations
u=xy and v=x+y and divide by 2 we get x=(1/2)(u+v). If we subtract
the first equation from the second and divide by 2 we get
y=(1/2)(vu).
We have parameterized the xyplane by
r(u,v)=(1/2)(u+v)i+(1/2)(vu)j+0k. There
is an area distortion factor which we computed as:
dA_{x,y}=r_{u}xr_{v}dA_{u,v}.
So let's compute:
r_{u}=(1/2)i(1/2)j+0k and
r_{v}=(1/2)i+(1/2)j+0k and:
( i j k ) det( 1/2 1/2 0 )=0i+0k+(1/2)k ( 1/2 1/2 0 )Therefore: _{v=0}^{v=1}_{u=2}^{u=2}u^{40}(x+y)^{50}dA becomes _{R}(xy)^{40}v^{50}(1/2)du dv. This can be evaluated exactly easily: (1/2)2·(2^{41}/41)(2^{51}/51).
What's going on?
If there is something common among the algebraic and geometric
specifications of a double (or a triple!) integral, then we can
sometimes take advantage. We can parameterize with
r(u,v)=x(u,v)i+y(u,v)j+0k. The functions
x(u,v) and y(u,v) are chosen (ideally!) so that description of the
region and the specifications of integrand are simpler. The "area
distortion factor", r_{u}xr_{v}, then turns out to
be the absolute value of the 2by2 determinant,
( x/u y/u ) ( x/v y/v )This is just the k component of r_{u}xr_{v}: the i and j components are 0 in this case. This determinant is called the Jacobian.
Another example
The following example could arise in thermodynamics or physical
chemistry. Suppose R is the region in the first quadrant bounded by
y=2x, y=4x, y=1/x, and y=3/x. Let's compute _{R}x^{4}y dA.
Here a neat "change of variables" is a bit hidden, but maybe you can see that the boundary curves of the region are y/x=2 and y/x=4 and xy=1 and xy=3. Then you might (!) think to define u=y/x and v=xy. If you do, then uv=(y/x)(xy)=y^{2} so that y=u^{1/2}v^{1/2}. Then v=xy becomes v=x(u^{1/2}v^{1/2}) so that x=u^{1/2}v^{1/2}. With the equations x=u^{1/2}v^{1/2} and y=u^{1/2}v^{1/2} the original integrand x^{4}y becomes u^{3/2}v^{5/2}. The Jacobian computation is:
( x/u y/u ) ( (1/2)u^{3/2}v^{1/2} (1/2)u^{1/2}v^{1/2} ) ( x/v y/v ) ( (1/2)u^{1/2}v^{1/2} (1/2)u^{1/2}v^{1/2} )and this is (1/4)u^{1}(1/4)u^{1}. We want the absolute value (the magnitude of r_{u}xr_{v}) so we have (1/2)(1/u). The double integral which results is _{1}^{3}_{2}^{4}u^{3/2}v^{5/2}(1/2)(1/u)du dv. The region of integration has become a rectangle, the integrand is not horrible, and the Jacobian factor is also not too bad. I won't compute this, but I hope that you see it is easy enough.
This material is in section 15.9 of the text, Changing Variables. It will not be tested on the final exam, but you should see that the technique may be useful.
And so we end what one student in the class,
Ms. Karanam, called, perhaps appropriately, a course in
Problem U from the first set of review problems
Suppose Q(x,y) is defined by the equation Q(x,y)=e^{xf(y)}
where f is a differentiable function of one variable with f(0)=A,
f´(0)=B, and f´´(0)=C. Use this information to compute these
quantities:
Q(0,0),
(Q/x)(0,0),
(Q/y)(0,0),
(^{2}Q/x^{2})(0,0),
(^{2}Q/xy)(0,0),
and
(^{2}Q/y^{2})(0,0).
Solution Q(0,0)=e^{0f(0)}=e^{0}=1.
The other computations need the Chain Rule.
Q/x=(/x)e^{xf(y)}=e^{xf(y)}f(y), so
(Q/x)(0,0)=
e^{0f(0)}f(0)=e^{0}A=A.
Q/y=(/y)e^{xf(y)}=e^{xf(y)}xf´(y), so
(Q/y)(0,0)=
e^{0f(0)}0·f´(0)=0.
^{2}Q/x^{2}=(/x)[e^{xf(y)}f(y)]=e^{xf(y)}f(y)^{2}, so (^{2}Q/x^{2})(0,0)=e^{0f(0)}f(0)^{2}=1·A^{2}=A^{2}.
^{2}Q/xy=(/x)[e^{xf(y)}f(y)]=e^{xf(y)}xf(y)^{2}+e^{xf(y)}f´(y), so
(^{2}Q/xy)(0,0)=e^{xf(y)}xf(y)^{2}+e^{xf(y)}f´(y)=e^{0f(0)}0f(0)^{2}+e^{0f(0)}f´(0)=B.
^{2}Q/y^{2}=(/y)[e^{xf(y)}xf´(y)]=e^{xf(y)}x^{2}f´(y)^{2}+e^{xf(y)}xf´´(y), so
(^{2}Q/y^{2})(0,0)=e^{0f(0)}0^{2}f´(0)^{2}+e^{0f(0)}0f´´(0)=0.
[Not discussed in lecture.]
In the last three computations (the second derivatives) we use the
formulas for the first derivatives (before they are evaluated).
Statement of the Divergence Theorem
Suppose E is a solid bounded region in space (R^{3}) and S is
the boundary of E, with n the outward pointing normal on
S. Suppose also that F is a vector field with differentiable
coefficients. Then:
_{S}F·n dS=_{E}div F dV.
The 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.  

Proving the Divergence Theorem for the unit cube
I tried to "demystify" the Divergence Theorem by explaining why it is
true for the unit cube in R^{3}.
The unit cube is a parallelopiped whose vertices (corners) have entries 0 or 1. There are 8 vertices: (0,0,0), (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1), and (1,1,1). There are 12 edges. Edges join at two vertices whose coordinates differ by one entry. There are 6 faces, each obtained by holding one coordinate equal either to 0 or to 1. By the way, the unit cube and its generalizations in higher dimensions turn out to be very interesting. One reason is the existence of Gray codes.
Now the triple integral side of the Divergence Theorem is _{The cube}(P/x)+(Q/y)+(R/z) dV. I will split this into three separate integrals, and analyze each part.
Let's look at _{The cube}(Q/y) dV. I chose to write dV here as dy dA_{x,z}. The reason for this is that the integration with respect to y will "undo" the
/y.
The innermost integral is then
_{y=0}^{y=1}(Q/y) dy. The Fundamental Theorem of Calculus immediately applies and we get Q(x,y,z)]_{y=0}^{y=1}=Q(x,1,z)Q(x,0,z). We then integrate both of these:
_{x,z between 0&1}Q(x,1,z) dA_{x,z}_{x,z between 0&1}Q(x,0,z) dA_{x,z}
Now look at the surface. The (x,1,z) part of the surface integral has
j as normal, and the (x,0
Let's look at _{The cube}(R/z) dV. I would write dV as dz dA_{x,y}. The
Fundamental Theorem of Calculus would apply to the innermost
integral:
_{z=0}^{z=1}(R/z) dz=Q(x,y,z)]_{z=0}^{z=1}=Q(x,y,1)Q(x,y,0). Again
integrate both of these:
_{x,y between 0&1}Q(x,y,1) dA_{x,y}_{x,y between 0&1}Q(x,y,0) dA_{x,z}
The minus sign comes from the Fundamental Theorem of Calculus and it comes from the +/orientation of the normals.
Finally the last term is _{The cube}(P/x) dV. I hope that you see dV here should be written as
dx dA_{y,z}. Then the Fundamental Theorem of Calculus
applies and we've got this:
_{x=0}^{x=1}(P/x) dx=P(x,y,z)]_{x=0}^{x=1}=P(1,y,z)Q(1,y,z).
Each of these terms needs to be integrated with respect to y and z
from 0 to 1. The + part (that is, at
(1,y,z) in the cube) has a normal vector of i and the  part (that is, at (0,y,z) in the cube) has a
normal vector of i. So this part of the triple integral, after
using the Fundamental Theorem of Calculus, gives the flux over the two
indicated pieces of the boundary of the cube.
If now we add up the three pieces of the triple integral we will get the surface integral of F·n over the boundary of the cube with the "correct" (outward) orientation. I wanted to tell you that the Divergence Theorem is a version of the Fundamental Theorem of Calculus, and that the signs checked out: algebraically, they occur because of FTC and ]. Geometrically, they come from the outward choices.
An old computation
We first introducted flux computations on April 18. And one of the examples given was this:
If F(x,y,z)= x^{2}i+yzj4zk, and
the surface is the sphere of radius 5 centered at the origin, what is
the total flux of F through this sphere (directed
outwards). Please look at what we did on April 18. We can also use the
Divergence Theorem:
The flux would be the (triple) integral over the whole sphere (its
inside!) of div F=((x^{2})/x)+((yz)/y)+((4z)/z)=2x+z4.
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 x^{33} and
y^{47} and z^{2003} over this sphere will be 0, for
the same reasons. I don't know what the integral of even powers would
be.
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)Pi·5^{3}. The result is (2000/3)Pi, 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.
16.9, #20
Here is a standard textbook problem in the Divergence Theorem section.
F(x,y,z)=z arctan(y^{2})i+z^{3}ln(x^{2}+1)j+zk. Find
the flux of F across the part of the paraboloid
x^{2}+y^{2}+z=2 that 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! The paraboloid is z=2x^{2}y^{2}: it "opens"
down. The vertex or top is at (0,0,2). The normals to the paraboloid
vary a great deal. While it might be possible to compute the flux
directly, the Divergence Theorem states that the integral of 1 (that's
div F) over the solid region above z=1 and below
z=2x^{2}y^{2} will equal the flux through the
parabolic cap plus the flux through the disc on the plane z=1. That
disc has radius 1, centered at the origin, since the boundary is
1=2x^{2}y^{2} or x^{2}+y^{2}=1. Also
the outward normal on the disc is constant because the disc is flat,
and the outward normal is k.
Let's compute the triple integral: _{The cup}1 dV. Probably this is
simplest to compute with cylindrical coordinates. will go from 0
to 2Pi, and r will go from 0 to 1. That's a polar description of the
base of the solid. What's the height? The bottom is at z=1, and the
top is at z=2x^{2}y^{2}, or (in "polar"),
z=2r^{2}. So we compute
_{=0}^{=2Pi}_{r=0}^{r=1}_{z=1}^{z=2r2}1 dz r dr d=_{=0}^{=2Pi}_{r=0}^{r=1}(2r^{2}1)r dr d=
_{=0}^{=2Pi}_{r=0}^{r=1}(rr^{3})dr d=
_{=0}^{=2Pi}(r^{2}/2r^{4}/4)]_{r=0}^{r=0}d=_{=0}^{=2Pi}(1/4)d=Pi/2.
Now the surface integral over the "bottom" disc. F·n is (z arctan(y^{2})i+z^{3}ln(x^{2}+1)j+zk)·(k) which is z. But z=1 on this disc, so we need to integrate 1 over a disc bounded by a circle of radius 1: the answer is Pi.
We now have: Pi/2 (the divergence integral) equal to the flux over the paraboloid plus Pi (the flux over the disc). Therefore the flux over the paraboloid must be (3Pi)/2.
Other uses
While textbook problems are (sometimes) nice, more interesting uses of
the Divergence Theorem include a discussion of heat transfer (Friday)
and maybe some mention of Gauss's Law. I tried to do Gauss's law and
screwed up in the lecture to sections 8, 9, and 10. Maybe I can do
better here.
Gauss's Law [Not discussed in lecture.]
Let's start with an inverse square force field in R^{3}
centered at the origin:
F=xi/(x^{2}+y^{2}+z^{2})^{3/2}yj/(x^{2}+y^{2}+z^{2})^{3/2}zk/(x^{2}+y^{2}+z^{2})^{3/2}
I want the divergence of F so I probably need to compute things
like this: (/x)(x/(x^{2}+y^{2}+z^{2})^{3/2})=I'm tired! I'll have a friend compute this:
> A:=diff(x/(x^2+y^2+z^2)^(3/2),x): > B:=diff(y/(x^2+y^2+z^2)^(3/2),y): > C:=diff(z/(x^2+y^2+z^2)^(3/2),z): > A+B+C; 2 2 2 3 3 x 3 y 3 z   +  +  +  3/2 5/2 5/2 5/2 %1 %1 %1 %1 2 2 2 %1 := x + y + z > simplify(A+B+C); 0So the divergence of F is 0. But now let me compute the flux of F across a sphere of radius A centered at the origin: x^{2}+y^{2}+z^{2}=A^{2}. We can get a normal to this sphere by taking the gradient: 2xi+2yj+2zk. A unit normal pointing "out" is n=(x/A)i+(y/A)j+(z/A)k. Then F·n is
Confusion!
But the divergence is 0 and the flux should be the integral of the
divergence, so ... what's going on? Again (we saw this earlier in an example with Green's Theorem) the vector field
F is not differentiable in all of the inside of the sphere: it
is not even defined at the origin. Here we have the flux equal to a
(nonzero) constant, not depending on the radius of the sphere.
Take any surface which completely surrounds the origin. Put a sphere centered at the origin with very small radius between that surface and the charge at the origin. The Divergence Theorem applies to the region between the sphere and the origin since F is differentiable away from the origin. But div F is 0 in that region, so the total flux integral must be 0. That means the flux integral outward through the surface is exactly balanced by the flux integral inward through the sphere. The flux integral through the sphere does not depend on the radius of the sphere, so the surface flux doesn't depend on anything except that the surface does wrap completely around the origin. And this is (a version of) Gauss's Law: the flux integral around an isolated point charge is always constant.
Gauss's Law is discussed in many physics books. Also you can look at pages 1130 and 1131 of the textbook. Again, my apologies for messing this up in lecture!
Problem Q from the second set of review problems Sketch the three level curves of the function W(x,y)=ye^{x} which pass through the points P=(0,2) and Q=(2,0) and R=(1,1). Label each curve with the appropriate function value. Be sure that your drawing is clear and unambiguous. Also, sketch on the same axes the vectors of the gradient vector field W at the points P and Q and R and S and T. The point S=(0,2) and the point T=(2,0).
Problem solution and discussion 
The ingredients for Stokes' Theorem
So this is a curve in space (R^{3}) with START=END and no other selfintersections.  
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. (We will do one of these below). Stokes' Theorem involved computing the work of a vector field around the closed curve, and then the flux of a related vector field over the 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 word "orientation" here means how to select t, the unit tangent vector on the boundary curve, and N, the unit normal on the surface. The boundary curve will be a parameterized curve. It has a unit tangent vector, t, pointing in the direction of increasing parameter value. If we "walk" along the boundary curve in this direction, the surface should be to our left. Now we have t and a direction to the left. Complete this to a righthanded coordinate system. The selection of N, the unit normal vector to the surface, is made so N points in the direction of the last entry of the righthanded coordinate system. I think in the accompanying picture to the right, the N would point "out" of the page, and towards the "inside" of the cupshaped surface. 
A textbook problem
Here is problem 13 from section 16.7 of our textbook:
Verify that Stokes' Theorem is true for ... the vector field
F(x,y,z)=y^{2}i+xj+z^{2}k
and the surface is the part of the paraboloid
z=x^{2}+y^{2} that lies below the plane z=1, oriented
upward.
Some discussion
The plane z=1 intersects the paraboloid in a circle. This is a circle
of radius 1 centered at (0,0,1). The paraboloid "overlays" a region
inside a circle of radius 1 centered at the origin in the xyplane. We
will compute both integrals in Stokes' Theorem and (I hope!) get the
same answers. If the paraboloid is "oriented upward" then I presume
that the N points up. Going around the blue circle in the
standard (counterclockwise/positive) direction will orient the
boundary curve "compatibly": the t, the leftish piece of
surface next to the boundary curve, and the up N form a
righthanded triple.
The work integral
So I need to compute _{The curve}y^{2}dx+x dy+z^{2}dz.
The curve is a circle, and can be parameterized as:
x=1cos(t) dx=sin(t)dt y=1sin(t) dy=cos(t)dt z=1 dz=0and the parameterization interval for the whole circle is [0,2Pi]. Then _{The curve}y^{2}dx+x dy+z^{2}dz becomes
The surface integral
Now we need to compute _{The paraboloid}curl F·N dS.
The curl
This is xF, so:
( i j k ) det( /x /y /z )=0i0j+(12y)k ( y^{2} x z^{2} )Parameterizing the surface, etc.
( i j k ) det( 1 0 2u )=2ui2vk+1k ( 0 1 2v )We discussed the magical cancellation in the last lecture. V·N dS became V·(r_{u}xr_{v}) dA_{u,v}. curl F here is (12y)k=(12v)k so that curl F·N dS=(12v)k·(2ui2vk+1k)dA_{u,v}=(12v)dA_{u,v}.
This instantiation (?) of Stokes' Theorem is verified: Pi=Pi.
Another textbook problem
Here is a slightly more vicious (viscous?) problem from the Stokes'
Theorem section of a calculus text by Robert A. Adams:
Find _{The surface}curl F·N dS where the surface is that part of
the sphere x^{2}+y^{2}+(z2)^{2}=8 which lies
above the xyplane, and N is the outward unit normal on the
surface, and F is
y^{2}cos(xz)i+x^{3}e^{yz}je^{xyz}k.
Since the problem occurs in the Stokes' Theorem section of the text we
should probably use Stokes' Theorem. The region of the sphere is shown
to the right. The sphere is centered at (0,0,2) and its radius is
sqrt(8)=2sqrt(2). So a portion of the sphere extends below the
xyplane.
The boundary of the top portion occurs if z=0 in the
equation
x^{2}+y^{2}+(z2)^{2}=8. Then
x^{2}+y^{2}+(2)^{2}=8 and
x^{2}+y^{2}=4. This is a circle of radius 2 centered
at the origin in the xyplane. We should establish the orientation of
this circle. If we look closely at a small piece of the surface near
the boundary curve, the outward unit normal points slightly down. We
must "walk" along the curve so that the surface is to the left. The
t direction is the standard counterclockwise direction on the
boundary circle. I hope the local picture to the right helps to
convince you of that.
Now Stokes' Theorem applies:
_{The spherical surface}curl F·N dS=_{The boundary circle}F·t ds.
But notice: this circle is also the correctly oriented boundary of the
disc of radius 2 centered at the origin in the xyplane. So I can use
Stokes' Theorem a second time to change the line integral to a
much simpler surface integral:
_{The boundary circle}F·t ds=_{The disc}curl F·N dS
This is simpler for several reasons. The region over which we're
integrating is flat, a disc in the xyplane. The correctly oriented
normal, N, is just k. I hope the picture convinces you
of that.
We should compute curl F. Wait, we just need to compute
the k part of curl F:
( i j k ) det ( /x /y /z )=Blah!iBlah, blah!j+[3x^{2}e^{yz}2ycos(xz)]k ( y^{2}cos(xz) x^{3}e^{yz} e^{xyz} )I further simplification occurs. We're on the xyplane, where z=0. So the k component [3x^{2}e^{yz}2ycos(xz)] becomes 3x^{2}2y because cos(0)=1 and e^{0}=1.
Green's Theorem [Not discussed in lecture.]
If the boundary curve is in R^{2} and the "surface" is a
region in R^{2} then Stokes' Theorem is Green's
Theorem. The only part of the curl that we see is the k, and if
the vector field is Pi+Qj+Rk, the normal N
is k and the k component of the curl of the vector field
is Q_{x}P_{y}. So the integral of Pdx+Qdy over the
boundary curve (properly oriented) equals the double integral of
Q_{x}P_{y} over the interior.
See page 1121 of the text.
What is curl?[Not discussed in
lecture.]
See page 1124 of the text. If F is a vector field, the text
shows how, using Stokes' Theorem, the vector curl F
represents how much the fluid flow
corresponding to F rotates.
Simply connected [Not discussed in
lecture.]
A region is simply connected if every simple closed curve can
be contracted to a point where the shrinking only occurs inside the
region. The picture below shows one simple closed curve shrinking to a point. In the torus, though, the red curve can't be shrunk to a point inside the torus.
If a closed curve can be contracted to a point, and if F is a
vector field with curl F=0, then you can use Stokes'
Theorem to change the work integral around the curve to a flux
integral over a surface (the surface is green in the lefthand part of
the picture), and that integral is 0 since curl F=0. This
means in a simply connected region, any vector field whose curl is 0
must be conservative, and therefore must be a gradient vector
field. This doesn't necessarily happen in a nonsimply connected
region, such as the torus. It turns out that such phenomena are
relevant to realworld applications.
This is on the bottom of page 1124 of the text.
Coordinate charts
The picture and the understanding I wanted from the last lecture are:
a coordinate chart is a vectorvalued function of two variables,
r(u,v). The u and the v live in some domain in R^{2},
and each pair (u,v) in that domain give (using
The advantage of (u,v) space is that things are easy to understand
there. On the surface itself, the coordinate curves need not be
orthogonal and the way they are put together can vary tremendously
from one point to the next.
dS
dS=r_{u}xr_{v}du dv.
Surface integrals
If we believe a piece of surface in R^{3} could be a
mathematical model of a thin curved plate, then the plate could have
varying density. We could add up the density multiplied by the area to
get mass. This is a surface integral:
_{The Surface}The "density" function dS.
This will be computed using a coordinate chart, much as we computed
line integrals with parameterizations. Below are two textbook
problems. The picture to the right is supposed to be showing a thin
plate with ... uhhh ... some heavier and some lighter regions. As we
will see, this model doesn't correspond too well with the surface
integrals people actually compute, but, well, we can practice a bit
first.
16.7, #8
Evaluate the surface integral.
_{The Surface}y dS,
S is the surface z=(2/3)(x^{3/2}+y^{3/2}), 0<=x<=1, 0<=y<=1
A picture of this textbook example is to the right.
The surface is a graph, and, for a graph, a first try at
parameterization would use the graph variables, and the graph
function:
x=u, y=v, and z=(2/3)(u^{3/2}+v^{3/2}) so
that
r(u,v)=ui+vj+(2/3)(u^{3/2}+v^{3/2})k. We
can compute r_{u}xr_{v}:
( i j k ) det( 1 0 u^{1/2} )=u^{1/2}iv^{1/2}j+k ( 0 1 v^{1/2} )The magnitude of this vector is sqrt(u+v+1). I hope that you realize the example has been "invented" quite carefully so that the dS term is not very horrible. Imagine if other powers had been used in the graph equation!
Evaluating the integral
_{u=0}^{u=1}v·sqrt(u+v+1) du=(2/3)v(u+v+1)^{3/2}]_{u=0}^{u=1}=(2/3)v(v+2)^{3/2}(2/3)v(v+1)^{3/2}
because the integrand is just (integrating du!)
Constant_{1}·sqrt(v+Constant_{2}).
Now we need to antidifferentiate (dv) the function (2/3)v(v+2)^{3/2}(2/3)v(v+1)^{3/2}. I'll do the first part:
(2/3)v(v+2)^{3/2}dv=(2/3)(w2)w^{3/2}dw if the
substitution v+2=w, dv=dw, and v=w2 is done. Then (2/3)(w2)w^{3/2}dw=(2/3)w^{5/2}(4/3)w^{3/2}dw=
(4/21)w^{7/2}(8/15)w^{5/2}=(4/21)(v+2)^{7/2}(8/15)(v+2)^{5/2}. Then the definite integral (Maple tells me!) is
(12/35)sqrt(3)+(64/105)sqrt(2).
Students should try the other piece! You may check your answer:
the value of _{v=0}^{v=1}(2/3)v(v+1)^{3/2}dv is
reported to be (8/105)(16/35)sqrt(2).
16.7, #11
Evaluate the surface integral.
_{The Surface}y dS,
S is the part of the paraboloid y=x^{2}+z^{2} that lies inside the
cylinder x^{2}+z^{2}=4
In this case I decided to create a "hand" drawn picture since I
couldn't find an angle that I liked for a Maple graph. The
cylinder x^{2}+z^{2}=4 is a right circular cylinder
over a circle in the xzplane. That circle has center at the origin
and radius 2. Then the cylindircal surface is gotten by pulling the
circle "out" parallel to the yaxis, which is the axis of symmetry of
the cylinder. The paraboloid y=x^{2}+z^{2} also has
the yaxis as axis of symmetry. It "opens" out to the positive
yaxis. The cylinder and paraboloid intersect where y=4. The
paraboloid over which we will integrate is a graph over the inside of
the circle x^{2}+z^{2}=4 in the xzplane.
Since the paraboloid is a graph, that suggests a parameterization:
x=u, y=u^{2}+v^{2}, and z=v, so that
r(u,v)=ui+(2/3)(u^{2}+v^{2})j+vk.
And now r_{u}xr_{v}:
( i j k ) det( 1 2u 0 )=2ui1j+2vk ( 0 2v 1 )The magnitude of this vector is sqrt(4u^{2}+4v^{2}+1). y is u^{2}+v^{2}. The surface integral we'd like to compute is
The computation
In this case, I hope that both the region of integration and the integrand suggest polar coordinates in the (u,v) plane. Then: dA_{uv}=r dr d and the u^{2}+v^{2} will be replaced by r^{2}. The region of integration can be described in polar coordinates in a straightforward manner: goes from 2 to 2Pi and r goes from 0 to 2. So we need to compute:
_{=0}^{=2Pi}_{r=0}^{r=2}r^{2}sqrt(4r^{2}+1) r dr d.
The most horrible thing in this antidifferentiation is
4r^{2}+1, so I will try to compute the antiderivative of
r^{2}sqrt(4r^{2}+1) r with the substitution
w=4r^{2}+1. This is a textbook problem, and things should
work! Also I see that I have a very helpful r along with dr. So:
r^{2}sqrt(4r^{2}+1) r dr=[(w1)/4]sqrt(w)(1/8)dw since:
w=4r^{2}+1. so dw=8r dr and (1/8)dw=r dr and
[(w1)/4]=r^{2}. I can compute this:
[(w1)/4]sqrt(w)(1/8)dw= (1/32)[w^{3/2})w^{1/2}]dw=(1/80)w^{5/2}(1/48)w^{3/2}=(1/80)(4r^{2}+1)^{5/2}(1/48)(4r^{2}+1)^{3/2}. And now we need to substitute, etc. Or:
> int(int(r^2*sqrt(4*r^2+1)*r,r=0..2),theta=0..2*Pi); / 1/2 1/2 1/2\ 1/2  3128 Pi 17 8 Pi  Pi     Â¥ 15 15 /   32 > simplify(%); 1/2 Pi (391 17 + 1)  60
Flux
The horrible factor r_{u}xr_{v} makes a "random" surface integral almost
impossible to compute in terms of antidifferentiations involving
familiar functions. It is very nice that the surface integrals of most
interest in physical and engineering problems are not "random" but
result from computations of flux, and it turns out that the horrible
factor disappears for such computations.
Suppose we have a vector field, F in R^{3}. We could
imagine a surface in R^{3}, and then try to see how the flow
of the vector field interacts with the surface. The picture to the
right is quite imaginary. I've never seen the arrows of a vector
field, and I want the surface, sort of like a net, not to give any
resistance to the imaginary arrows. It is, of course, an imaginary
surface.
The flux is the net flow through the surface.
What do the words mean?
net The direction the fluid flows means something. It is
possible that at some points the fluid crosses the surface in
different directions. We should have some way of giving a sign to the
flow, left to right/right to left, inward/outward, and then totaling
the different contributions, with signs, to see whether the net
flow is positive or negative.
through The flow through the surface is important. The same
piece of surface ("dS") can have different flux, even if the vector
field is constant  always the same direction and magnitude. What can
then change can be 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.
The whole surface If we want to
compute the net flux through the whole surface, then we will need to
assign unit normal vectors at every point of the surface. There are
some surfaces which can't accept such assignments. The simplest
example is the Mobius strip (take a long rectangle, make a halftwist
in the long direction, and attach the short edges together). If you
give an N at any one point, and then follow around the
assignment continuously, when you get back to the point, you'll
discover that you have reversed the normal! So there will be no nice
way to define and compute flux through surfaces which don't permit
nice "assignments" of normals. I will assume such a problem will not
occur in the remainder of this course (hey, it doesn't for planes and
spheres and toruses and ... almost anything you will encounter in
applications).
Magical cancellation!
If our surface is parameterized there is a natural way to get a
unit normal. Just take r_{u} and r_{v}:
these are velocity vectors for curves on the surface, and are tangent
ot the surface. Their crossproduct will be perpendicular to the
surface. If we then normalize (divide by the magnitude) we'll get an
acceptable N. So we can take N to be
r_{u}xr_{v}
divided by the scalar r_{u}xr_{v}. If Flux=_{The surface}F·N dS this will be the same as
_{The surface}[F·(r_{u}xr_{v})/r_{u}xr_{v}] r_{u}xr_{v} dA_{uv}
Look at the marvelous cancellation (much the same as what occurred in
the line integral case).
The flux integral is _{The surface}F·(r_{u}xr_{v}) dA_{uv}.
While a "plain" surface integral needs to be very carefully prepared
to be "computable" (as in the two previous examples), the cancellation
here means no horrible square root terms, and many flux integrals
should be computable.
An example on the sphere
Suppose F is x^{2}i+yzj4zk and I
would like to know the flux through the sphere of radius 5 centered at
the origin. We considered this surface before and
there we learned
r_{u}xr_{v}=25cos(u)[sin(v)]^{2}i25sin(u)[sin(v)]^{2}j25sin(v)cos(v)k
and the parameterization itself was
x(u,v)=5cos(u)sin(v); y(u,v)=5sin(u)sin(v); z(u,v)=5cos(v). This means that F described in (u,v) terms becomes
25sin(u)^{2}sin(v)^{2}i+25sin(u)sin(v)cos(v)j20cos(v)k
We need to integrate F·N dS, but
this, because of the cancellation of the horrible factor r_{u}xr_{v} becomes just F·(r_{u}xr_{v}) dA_{u,v}. Let me match up the
components and get the integrand:
625sin(u)^{2}sin(v)^{4}cos(u)625
sin(u)^{2}sin(v)^{3}cos(v)+500cos(v)^{2}sin(v).
This must be integrated from 0 to 2Pi in u and from 0 to Pi in v. I
emphasized in class that this integration is not difficult, 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!
HOMEWORK
Read and do
problems in Chapter 16. Do problems! Do many problems!!
The five integrals you meet in Math 251
Here's the list.
How to describe a plane
Suppose I give you a point, p=(3,1,2), and two vectors,
A=<4,1,3> and B=<1,5,2>. I would like to
describe algebraically the plane which contains the point p in the
direction of the vectors A and B. Here is how we did
this problem earlier in the course.
An implicit description of the plane
Compute the crossproduct of A and B:
( i j k ) det( 4 1 3 )=17i5j+21k. ( 1 5 2 )It is easy to check (I just did!) that the resulting vector is perpendicular to both A and B and is a normal vector to the plane we'd like to describe. A point with coordinates (x,y,z) is on this plane if
An explicit description
Change the point p=(3,1,2) to a position vector,
P=<3,1,2>. Then every point on this plane has a unique
description as P plus some multiple of A plus some
(possibly other) multiple of B. Your text calls these
multiplies u and v, so I will also. So the plane is anything which can
be written as P+uA+vB. We can write this with
more details:
<3,1,2>+u<4,1,3>+v<1,5,2>=<3+4u+1v,1u+5v,2+3u+2v>. The
textbook writes this as a vector position function:
r(u,v)=(3+4u+1v)i+(1u+5v)j+(2+3u+2v)j
The i component is called x(u,v), the j component is
called y(u,v), and the k component is called z(u,v), Each point
on the plane has a unique "address" in terms of the pair of numbers
(u,v). This is like the central Manhattan street grid: streets and
avenues. So 3^{rd} Avenue and 47^{th} Street is a
unique point, and every intersection has a unique address.
A sphere
Creating a "mapping" from a piece of the (u,v) plane to a surface in
R^{3} is frequently called a coordinate chart (although
not in your book). Getting a coordinate charts for a plane is not
difficult. Getting (useful!) coordinate charts for curved surfaces can
be much harder. Here's an example which is only moderately difficult
because we have studied spherical coordinates: a sphere of radius 5
centered at the origin in R^{3}.
Now let's use spherical coordinates, with rho fixed at 5. We see that
a point on the sphere will be described by (I use u for and v for phi):
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<=2Pi and 0<=v<=Pi.
Horizontal lines in the (u,v) domain become circles parallel to the
xyplane on the sphere (latitudes?). Vertical lines in the (u,v)
domain, where v varies and u is fixed, become half great circles on
the sphere, going from the "North Pole" to the "South Pole". The
region in the (u,v) domain where 0<=u<=Pi and Pi/2<=v<=Pi
becomes the lower (z<=0) and "forward" (y>=0) quarter sphere.
A torus
I'll try now to give a parametric description of a torus. So the torus
(the surface of a doughnut) will be a surface which is gotten when a
circle perpendicular to the xyplane, and radially oriented, is
revolved around the zaxis. One view of what I'm trying to describe is to the right.
The center of the circle to be revolved around the zaxis is moved so that it describes a circle in the xyplane centered at the origin. There are two more views below of the surface. One is from "above", looked down the zaxis. The other is from the "side", looking along the yaxis.
I can give points on the torus a unique "address" in terms of two
numbers. These two numbers will represent angles. One angle will be
, but I'll call it u to agree with the text.
I'll make this specific torus more definite: the length of the vector
from the zaxis to the center of the circle will be 4, and the radius
of the circle being revolved around the zaxis will be 2. The left
picture below shows the angle u. The right picture shows a slice
perpendicular to the xyplane, through the vector of length 4. I will
specify a point on the torus by letting v be the angle made by a line
from the point on the torus to the center of the circle compared to
the xyplane itself.
The z coordinate of the point only involves v. The z coordinate is the
"opposite" side of a right triangle with acute angle v and hypoteneuse
2. So z=2sin(v). The vector from the origin to the center of the
circle has length 4. The angle v adds on another 2cos(v) to that
length. But then we use this total length, 4=2cos(v), along with the
angle u (secretly, ) to get the x and y coordinates. So
x=(4+2cos(v))cos(u) and y=(4+2cos(v))sin(u).
Therefore the torus will be given by the following vectorvalued function:
r(u,v)=x(u,v)i+y(u,v)j+z(u,v)k with
x(u,v)=(4+2cos(v))cos(u) and
y(u,v)=(4+2cos(v))sin(u) and z(u,v)=2sin(v).
The domain of the coordinate chart, which will give each point on the
torus surface a unique (u,v) address, will be 0<=u<=2Pi and
0<=u<=2Pi.
In the 21^{st} century I can check what I've just
written. Below is a Maple command and its output. I should
remark, if you've never been in my office watching me try to
use Maple, that typing this command and looking at its output
took 5 attempts before I was successful. Such wonderful effort may
explain why I wouldn't trust myself to use Maple in class
while people watched and giggled, and we all wasted time.
plot3d([4+2*cos(v))*cos(u),4+2*cos(v))*sin(u),2*sin(v)], u=0..2*Pi,v=0..2*Pi, scaling=constrained,axes=normal);
Finally, we can consider horizontal lines across the domain of the
coordinate chart, r(u,v), and ask what kinds of curves these
create on the torus. These lines, where u changes and v is unchanged,
become circles around the zaxis. Similarly, the vertical lines in the
domain of r become lines "around" the torus, for fixed u=.
If we restrict to region where 0<=u<=Pi and Pi<=v<=2Pi,
then the part of the torus which results is "forward" of the xzplane
(where y>=0) and below the xyplane (where z<=0). This is a
quarter of the torus.
A graph
Here I considered a rather simple surface given by a graph:
z=3x^{2}+5y^{4}. Certainly all I expected people to
see and say was this this surface is (vaguely) cupshaped, and its
lowest point is at (0,0,0). There's a really simple way to
parameterize such surfaces: just use x and y. So
r(u,v)=ui+vj+3u^{2}+5v^{4}k. The
parameterization geometrically consists of pushing up a grid from the
xyplane until it hits the graph.
Surface area
Here once again was used the wonderful integral calculus slogan:
Cut up, approximate, sum, limit
This is very brief, since my eyes are tired! Please look in the
text for more details.
Small rectangles in the uv domain, of
dimension u by v were changed to approximate
parellelograms, and they were magnified by r_{u} and
r_{v}, respectively. The resulting area is obtained by
taking a cross product, and we get r_{u}xr_{v}uu. Then add these up and take the
limit. The surface area for a part of the surface which is
parameterized by a domain, D, in the uvplane turns out to be
_{D}r_{u}xr_{v} du dv. The
textbook calls r_{u}xr_{v} du dv by the
name, dS, for a piece of surface area.
The surface area of the sphere
Here
r(u,v)=5cos(u)sin(v)i+5sin(u)sin(v)j+5cos(v)k
Therefore
r_{u}=5sin(u)sin(v)i+5cos(u)sin(v)j+0k
and
r_{v}=5cos(u)cos(v)i+5sin(u)cos(v)j5sin(v)k
( i j k ) r_{u}xr_{v}=det(5sin(u)sin(v) 5cos(u)sin(v) 0 ) ( 5cos(u)cos(v) 5sin(u)cos(v) 5sin(v) )The i component is just 25cos(u)[sin(v)]^{2} and the j component is 25sin(u)[sin(v)]^{2} and the k component is 25[sin(u)]^{2}sin(v)cos(v)25[cos(u)]^{2}sin(v)cos(v). The last component simplifies (sin(u)^{2}+cos(u)^{2}=1) and the result is
The surface area of the torus [Not discussed in lecture.]
You could try this yourself. The answer is 32Pi^{2}, and a
detailed explanation is below.
Let's try this:
r(u,v)=(4+2cos(v))cos(u)i+(4+2cos(v))sin(u)j+2sin(v)k.
Then
r_{u}=(4+2cos(v))sin(u)i+(4+2cos(v))cos(u)j+0k
and
r_{v}=(2sin(v))cos(u)i+(2sin(v))sin(u)j+2cos(v)k
( i j k ) r_{u}xr_{v}=det((4+2cos(v))sin(u) (4+2cos(v))cos(u) 0 ) ( 2sin(v)cos(u) 2sin(v)sin(u) 2cos(v) )The i component is 8cos(v)cos(u)+4cos(v)^{2}cos(u)=4cos(v)(2cos(u)+cos(v)cos(u)). This is 4cos(v)cos(u)(2+cos(v)).
The surface area of a piece of the graph
Here the integral turns out to be _{0}^{1}_{0}^{1}sqrt(1+36x^{2}+400y^{6}) dx dy. Maple
can't "do" this integral, and gives the approximate value 6.8005 when
asked. Most surface areas can't be computed explicitely in terms of values of familiar functions.
Vector fields in R^{3}
Some simple examples of a threedimensional vector
field. Again, fluid flows and force fields are examples.
If F=x^{2}i+yj+yzk, compute the work done
along the twisted cubic x=t, y=t^{2}, z=t^{3} for t
going from 0 to 1. What's this? Well, dx=dt and dy=2t dt and
dz=3t^{2} dt, so that
_{Twisted cubic}x^{2}dx+y dy+yz dz=_{t=0}^{t=1}(
t^{2}+(t^{2})2t+(t^{2})(t^{3})3t^{2})dt=
t^{2}+2t^{3}3t^{7}dt=(1/2)+(2/4)+(3/8).
Now with the same vector field, the work done along a straight line
segment from (0,0,0) to (1,1,1). So one parameterization is x=t and
y=t and z=t with dx=dt and dy=dt and dz=dt. Then
_{Line segment}x^{2}dx+y dy+yz dz=_{t=0}^{t=1}t^{2}+t+t^{2}dt=
2t^{2}+t dt=(2/3)+(1/2).
The results are not the same, so we can again look for path
independence. A vector field is conservative if it is a gradient vector field. If P(x,y,z)i+Q(x,y,z)j+R(x,y,z)k has a potential f(x,y,z), then
f_{x}=P and f_{y}=Q and f_{z}=R.
If there is a potential, then (just as in two dimensions and
for exactly the same reasons (the chain rule!) a line integral can be
evaluated with this simple idea:
_{C}P(x,y,z)i+Q(x,y,z)j+R(x,y,z)k=Potential evaluated at the END of CPotential evaluated at the START of C.
Getting a potential from a conservative vector field
You can try to find a potential by integrating in each variable
separately and comparing the resulting candidate descriptions of
f(x,y,z). Here is an example.
We will consider:
[2x·cos(2y)+3x^{2}]i+[2x·cos(2y)+3x^{2}]j+[2x^{2}sin(2y)+3y^{2}e^{3z}]k.
The coefficients of i and j and k will be called
P and Q and R, respectively, all functions of x and y and z.
So f(x,y,z)=x^{2}cos(2y)+x^{3}+C_{1}(y,z)
and
f(x,y,z)=x^{2}cos(2y)+y^{2}e^{3z}+C_{2}(x,z)
and f(x,y,z)=y^{3}e^{3z}7z+C_{3}(x,y).
If I look through these three descriptions, I can see where each piece
is "hiding" in each description and I can detect an f(x,y,z) which has
all of the desired features:
f(x,y,z)=x^{2}cos(2y)+x^{3}+y^{2}e^{3z}7z.
Most important, look back: this formula satisfies all of the three
descriptions.
A quick way to check that there can't be a potential is to consider the compatibility conditions which follow.
Compatibility conditions
If P(x,y,z)i+Q(x,y,z)j+R(x,y,z)k is a gradient vector field, then
P_{y}=Q_{x} and P_{z}=R_{x}
and Q_{z}=R_{y}
as a consequence of Clairaut's Theorem (equality of mixed partials)
applied to the mixed partials (the second derivatives) of the
potential function. To me this says that being a gradient vector field
is more difficult in 3 dimensions than in 2. In 2 dimensions there's
only one compatibility condition. In 3 dimensions there are 3. Things
only get worse as the dimensions increase. In classical physics,
describing the dynamics of just one particle means giving the position
(3 numbers) and the momentum (3 more numbers): you're already in
R^{6}.
In R^{6} there are 6·5/2=15 equations required for the
compatibility conditions. So ... a gradient vector field there is
really rare.
Some notation which many people think is helpful
The conditions P_{y}=Q_{x} and
P_{z}=R_{x} and Q_{z}=R_{y} may be
difficult to remember. Some new notation which is supposed to help
follows. Here is a quote from a web page entitled Earliest Uses of
Symbols of Calculus:
The vector differential operator, now written as an upsidedown delta and called nabla or del, was introduced by William Rowan Hamilton (18051865). 
So "del" is given by: =(/x)i+(/y)j+(/z)k. Here are some uses:
( i j k ) det ( /x /y /z ) ( P(x,y,z) Q(x,y,z) R(x,y,z) )It turns out this is equal to (if I don't foul up the signs!) the following vector field:
Here is a version of the big picture, sort of.
=gradient (x)=curl (·)=div Functions become Vector fields become Vector fields become FunctionsSome examples computed
( i j k ) det ( /x /y /z ) ( y^{2}z^{3} 2xyz^{3} 3xy^{2}z^{2} )I think that the i component is (/y)(3xy^{2}z^{2})(/z)(2xyz^{3})=6xyz^{2}2xy(3z^{2})=0. The other components are also 0 because of Clairaut! So curl(grad of anything) is 0. Some people call vector fields whose curl is 0 irrotational. A gradient vector field is always irrotational.
Let me try another vector field: F=y^{2}zi+x^{2}yzj+x^{3}zk. The first component of the curl of F is (/y)(x^{3}z)(/z)(x^{2}yz)=0x^{2}y=x^{2}y. But please look at this:
> with(VectorCalculus): > SetCoordinates('cartesian'[x,y,z]); cartesian[x, y, z] > F:=VectorField(I did not know (but I suspected!) that Maple had a curl command. I typed help(curl) and learned just a little about the command. So the curl of our F is a vector field I will call G, and G is x^{2}yi+(y^{2}3x^{2}z)j+(2xyz2yz)k.); 2 _ 2 _ 3 _ F := y z e + x y z e + x z e x y z > Curl(F); 2 _ 2 2 _ _ x y e + (y  3 x z) e + (2 x y z  2 y z) e x y z
I'll compute the divergence of G. This div G is a function, a scalar function. It is (/x)(x^{2}y)+(/y)(y^{2}3x^{2}z)+(/x)(2xyz2yz)=(2xy)+(2y)+(2xy2y)=0. The divergence of a curl is always equal to 0: div(curl of anything)=0. If the divergence of a vector field is 0, it is called incompressible. So the curl of a vector field is incompressible.
The vector field x^{2}i+yj+z^{3}k has divergence equal to 2x+1+3z^{2}, so it is not incompressible.
HOMEWORK
Read on in
chapter 16. The next lecture will be about parametric surfaces (15.6)
and surface area (16.6).
Unfortunate event
I have had a detached retina which has been surgically
repaired. Convalescence from this has made me work at approximately 50%
efficiency. Consequences for the course include:
Rectangular Green's Theorem
During the last lecture (which may feel as if it were given a century
ago, due to the intervention of an exam), I verified a result called
Green's Theorem for a rectangle with corners at (0,0), (a,0),
(a,b), and (0,b).The verification amounted to some juggling with the
Fundamental Theorem of Calculus. The result was the following:
_{Rectangular bdry}P(x,y)dx+Q(x,y)dy=_{Inside of rect}Q(x,y)/xP(x,y)/y dA.
Silly example
I first tried a rather silly example which was something like the
following. I think I took a=3 and b=2, and specified that
P(x,y)=5y+Junk_{1}(x) and
Q(x,y)=x^{2}+Junk_{1}(y). Here the
Junk functions were some absurd formulas:
arctan(y^{3}) or e^{cos(x)}. I wanted to compute the
line integral of P(x,y)dx+Q(x,y)dy around the boundary of the
rectangle. Of course I used the version of Green's Theorem stated
above, and the two Junk terms disappeared after the partial
differentiations, and the double integral had integrand 2x5, which is
rather simple to compute over a rectangle.
Computation on a semicircle
This is another example created for the classroom, but it is more
complex, and more resembles some real uses of Green's Theorem. Here I
considered the line integral _{S}y dx+x^{2}dy
where S represents the upper half of the
circle of radius 3 centered at the origin. I think this integral
certainly can be computed straightforwardly by a parameterization but
I'd like to show another technique. First, what is the value of _{L}y dx+x^{2}dy where L is the line segment along the xaxis stretching
from (3,0) to (3,0)? Since y doesn't change on L, dy=0. Also notice that y=0 on all of L. Therefore y dx+x^{2}dy is 0 on all
of L.
So _{S}y dx+x^{2}dy=_{S}y dx+x^{2}dy+_{L}y dx+x^{2}dy=_{S+L}y dx+x^{2}dy, and the curve S+L, which is S followed by L, encloses the upper half of the area inside a circle of radius 3 centered at the origin: a semidisc. I'll call it Q.
Then a version of Green's Theorem allows us to replace _{S+L}y dx+x^{2}dy by _{Q}(x^{2})/xy/y dA. The
integrand is 2x1, and we need to integrate it over Q.
The integral _{Q}2x dA is actually 0, because
the region Q is symmetric with
respect to the yaxis, and 2x takes equally positive and negative
values on this region. So the values cancel each other. And the integral
_{Q}1 dA is minus onehalf the area of a circle of radius 3. Therefore the original line integral,
_{S}y dx+x^{2}dy, has the
value 9Pi/2.
I would agree that this is an awkward, even ludicrous way of evaluating the line integral. 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 oriented piecewise smooth simple closed
curve, and D is the region bounded by C. Suppose that P(x,y) and
Q(x,y) are functions with continuous first partial derivatives in
D. Then:
_{C}P(x,y)dx+Q(x,y)dy=_{D}Q(x,y)/xP(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)/xP(x,y)/y just
"happens" to be 1. 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 try to make this even sharper. The curve C should be
parameterized. So x=x(t) and y=y(t) for t between a and b. The line
integral gets translated into the following Riemann integral:
_{t=a}^{t=b}(1/2)y(t)x´(t)+(1/2)x(t)y´(t) dt.
So suppose you are in a race car, driving around a "closed
course". Your position could be measured with a GPS device, so you
would know <x(t),y(t)> (the position vector). You could also
imagine that you have some idea of your velocity:
<x´(t),y´(t)>. So from this "boundary data" you could
compute (at least approximately) the Riemann integral _{t=a}^{t=b}(1/2)y(t)x´(t)+(1/2)x(t)y´(t) dt
and this must be the area of enclosed by the race track. This is
astonishing to me.
People designed interesting mechanical linkages to compute areas based
on such results. These instruments are called planimeters. People can be very clever.
Analogous results in higher dimensions would allow some knowledge of
the heart based upon electrical readings on the skin. Weird,
wonderful, terrific!
Another example, somewhat paradoxical
Here I will tell you what P(x,y) and Q(x,y) should be. These are
certainly not randomly chosen but are specific functions which are
used to model important physical phenomena.
P(x,y)=y/(x^{2}+y^{2}) and
Q(x,y)=x/(x^{2}+y^{2})
It just happens (!!) that:
P/y=[(x^{2}+y^{2})+y(2y)]/(x^{2}+y^{2})^{2}=[y^{2}x^{2}]/(x^{2}+y^{2})^{2}
and
Q/x=[(x^{2}+y^{2})2x(x)]/(x^{2}+y^{2})^{2}=[y^{2}x^{2}]/(x^{2}+y^{2})^{2}.
Therefore in this case, P_{y}=Q_{x} so that
Q_{x}P_{y} is 0.
Now let me compute the line integral _{C}P(x,y)dx+Q(x,y)dy, where C is, say, the circle of
radius 3 centered at (0,0). The first parameterization you should
consider is this: x=3cos(t) and y=3sin(t) so that dx=3sin(t) dt
and dy=3cos(t) dt. Then x^{2}+y^{2}=3^{2}
because sin^{2}+cos^{2}=1. Thus
_{C}y/(x^{2}+y^{2})dx+x/(x^{2}+y^{2})dy
becomes _{t=0}^{t=2Pi}([3sin(t)/3^{2}](3sin(t))+[3cos(t)/3^{2}]3cos(t))dt. If the arithmetic is done correctly we need to
integrate 1 from 0 to 2Pi (almost everything cancels!) and the result
is 2Pi.
But but but ... the integrand on the "other side" of Green's Theorem is Q_{x}P_{y} and we computed that to be 0. A circle certainly is a simple closed curve, positively oriented, etc. We haven't made any errors. Maybe it is true that 2Pi=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 x^{2}+y^{2} in the denominator (the bottom of the
fraction). That expression is 0 at the origin, (0,0). In fact, P(x,y)
and Q(x,y) and their first partial derivatives are not
continuous at (0,0), which I will call the "bad point". This bad point
is inside the circle. The example shows that even if the hypothesis
"P(x,y) and Q(x,y) are functions with continuous first partial
derivatives" fials at just one point of D, the equality predicted by
Green's Theorem can fail very very emphatically (I think "zero equals
something nonzero" is a fairly emphatic failure).
Information transported from one curve to another, using the inside
The example considered is, however, very important in practice. Let me
show why. Let's consider another path, W, which is a quadrilateral. W
is a simple closed curve made of four straight line segments. It
starts at (8,0), goes to (2,6), then to (7,1), then to (0,5), and
finally back to (8,0). I would like to compute _{W}y/(x^{2}+y^{2})dx+x/(x^{2}+y^{2})dy. Here
I am not so sure I could "easily" compute the integrals which would
come out of parameterizing the four line segments, and changing the
x,y language to t language. At least the task would be lengthy. So let
me show you another way. Again, I can't just apply Green's Theorem to
W and the region inside W, since W encloses the same bad point. But I
can do something very clever using Green's Theorem.
Crosscut
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, P_{y}=Q_{x} and therefore the double
integral side of the Green's Theorem equation is 0. The line integral
side is also 0. This will work for any separation, no matter how
small. Now the line integral will vary quite 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 2Pi!
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 2Pi for similar reasons.
A physical model?
I think the computations I've just done are difficult to understand. I
mentioned the following physical model to some students after
class. Line integrals can also be thought of as measuring flux,
the net flow through the curve. We will investigate this situation in
R^{3} 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 P_{y}Q_{x}=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 out 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. So: 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.
Why?
Take two points, A and B, in the plane and suppose that C_{1}
and C_{2} are two paths going from A to B. If the line
integrals along C_{1} and C_{2} are equal, then if you
first integrate along C_{1} and then backwards along
C_{2} (this is usually called, not surprisingly,
C_{2}) the net result will be 0. So the integral along any
closed curve will be 0. On the other hand, if you know the integral
around all closed curves is always 0, then you can compare integrals
around two paths, say C_{1} and C_{2}, from A to B by
just looking at the work done around
C_{1}C_{2}. Since, by assumption, the integrals
around closed curves are 0, the integral along C_{1}
must exactly be canceled by the integral along
C_{2}. Therefore the integral along C_{1} is the same
as the integral along C_{2}: this is path independence.
More restatement and summary
We showed that a gradient vector field is actually path
independent. Even further, if we happen to know a potential,
that is, a function whose gradient is that vector field, then the
integral from START to END will just be the
difference: the value of the potential at the END minus the
value of the potential at the START.
How to answer the previous QotD
So V is
(x^{3}Ax^{2}yy^{3})i+(Bxy^{2}2x^{3})j. We're
told that there are values of the constants A and B which make
V a gradient vector field. What are these constants? Here is
one approach, which also "creates" the potential function. The reason
to use this approach is that (looking ahead in the problem) we'll need
to use the potential function to evaluate a line integral. Suppose
f(x,y) is a potential function for V. Then:
f/x=x^{3}Ax^{2}yy^{3}
f/x=Bxy^{2}2x^{3}
Let's integrate the first equation with respect to x (I put
that in bold type because it is easy to get confused about what is
variable and what is constant  here x is the variable,
and A and y are constants):
f(x,y)=(1/4)x^{4}(A/3)x^{3}yxy^{3}+C_{1}(y). The function C_{1}(y) is an antidifferentiation "constant". It could contain any function of y (including just ordinary constants).
The second equation, integrated with respect to y, gives:
f(x,y)=(B/3)xy^{3}2x^{3}y+C_{2}(x). The
function C_{2}(x) is similar to the previous "constant": it
can contain any function of x.
Now we have two descriptions of the potential, f(x,y):
Description #1
(1/4)x^{4}(A/3)x^{3}yxy^{3}+C_{1}(y)
Description #1
(B/3)xy^{3}2x^{3}y+C_{2}(x)
We can compare and crosscheck. Everything should appear exactly once
and only once in f(x,y). So the (1/4)x^{4} appearing in #1
must be somewhere in #2: ahhh, it is hiding inside
C_{2}(x). The (A/3)x^{3}y in #1 must be
2x^{3}y in #2, and therefore A/3=2 so that A=6. The
xy^{3} in #1 is (B/3)xy^{3} in #2, and we infer that
1=B/3 so that B=3. There's nothing corresponding to C_{1}(y)
in #2, so I might as well (to make my life simpler!) take
C_{1}(y) to be 0. If I were working on an exam (or any time,
actually), I think I would add suspenders to my belt and check the
other way: that is, look at all the terms in #2 and make sure I can
find them in #1. That is indeed true here. I combine all the
information about my description and tell you that a potential for
V is f(x,y)=(1/4)x^{4}2x^{3}yxy^{3}.
The second part of the QotD told us that C was a curve starting at
(0,2) and ending at (3,1). We were requested to find the line integral
_{C}V·T ds. We have a potential, so
this is just
f(END)f(START)=f(3,1)f(0,2)=(1/4)3^{4}2(3^{3})13(13)0
(all the terms of f(x,y) have an x in them and x=0 at the
START).
Comment If this problem occurred on an exam I gave, I would
want students to leave the answer as it is. Don't touch it  you
might break it. If you break it, then you "buy" it, and I should
reduce the score.
Another way to find A and B
If we just wanted to find A and B, well, differentiation is almost
always easier than antidifferentiation (that doesn't occur here, with
polynomials). We can use Clairaut's Theorem (equality of mixed partial derivatives) to find A and B. Since
f/x=x^{3}Ax^{2}yy^{3}
f/x=Bxy^{2}2x^{3}
we know that (/y)f/x=(/x)f/y
and therefore (/y)(x^{3}Ax^{2}yy^{3})
must equal (/x)(Bxy^{2}2x^{3}),
so that
Ax^{2}3y^{2}=By^{2}6x^{2}. Again we
compare these two descriptions of polynomials. If these
descriptions are to be the same, then A=6 and 3=B. This leads to
the same values of A and B as we got before. I didn't use this method,
which is frequently quicker than integrating, because I believe the
person who wrote the problem: that there would be values of A and B
making this vector field conservative, and that I would need the
potential to complete the problem. If I just done the differentiations
here, then I still would have had to integrate to get a potential.
Another way to find a potential
Sometimes in physics and engineering applications there are reasons to
"prefer" one point in the plane to another. This sounds awfully silly,
but ... it does occur. When this happens, "construction" of a
potential can be done by paying attention to this specified
point. I'll call the specified point, "the ground state". Here in this
example I'll use (1,2) as the ground state. This is certainly
artificial, but the choice will illustrate the method.
If someone gives me a vector field which is guaranteed path independent, then I might want to believe the potential could be revealed by making measurements relative to the ground state. I might take paths from (1,2) to (x,y) and call the work done f(x,y). Since the vector field is supposed to be path independent, the choice of path won't matter. I don't think I would take the silly curvy path shown. That's hard to work with. Indeed, as I mentioned in class, almost all of the paths you're likely to see in "real life" will be line segments and arcs of circles. I'd probably take the path shown, a line segment from (1,2) to (1,y) and then a line segment from (1,y) to (x,y). An alternative suggested by students is the direct trip: from (1,2) to (x,y) and that is certainly used some of the time, but I am lazy (I don't want to compute the darn tilted line!).
I will write this as a sum of two integrals:
f(x,y)=_{Vertical segment}(x^{3}6x^{2}yy^{3})dx+(3xy^{2}2x^{3})dy
+ _{Horizontal segment}(x^{3}6x^{2}yy^{3})dx+(3xy^{2}2x^{3})dy
What's going on?
Why shouldn't the integral over closed curves always be equal to 0?
A computation
A realization
Example
Green
z/x
Use the chain rule. The result is f´(5cos(x)+3e^{y})(5sin(x)).
z/y
Use the chain rule. The
result is f´(5cos(x)+3e^{y})(3e^{y}).
^{2}z/x^{2}
Use the chain rule and the product rule beginning with the very first
result about z: differentiate z_{x} with respect to x.
So we get
f´´(5cos(x)+3e^{y})(5sin(x))^{2}+f'(5cos(x)+3e^{y})(5cos(x)).
A more difficult example might be f(7cos(x)e^{2y}).
A second example
I think I then did an almost useful two variable example. Suppose that
z=f(x,y), where f is a differentiable function of two variables, and
x=r cos() and
y=r sin().
z/r
This is (f/x)·(x/r)+(f/y)·(y/r)= (f/x)cos()+(f/y)sin()
z/
A line integral
Let's look at the curve
x(t)=cos(t)
y(t)=sin(t)+4cos(t)
for t in the interval [Pi/2,Pi]. This is the curve C.
Let's look at the "force field" (vector field)
F=3ixj.
To the right is a picture of the curve. It goes from the START at (0,1) (plug in t=Pi/2 in the equations for x(t) and y(t)) to END at (1,4). It seems to be and is threequarters of a tilted ellipse.
The work integral _{C}F·T ds which I immediately
translate to _{C}3 dxx dy. Since we are given a
parameterized C, I retranslate to tland:
x(t)=cos(t) leads to dx=sin(t)dt
y(t)=sin(t)+4cos(t) leads to dy=[cos(t)4sin(t)]dt
The dictionary gets us
_{t=Pi/2 [START]}^{t=Pi [END]}(3cos(t){cos(t)}[cos(t)4sin(t)])dt
and we can multiply etc.:
_{t=Pi/2}^{t=Pi}(3cos(t)+cos(t)^{2}4cos(t)sin(t))dt
Another line integral
The path I wanted here was a line segment from (0,1) to (1,4). This
path is shown to the right. In vector notation, we want
(1t)<0,1>+t<1,4>. This takes us from t=0 at (0,1) to
(1,4) at t=1.
Understanding what we did
We showed that work done by the same vector field between the same
pair of points can depend on the path traveled. This is actually
realistic. If you push a marble from point A to point B, then the
local conditions of the path (friction, slipperiness, etc.) can matter
a great deal.
Computationally what we did was really fairly
straightforward. We took the work, turned it into a
P dx+Q dy integral, then parameterized and turned everything
into tland. And we computed.
Conceptually I would like to try to understand why there is a
difference in the work done. I would like to understand when the work
is the same.
Abstract version of the computation
What if ...
the vector field involved was a gradient vector field?
Consequences, prose version
A conservative vector field is one where the work done along
paths depends only on the start and end points. Such a vector field is
also called path independent. This condition is the same as
asking that the integrals along closed curves (where the
starting point and ending points agree) must always be zero.
If a vector field is a gradient vector field so that it is the gradient of a function called a potential of that vector field, then that vector field is conservative. The work done along a curve is equal to the value of the potential at the end of the curve minus the value of the potential at the start of the curve.
Gravitation
For example, the inverse square law we introduced is conservative
because it is a gradient vector field. We can verify this by
exhibiting a potential, and the validity of being a potential can be
checked with two derivative computations. Then all work integrals will
be "easy".
Vocabulary
Conservative
Closed curves
Gradient
Potential
The earlier example
is not a gradient vector field.
Reason #1: it is not conservative
Reason #2: try to create a potential
An exam question as the QotD
HOMEWORK
Prepare the
Maple assignment. Read and do the syllabus problems for
16.3. Look at the review questions and select one or two at random and
write out solutions. If I don't have the solutions posted already,
then send them to me, please. I will check them and post them, thank
you then.
Maple
If you did not get information about the third Maple
assignment, please send me
email immediately. Messages were sent out by midday
Saturday. This assignment is somewhat longer than the previous two
assignments and likely will need more time.
Exam
There will be an exam on Friday, April 7. A draft formula sheet has been
prepared. Please look at it. I don't want any further mistakes to
occur, and you could also advise me if you think something important
is not there. Some review problems
have also been prepared. I hope that students will write solutions to
some of these problems. These solutions can be sent to me. I will
proofread them, and post them. Everyone can benefit from this. There
are a number of review problems, and they probably can't all be done
the night before the exam! The sheet of review problems also contains
information about what will be tested by the exam.
Irrelevance awards
I asked in the last diary entry for the identity of the individual
pierced by arrows and for the painter. The answers are, respectively,
St. Sebastian and Andreas Mantegna
(c. 14311506). St. Sebastian was martyred c.288 at Rome. I found the
following information which I did not know before:
During the 14th century, the random nature of infection with the Black Death caused people to liken the plague to their villages being shot by an army of nature's archers. In desparation they prayed for the intercession of a saint associated with archers, and Saint Sebastian became associated with the plague. 
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 F(x,y) of two variables. Then the
gradient of F is F=[F/x]i+[F/y]j is a vector field: it is
a function multiplied by i plus a function multiplied by
j.
An example
I looked at F(x,y)=x^{2}+y^{2}. Here I recalled that
the contour lines were x^{2}+y^{2}=Constant. These are
all circles centered at the origin. If we draw a collection of contour
lines for equally spaced constants (for example, for 1 and 2 and 3 and
4 and ...) then it turns out that the lines 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 and point away from the origin in the direction of
increase of F. 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 F:=x^2+y^2;. The
picture displays the output of these Maple
commands:
> contourplot(F,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. The function is x^{2}y2x+y^{2}+3y. The level
curves are shown for the values 12, 10, 8, 6, 4, 2, 0, 2, 4, 6,
8, and 10: the even integers from 12 to 12. Notice that the arrows
seem to be perpendicular to the level curves, and that the arrows are
longer when the spacing between the contour lines (corresponding to
equal differences in the constants) gets closer.
Comment
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: F_{x}=2xy2=0
and F_{y}=x^{2}+3y+3=0. The first equation
tells me that y=1/x and then the second equation becomes
x^{3}+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
R^{3}). That is, the gradient vector field "arrows" are
perpendicular to the level curves. They "point" towards increasing
values of the function. Their magnitude (the length of the arrows) is
a measure of the density (?) of the contour lines: if the function is
rapidly changing then the length will be longer.
By the way, if F=V,
then F is called a potential of the vector field,
V. The word and concepts connected with it are important in
physics. I'll discuss more about this during the next class.
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? If it is not, can we
explain why? This turns out to be a serious and interesting question,
because of things we will learn very soon (Friday). Also it has some
interesting physical consequences. Well, here is one way to decide the
answer.
Suppose F(x,y) is a potential for yi+xj. That
means
/y F_{x}=y > F_{xy}=1 /x F_{y}=x > F_{yx}=1But "mixed" partial derivatives are supposed to be equal (that is, if some hypotheses are satisfied, but these are all very reasonable functions and nothing weird happens). Since 1 is not equal to 1 (this week?) we know that the vortex flow is not a gradient vector field. I don't think this is an obvious fact.
Line integrals
We need an additional technical tool to investigate vector fields:
line integrals. Line integrals will allow the computation of work for
force fields, and flux for fluid flow, and something for gradient
vector fields. Work and flux and other things are important physical
quantities and turn out to have nice mathematical properties. So here
we go.
Mass of a wire
I'll introduce line integrals using the metaphor of the mass of a wire.
The integral calculus mantra is:
chop up, approximate, sum, limit.
Here I'll assume that there is a wire sitting in R^{2}. Its
geometry is simple, with a constant crosssectional area (assumed to
be 1 in the measurement system used). The density of the wire will
vary according to the position on the wire: at some points the density
will be high, and the wire heavy, and at other points, the density
will be low, and the wire rather light. The task is to create some
technical tool which will represent the total mass of the wire. By the
way, the accompanying picture looks like a particularly ugly
worm or snake. I am sorry.
Suppose I cut the wire up into lots of little chunks. How little?
Well, the length of each chunk will be ds, a tiny piece of arc length
(we discussed this in several lectures given late January, long, long
ago). If I assume that the density D(x,y) varies only a small amount
because the chunk of the wire is very small, then dm, the mass of this
piece, is nearly D(x,y) ds (remember I made the crosssectional
area equal to 1). Further, I can add up these pieces of mass to get
the total mass. I can take the limit as the number of pieces gets
large and the length of the pieces gets small. The result is that the mass
of the wire should be given by
_{The wire}D(x,y) ds.
This integral is called a line integral, where the word
line doesn't mean "straight line" but is used in the sense "A
thin continuous mark, as that made by a pen, pencil, or brush applied
to a surface." Now I need to show you what this means. So I will
compute a very artificial example.
Silly example
My example was the following: the wire follows the part of the circle
of radius 3 which is in the first quadrant. The density of the wire at
the point (x,y) is 7y+5. Find the mass of the wire. Well, I will take
the integral _{The wire}D(x,y) ds and parameterize
everything in sight. To me the natural parameterization of an arc of a
circle uses essentially the angle from the center. The text likes to
use t here, so I will parameterize this quarter circle with x=3cos(t)
and y=3cos(t). The interval of this parameterization is [0,Pi/2]. Now
ds is sqrt([dx/dt]^{2}+[dy/dt]^{2}) dt. The
square root stuff is the magnitude of the velocity vector, the
speed. And ds=SPEED·dt is a translation of
distance=rate·time of course. In this case, dx/dt=3sin(t) and
dy/dt=3cos(t) so that
sqrt([dx/dt]^{2}+[dy/dt]^{2}) just happens (!!!) to
simplify to 3. And ds=3 dt. What about the density? Since
D(x,y)=7y+5, we know that the density is 7(3sin(t))+5. And the
integral should go from 0 to Pi/2. Therefore
The mass=_{The wire}D(x,y) ds=_{t=0}^{t=Pi/2}{21sin(t)+5}3 dt=63cos(t)+15t]_{t=0}^{t=Pi/2}=63+(15/2)Pi.
This is a totally insignificant, physically unrealistic (to me) computation. The only thing I had fun with is drawing the picture, which with its varying colors is supposed to suggest the increasing density of the wire as y increases. There is one very important fact which should be mentioned now:
Independence of parameterization
In this line integral and in all line integrals, the result will be
the same no matter which parameterization is used. Verification of
this uses the one variable chain rule and is not too interesting right
now. So let me go on.
Random example
I then featured a really random example. I asked students for some
random (positive) integers which I then used to construct an example
very much like the one presented in what follows. The previous example
was arranged so that ds was nice. I now tried to compute something
like the mass of a wire where the wire was sitting on the curve
y=x^{4} from, say, (0,0) to (2, 16), and the density is
D(x,y)=x^{4}y^{7}. I not being totally idiotic
here. Any computational strategy in calculus should be able to handle
lowdegree polynomials fairly efficiently, even by hand. So here's
what we did.
A simple parameterization of a graph of a function is just to use the
independent variable as the parameter. So I used x=t and
y=t^{4} and then the density x^{4}y^{7} became
t^{4}(t^{4})^{7}=t^{28}. This isn't
too bad, but I am saving the worst for last, and in this computation
the worst is ds. So dx/dt=1 and dy/dt=4t^{3},
and therefore ds/dt=sqrt(1+16t^{6}). The line integral for
this "mass" translates in tland as follows:
_{The wire}D(x,y) ds=_{t=0}^{t=2}t^{28}sqrt(1+16t^{6})dt.
This integral cannot be computed exactly in terms of standard
"familiar" functions. I found to my amazement that Maple does
have a storehouse of weird functions which it can use to "evaluate"
this integral. But then I have no feeling for the functions it used
and certainly not for the values of the functions.
ds is the obstacle
What is horrible about most line integrals is ds. Almost no ds except
for those included in textbook problems lead to familiar
antiderivatives. At the same time, important quantities such as work
and flux are defined as mathematical objects in terms of line
integrals, and the general expectation is that these quantities
can be computed exactly in terms of familiar functions.
To me this is an excellent example of the psychological phenomenon
known as cognitive
dissonance:
Cognitive dissonance arises from conflicting cognitions. Cognitive dissonance is the perception of incompatibility between two cognitions, which for the purpose of cognitive dissonance theory can be defined as any element of knowledge, attitude, emotion, belief or value, as well as a goal, plan, or an interest. In brief, the theory of cognitive dissonance holds that contradicting cognitions serve as a driving force that compels the mind to acquire or invent new thoughts or beliefs, or to modify existing beliefs, so as to minimize the amount of dissonance (conflict) between cognitions. 
Work
Let's consider the physical concept called work. This is
"force" times "distance", loosely, but we're going to need to be a bit
more specific. For example, consider a mass being pushed up a
frictionless triangle. If the triangle is steeper, then more force is
needed. In the picture, the force of gravity is directed down. The
blue arrows on each inclined plane (triangle) show the force component
needed to move the box up that triangle.
So instead of dieting, avoid steep triangles and your weight
will decrease ... no no no
... this is just more bad information from the lecturer!
Anyway, what really matters is the component of the force in the
direction of motion (we could take the dot product of the force and
the displacement also).
Work with a varying vector field along a curve
Again, chop up, approximate, sum,
limit.
I would like to describe how to compute the work done against a
varying force field F as we travel from p to q (in
R^{2} but the same ideas will work in R^{3}, also)
along a curve, C. We chop up C into small pieces (the red lines are
the chopping places). One of them is displayed under a "magnifying
glass" (sort of) in the picture. The piece is so small that it is
almost a straight line, and its length is ds. The piece is also so
small that the vector field, F, is almost constant near the
piece. Remember that a unit tangent vector, pointing in the direction
of the piece, is called T (go back and think of January!). Then
the part of the force field along the curve is F·T and
the piece of the work will be approximately F·Tds. All
of the work will be the integral along C of this quantity. Therefore
the work is _{C}F·Tds.
A computation
Let us "test" this definition with a random computation: well,
sort of "random". C will be the curve y=x^{4} and p will be
the point (0,0) and q will be the point (2,16). I will have the force
field be
x^{2}y^{3}i+xy^{5}j. Now let
us compute. We need to change everything
into tland, where I will choose a most routine parameter: x=t so that
y=t^{4}.
ds
As before, the speed becomes
sqrt([dx/dt]^{2}+[dy/dt]^{2})=sqrt([1]^{2}+[4t^{3}]^{2})=sqrt(1+16t^{6})
which is horrible enough. Thus ds=sqrt(1+16t^{6})dt.
F
At the point (x,y), F is
x^{2}y^{3}i+xy^{5}j so that at
(t,t^{4}) on the curve, F is
t^{2}(t^{4})^{3}i+t(t^{4})^{5}j=t^{14}i+t^{21}j.
T
Now comes almost the miracle. In the movie Shakespeare in Love,
one character states, "The natural condition is one of insurmountable
obstacles on the road to imminent disaster." He then says almost
immediately, "Strangely enough , it all turns out well." When asked
"How", the character replies, "I don't know. It's a mystery." So, if
not a miracle, let me show you the little mystery here.
The unit
tangent vector, T, is a vector in the direction of the
curve. The position vector of the curve is <t,t^{4}> so
that the velocity vector is <1,4t^{3}>. But we need a
unit vector to get the projection of F in the direction
of the curve. Divide by the magnitude of
<1,4t^{3}>. Therefore,
T=<1,4t^{3}>/sqrt(1+16t^{6}).
Assembling the work integral
_{C}F·Tds=_{t=0}^{t=2}(t^{14}i+t^{21}j)·
(<1,4t^{3}>/sqrt(1+16t^{6}))sqrt(1+16t^{6})dt. This is
_{t=0}^{t=2}t^{14}+4t^{24} dt.
which with totally routine polynomial calculations can be evaluated:
it is 2^{15}/15+(4/25)2^{25}.
What happens?
The speed comes in to squeeze down the velocity vector to get the unit
tangent vector. The speed comes in as the factor which multiplies dt
to get ds. The two appearances of the speed cancel. They will always cancel!.
Using the notation to help
Here is how people use notation to guide their way through the
computation. No one computes the speed (the square root stuff) when
computing work because it will cancel. So:
Problem statement
Compute the work if
F(x,y)=x^{2}y^{3}i+xy^{5}j
and the curve is y=x^{4} going from (0,0) to (2,16).
A solution
So I initially write _{C}F·Tds but then I immediately change
to _{C}x^{2}y^{3}dx+xy^{5}dy. I
evaluate this line integral again by changing everything to tland. So
x=t and y=t^{4} and dx=1dt and dy=4t^{3}dt, Also,
x^{2}y^{3}=t^{2}(t^{4})^{3}=t^{14}
and
xy^{5}=t(t^{4})^{5}=t^{21}. Therefore
x^{2}y^{3}dx+xy^{5}dy=t^{14}dt+t^{21}4t^{3}dt.
Therefore _{C}x^{2}y^{3}dx+xy^{5}dy=
_{t=0 [START]}^{t=2 [END]}
t^{14}+4t^{24}dt. And the result will be the
same. Almost everyone uses this notation, and never bothers with
computing T and ds and then canceling, etc. Of course, the
ideas are important: the physical quantity we are computing has
certain properties (wait until Friday!) which make this computation
extremely interesting.
QotD
What is the work done if F is the vector field
x^{2}i+yj is and the curve is x=t and
y=t^{3}t as the parameter goes from t=0 to t=1? The answer
turns out to be 1/3.
Caution!!!
I just read students' answers to the QotD. Many people did this
problem completely correctly. Many others clearly had the ideas
correct but made one or two or three algebra or
arithmetic errors. This sort of thing can really hurt you on
exams!
The word iatrogenic refers to illnesses or other problems "caused by medical examination or treatment." Please don't let your solutions make the problems worse!
HOMEWORK
Keep reading
chapter 16. I will discuss conservative vector fields, etc. (16.3) on
Friday. Please be there.
Maple?
I hope that either today or tomorrow information about the third
Maple lab will be sent to students. This lab will be worth
20 points (!). Please watch for email.
Spherical coordinates
I drew the same picture I showed last
time. I deduced the following formulas (with errors but they were
corrected):
x=(rho)sin(phi)cos()
y=(rho)sin(phi)cos()
z=(rho)cos(phi)
I remarked that it was useful to know that such formulas existed, but
that I had rarely used them. One result that I have used frequently
comes from the fact that rho represents the distance from (x,y,z) to
the origin.
x^{2}+y^{2}+z^{2}=rho^{2}.
Standard restrictions on spherical coordinates
Because the angles sort of fold over when Pi's and 2Pi's are added,
most people who use spherical coordinates put some restrictions on how
big/small and phi can be. If we only allow to be between 0
and 2Pi and only allow phi to be between 0 and Pi, then there will be
unique spherical coordinates for every point in R^{3}. So I
will generally work with these restrictions.
Some shapes in spherical coordinates
rho=constant gives a sphere centered at the origin. So, for example, rho=5 is a sphere centered at the origin of radius 5.  phi=constant gives a right circular cone whose axis of symmetry is the zaxis. For example, phi=Pi/6 is a cone with vertex at the origin and whose axis of symmetry is the positive zaxis. The angle between the positive zaxis and any of the cone's "generators" (lines from the vertex on the surface of the cone) iw Pi/6 (yes, 30^{o}). The bottom half of the cone is not included because that is where phi is between Pi/2 and Pi.  =constant gives a halfplane, with the zaxis being the edge of the halfplane. For example, =Pi/4 gives a halfplane which is perpendicular to the halfline y=x (x>0) in the xyplane. The other half of the plane is where is 3Pi/2, and so it is not included in this obhect. 
Integral #1
A spherical region of radius R is filled with material whose density
is directly proportional to the distance from the origin. What is its
mass?
This is not very realistic. The center is light and
fluffy and the outer edge is heavy and tough (my kind of
cooking?). The density is supposed to interpolate linearly between
these extremes. Maybe the appropriate assignment would be to build an
object of this type.
The math setup
Take a small piece of volume, dV, in the sphere. The corresponding
piece of mass, dm, is related to dV by dm=(density)dV. We know that
the "density is directly proportional to the distance from the
origin." Place the origin of the coordinate system at the center of
the sphere. So there is some constant C>0 so
density=C rho. And the total mass is the sum of the dm's. This
"sum" should be a triple integral:
Total mass=_{The whole ball}C rho dV.
In spherical coordinates, a description of a sphere of radius R
centered at the origin is easy: rho goes from 0 to R, goes from
0 to 2Pi, and phi goes from 0 to Pi. We just use the agreed upon
ranges for the angles to sweep out a whole sphere. There is one sticky
point, however.
dV in spherical coordinates
We need to convert dV to spherical coordinates. In fact,
dV=(rho)^{2}sin(phi)d(rho)dd(phi)
I know this is true. First, there is a diagram which is supposed to be
convincing in the text (figure 8, page 1035). Second, I said it in
class. Third, I actually can give an understandable argument if there
is enough time later in the course. In any case, when I use spherical
coordinates, I almost never bother thinking about this weird mess, but
I just write it.
Vocabulary Several students commented that the "weird mess" is called
the Jacobian, which is how area or volume is distorted with a change
of coordinates. Again, if there is time I'll discuss Jacobians later
in the course (see 15.9 for more information right now, if you'd
like).
The computation
So we have
Total mass=_{phi=0}^{phi=Pi}_{=0}^{=2Pi}_{rho=0}^{rho=R}(C rho)(rho)^{2}sin(phi)d(rho)dd(phi).
The inner integral
_{rho=0}^{rho=R}(C rho)(rho)^{2}sin(phi)d(rho)=_{rho=0}^{rho=R}C(rho)^{3}sin(phi)d(rho)=Csin(phi)(rho)^{4}/4]_{rho=0}^{rho=R}=Csin(phi)R^{4}/4.
The middle integral
_{=0}^{=2Pi}Csin(phi)R^{4}/4 d=(2Pi C)sin(phi)R^{4}/4=[(Pi C)/2]sin(phi)R^{4}.
(Just multiply by 2Pi, since there is no in the integrand.)
The outer integral
_{phi=0}^{phi=Pi}[(Pi C)/2]sin(phi)R^{4}d(phi)=[(Pi C)/2]cos(phi)R^{4}]_{phi=0}^{phi=Pi}=(Pi C)R^{4}.
I don't know any way to check this answer. Build a model? Weigh it?
Is this silly?
Well, yes, it is silly. The problem is invented and certainly designed
exactly for spherical coordinates. But I would not use spherical
coordinates, which definitely have peculiarities (look at the pictures
above and look at the expression for dV) unless both the region and the integrand
can both be described in a nice fashion with spherical coordinates. I
won't use this coordinate system otherwise. (Could you imagine using
spherical coordinates to describe a cube?)
Integral #2
Consider the region in the first octant consisting of points whose
distance to the origin is at least 1. Imagine that this is filled with
material whose density is inversely proportional to the fifth power of
the distance to the origin. What is the mass of this object?
Translating
All of R^{3} is divided into eight parts by the coordinate
planes: x=0, y=0, and z=0. Each part is called an octant. While
the corresponding regions in the plane (the quadrants) have individual
designations, the
only octant that is named is the first: the octant where
x>0 and y>0 and z>0. In this first octant, I'm
excluding points whose distance to the origin is less than
1. What does the remaining region look like? I think the best picture
I drew in class was one looking at the octant from the back, so here
is a version of that picture. In this picture (sort of the corner of a
rectangular box), a spherical "bite" has been taken out of the
corner. The bite is centered at the vertex (the origin) and has radius
1. Wow!
To the right is a more oblique view of the octant with the bite.
The nice thing about this region is that it can be described very briefly
in terms of spherical coordinates. Certainly, rho will go from
1 (as close to the origin as the bite will let us get) out to ... out
to ... infinity (an improper integral!). What about and
phi? Here students should look closely at the definitions of and
phi. Each of them will go from 0 to Pi/2. This is best confirmed by
taking "angles" with vertex at the origin and a side along the x
(respectively, zaxis) and then opening the second side of the angle
to an aperture of Pi/2 (I think "aperture" means the angle's opening).
By the way, as I remarked in class, this is actually a more realistic
example than the first integral. Things like this do occur in
electricity and magnetism.
The computation
Again dm=(density)dV=[C/(rho)^{5}]dV because "density is
inversely proportional to the fifth power of the distance to the
origin." And we know the limits from the discussion above, so the
total mass is
_{phi=1}^{phi=Pi/2}_{=0}^{=Pi/2}_{rho=1}^{rho=infinity}[C/(rho)^{5}](rho)^{2}sin(phi)d(rho)dd(phi).
The integrand is [C/(rho)^{3}]sin(phi) after cancelling some powers.
The inner integral
This is an improper integral, so I will be careful.
_{rho=1}^{rho=BIG}C/(rho)^{3}sin(phi)d(rho)=C/(2(rho)^{2})sin(phi)]_{rho=1}^{rho=BIG}=C/(2(BIG)^{2})sin(phi)
+C/(2(1)^{2})sin(phi). As BIG>infinity, the
term C/(2(BIG)^{2})sin(phi)>0 so the improper
integral
_{rho=1}^{rho=infinity}C/(rho)^{3}sin(phi)d(rho) converges and its value is
(C/2)sin(phi).
The middle integral
_{=0}^{=Pi/2}(C/2)sin(phi)d=(C/2)sin(phi)(Pi/2)=[(C Pi)/4]sin(phi).
The outer integral
_{phi=0}^{phi=Pi/2}[(C Pi)/4]sin(phi)d(phi)=[(C Pi)/4]cos(phi)]_{phi=0}^{phi=Pi/2}=[(C Pi)/4]
Again, I will admit that I don't know any way to check this
answer. When such an integral comes from a real physical problem,
there is frequently some way to see if the final answer is reasonable.
Further defense of silly (the same defense)
I would only use this technique, I hope!, where both the region
and the integrand are suitable. So, although the problems may have
seemed silly, they are the sort of applications which might occur. We
will need integration in spherical coordinates a few times later in
the course.
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
many complicated "things" 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 some neat way. We'll try to understand this.
The Fundamental Theorem of Calculus, one of the wonderful results of
calc 1, is something like this:
If F´=f, then _{x=a}^{x=b}f(x) dx=F(b)F(a)
This is a wonderful result. Think about it. It states that some sort
of stuff inside the interval [a,b] (we could think about height times
width, f(x)dx, or some accumulation of "stuff": f(x) inside [a,b] and
added up over the whole interval, [a,b]) is equal to an edge
quantity. There isn't much "edge" to an interval. Here the edge
quantity is F(b)F(a). You can, with some effort, see things as some
sort of balance between interior accumulation and stuff in and out
the edges. The remainder of the course is an effort to generalize this
to dimensions 2 (Green's Theorem) and 3 (Divergence Theorem) and
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 learned only
recently (there are thousands of commands, and I believe I know less
than a tenth of one percent of them). This command is in the
plots package. The picture was produced by
fieldplot([x^2+10*y,3*xy^2], x=1..5,y=1..5, arrows=slim,
grid=[10,10], fieldstrength=maximal(.9), thickness=2);
This vector field is the vector
(x^{2}+10y)i+(3xy^{2})j at the point (x,y). The
Maple command sketches a few of the arrows ([10,10] gives 100
of them). Sometimes such a picture of a vector field can be very
helpful.
Models?
Vector fields are mathematical models of some basic physical
phenomena. Two that are immediate are force fields and fluid flows.
The force field (gravity, electromagnetism, etc.) might be the
simplest. The "arrow" at every point denotes the presence of a "force"
which can act on the proper kind of object (gravitation: an object
with mass, magnetism, an object with magnetic "stuff", etc.). In
Newton's law of gravitation, we could think of a powerful mass, M,
(the sun?) at the center of the universe, and other very small masses, m's,
scattered around. Ignore interactions between the m's. Each m will be
attracted to the M with a force of magnitude GmM/(dist)^{2}
where "dist" is the distance from m to M. The direction of the
attraction is from m to M. To get a vector field, just sketch an arrow
in the direction of the force, with a length proportional to the
magnitude of the force. Then to complete the creation (?) of the force
field, just remove all the little m's and leave the arrows. This is
quite an idea, really a big leap of imagination. I haven't seen any of
these arrows, but imagining them is sometimes quite helpful. A quote
from Albert Einstein describing radio
may be appropriate here:
You see, wire telegraph is a kind of a very, very long cat. You pull his tail in New York and his head is meowing in Los Angeles. Do you understand this? And radio operates exactly the same way: you send signals here, they receive them there. The only difference is that there is no cat. 
More detail about the inverse square law Suppose we stand at (x,y) with a mass m, and the origin has a mass, M. Then the gravitational attraction is a force of magnitude GmM/(x^{2}+y^{2}). What's on the bootm is the square of the distance from (x,y) to (0,0). I'll forget the m now (erasing the mass m and leaving the arrow). So we need a force of magnitude GM/(x^{2}+y^{2}). The direction should be from (x,y) to (0,0): the direction of xiyj. But that vector has length sqrt(x^{2}+y^{2}). So to get a vector of the correct magnitude and direction we need to divide by sqrt(x^{2}+y^{2}) (that creates a unit vector pointing in the correct direction) and then multiply by GM/(x^{2}+y^{2}). So an inverse square law pointing at the origin is [x/(x^{2}+y^{2})^{3/2}]i[y/(x^{2}+y^{2})^{3/2}]j. Is this complicated enough? It can be difficult to see the inverse square law under all this algebra. To the right is a picture of this inverse square vector field in the first quadrant. The magnitude of the vector field decreases rapidly away from the origin.  
Fluid flow: vortex flow 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 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 gave an example, which turned out to be difficult to explain (this means "I didn't do it well"), of a vortex: a counterclockwise velocity field of a fluid spinning with uniform angular velocity. Suppose the vector field was Ai+Bj at the point (x,y). This "vortex" would need to be perpendicular to the circle centered at (0,0) going through (x,y). That means the vector field would be perpendicular to the radius vector, xi+yj. We can check perpendicularity with the dot product: Ax+By must be 0. But the magnitude of the vector, sqrt(A^{2}+B^{2}), also should increase directly in proportion to the distance of (x,y) to the origin, so the fluid flow will have constant angular velocity. After some computation we saw that this would have to be Kyi+Kxj, with K positive to keep this counterclockwise.  
Other flows: 3i+0j (uniform flow to the right) Please note that adjustments were made to the fieldstrength option in this fieldplot command and others to avoid having the arrows overlap each other and to display better some of the other vector fields.  
3xi+0j, a flow which doesn't move on the yaxis, but flows away from the yaxis otherwise, and faster, the farther the fluid was away from the axis.  
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.
So, what is a vector field?
HOMEWORK 

Maintained by greenfie@math.rutgers.edu and last modified 3/24/2006.