In this section we define and illustrate the very important ideas of line and sur- face integrals of vector fields.
Line integrals
Let F be a vector field. It assigns to each point in space a vector. Take for definiteness the following field in two dimensions:
(7.4.1) which assigns, for example, the value 8i + j to the point (1, 2). In a physics problem, this could be a force field. Let us now pick a path P joining end points r1 := 1 and' r2 := 2 as in Fig. 7.3.
We imagine this path as made up of tiny arrows [drili = 1 ... n] laid tip to tail. We then form the dot product F(ri) ã dri at each segment. (The vector F is assumed to be a constant over the tiny arrow.) In the physics example this would be the work done by the force over this tiny segment.
Definition 7.1. The Line Integral ofF along P joining points 1 and 2 (where
y
Figure 7.3. Line integral of F along path P.
1 and 2 are short for r1 and r2 respectively) is defined as
2 n
{ F(r) ã dr = lim ""F(ri) ã dri,
}1 n_.oo L.,
i=l
(7.4.2) i.e., in the limit in which each segment is of vanishing size.
The evaluation of line integrals is very similar to the evaluations of inte- grals in the complex plane: one must parametrize the path and reduce every- thing to an integral over the parameter t. Here is an example from two di- mensions.
Let once again
(7.4.3) be the vector field and let us integrate it from the origin to the point (1, 1) along the curve x = y2•
The first step is to parametrize the path with a variable t, i.e., make up functions x(t) and y(t) such that as t varies, the locus of points traces out the curve in question. If you wish you may pretend that t is really the time and the curve describes the trajectory of some fictitious particle moving in the x-y plane. In our problem the following choice will do:
y = t IO:::; t :::; 1) (7.4.4)
which describes motion along the curve x = y2• This choice now leads to
r(t) = lt2 + jt (7.4.5)
dr dt = 2tl +j (7.4.6)
1r2 Fãdr = 1t2 Fã-dt dr (7.4.7)
r1 tl dt
= 1t=l (2t2t2i + t4j) . (2ti + j)dt (7.4.8)
t=O
= ~oã (4t5 + t4)dt (7.4.9)
13 (7.4.10)
= 15"
Note that even though the curve lives in high dimensions, it is still one di- mensional and the line integral can therefore be mapped into an ordinary inte- gral in the variable t. It is the integrand that is calculated from higher dimen- sional considerations, i.e., by evaluating the dot product of the vector field with the displacement. In the physics case the parametrization has a direct meaning.
We are ttying to find the work done by tbe force as it drags some particle from 1 to 2 along P. If we imagine the motion as taking place in real time, F ã i stands for the power expended by the force and its integral over time is the work done. Thus the line integral becomes the ordinary integral over t of the power F ã ~~ between t 1 = 0 and t2 = 1.
The extension of this notion to vector fields in three dimensions is straightforward.
As with integrals in the complex plane, integrals over a closed curve will be denoted by fp-
The line integral is generally path dependent. For example if we join the same end points by a line at 45 ã, i.e., the line x = y, the result will be 5/6.
The details are left to the following exercise.
Problem 7.4.1. ~rw that the line integral ofF = 2xy2i + x2j between the origin and (1, 1) along the line x = y is 516. Note that you can use x itself as a parameter. Show the integral along a path that first goes horizontally (i.e., dr = idx, y = 0) to (1,0) and then straight up to (1, 1), (dr = jdy, x = 1) is 1. Show the answer is the same for a path that first goes up to (0, 1) and then horizontally to (1, 1).
Problem 7.4.1. Show that the line integral ofF= iy2+jxy+kxz from the ori- gin to the point (1, 1, 1) along a straight line connecting the end points equals unity and along the curve x = t, y = t2, z = t3 equals ;.
---~-~-! __ fJtf __ !_ - X
Figure 7.4. The flux.
Deflnldon 7.2. Given any vector field F. if the line integral is dependent on just the end points and not the path we say the field is conservative. An equiv-
alent definition of a conservative field is that
fFãdr=O (7.4.11)
for any closed path.
The equivalence of the two definitions is exactly as it was in the case of inte- gration in the complex plane.
Problem 7.4.3. (i) Calculate the line integral ofF= i2xy2+jx2 between (0, 0) and (1, 1) but along y = x2• (ii)What is the line integral around a unit circle centered at the origin? First make a sensible choice for the parameter t that will move the coordinates along the circle. Note that even for a nonconserva- tive field the line integral along certain closed paths can vanish.
Surface Integrals
Imagine a steady stream of particles moving via a pipe of rectangular cross section, along the x-axis with a velocity v = iv:r:, as shown in Fig. 7.4.
Let the density be p particles per unit volume. The current density is de- fined as
j=pv, (7.4.12)
and is given by
j = li:r: (7.4.13)
in our problem. Consider an area (denoted by S in the figure) of size S, that lies transverse to this flow, i.e., in the y - z plane. The number of particles going through the area in one second is called 4>, the flux through the area.
To find it, let us follow the particles that cross it at t = 0. At the end of this period, they would have gone a distance L = v:r:. Thus all the particles that crossed the area in the last one second lie inside a parallelepiped of base s
and length L = v:r:. The flux is then
(7.4.14)
Now the flux is a scalar, a number. We are obtaining it from a vector, j, and an area. Since one usually gets a scalar from taking the dot product of two vectors, we ask if we can represent the area as a vector. {I have already hinted at this possibility earlier.) Given a planar area as in this case, we can asso- ciate a vector with it of magnitude equal to the numerical value of the area and direction perpendicular to the plane of the area, i.e., the normal. But we can draw two normals to any plane. So we assign an orientation to the area as follows. We draw arrows around the perimeter of the area in such a way that as a person walks around the area along the arrows, she sees the area al- ways to her left or right. For each choice, we choose the normal vector to point as per the screw rule: the nonnal points along the direction of advance of a screw as the screw driver is turned in the same sense as the arrows along the perimeter.1 Thus the area vector in the figure is given by an arrow paral- lel to the current for the given choice of arrows along its edges. {If you stand upstream of the area and look at it, the arrows along the edges will run coun- terclockwise.) Using this convention for representing areas as vectors, we can write the area vector as
S =iS (7.4.15)
and the flux as
~=Six= S ãj (7.4.16)
using the fact that Six is really the dot product of two vectors which happen to lie entirely in the x-direction. But the above equation implies more than that. Suppose we tilt the area so that it no longer lies transverse to the flow of particles, i.e., so that its area vector does not lie parallel to the velocity, but enlarge it so that it still intercepts all the particles that flow in the pipe. The new area called S' in the figure is bigger by a factor 1/ cos8, 8 being the tilt. By construction the flux through it is the same as through S. Equation (7.4.16) would say the flux through S' is S' ã j, which is correct, since the increased area of S' is exactly offset by the cosine of the angle between the area vector and the current vector.
Thus the dot product of a cu"ent density vector and the area vector gives the flux of particles penetrating the area. We use the word flux for the dot product of any vector with an area even if it does not stand forã the flow of anything. For example if B is the magnetic field vector, its dot product with an area is called the magnetic flux.
1 An area without this orientation specified is like a vector from which we have erased the head and tail. Note also that only in three space dimensions can we represent an area as a vector, using that fact the area (assumed planar) has a unique direction perpendicular to it (up to a sign) which we can assign to the vector. In four dimensions for example an area in the x - y plane can have an infinity of distinct vectors perpendicular to it. An analogous statement is that in two dimensions we can define a unique normal to a curve at each point (up to a sign) while in three dimensions the normal can rotate in the plane perpendicular to the line at that point.
Figure 7.S. An undulating surface built out of smaller planar patches. The orientation of the patches is specified by the nonnals instead of arrows running along the perimeter. If you looked at the surface from above, these arrows would run counterclockwise.
Note that the flux is given by the above fonnula only if the area is planar and the current a constant over its extent. But we can generalize the notion of flux as follows. First, we note that any infinitesimal area can be approxi- mated by a plane and specified by the above convention. It will be referred to as dS. For example an area of size w-8 m2 , sitting at the north pole of a sphere centered at the origin, with its nonnal pointing outward, will be de- noted by dS = w-8k. Next we learn how to build up a macroscopic area by patching together little ones. Recall how the path P was built out of oriented line segments placed end-to-end. Each arrow was a directed segment with a beginning (tail) and an end (tip), i.e., with two boundaries. Whenever two ar- rows were so joined tip-to-tail, one boundary of the first arrow (its tip) was neutralized by one of the next (its tail) leading to a segment ~hose boundaries were the tail of the first and the tip of the second. We similarly glue areas, each with its boundaries marked by arrows running around in a definite sense.
We bring together two tiny areas and place them next to each other so that on the adjacent edges the orienting arrows run oppositely as in Fig. 6.28. (That is, the area vectors are roughly parallel and not antiparallel.) Thus the com- mon perimeter gets neutralized and we get a bigger area with a new boundary that is the "sum" of the old boundaries (where in the sum the common edges have canceled). In this fashion we build up an undulating surface in three di- mensions whose boundary is made up of the exposed edges of the pieces that tile it. A sample is shown in Fig. 7.5.
We now define the surface integral:
z
Figure 7.6. The unit cube centered at the origin. The faces we can see are at x = 1/2, y = 1/2, or z = 1/2. The other three are at minus these values.
Definition 7.3. The surface integral of a vector field V over the surface S is given by
f V. dS = lim tv(r1) ã dS;
J~s n-oo i=l (7.4.17)
where i labels the patches, all of which become vanishingly small as i -+ oo.
Note that even though the integration is over a two-dimensional region we use only one integration symbol. This is a simplification that should not cause any confusion. If the surface in question is closed (like a sphere) we will use the symbol f8 .
Imagine a point source of alpha particles at the origin. If we surround it with a surface and calculate the particle flux, i.e., the surface integral of the current density, (with area vectors pointing outward) we will get a nonzero number, determined by the emission rate. On the other hand, if we calculate the net flux out of a surface not enclosing the origin, we will get zero, since what comes in goes out. Likewise if the origin has a sink which absorbs par- ticles, we will get a negative surface integral.
Let us now work out an example: the surface integral of
(7.4.18) over a unit cube centered at the origin, as in Fig. 7.6.
First consider the two faces parallel to the y - z plane. These have coor- dinates x = ±1/2 and the area vectors are ±i. On these faces we need just the x-component of W which is x3y = ±h. On the face with area l, the contribution is
1 1/2 11/2 111/2 11/2
dy dz W ã i = - ydy dz = 0,
-1/2 -1/2 8 -1/2 -1/2 (7.4.19)
where the zero comes from the odd integrand in the y-integration. Likewise the opposite face also gives zero. If we consider the faces with areas ±j, we get a zero for similar reasons of symmetry, this time due to the x-integration.
Finally consider the two faces with areas ±k. On the face with area up the z-axis, we have z = 1/2 = Wz and the surface integral is
1 1/2 11/2 1 1
dx dy ã-= -.
-1/2 -1/2 2 2 (7.4.20)
On the opposite face, we get the same contribution because there are two extra minus signs that compensate, one due to the area vector and the other due to the change in sign of W z. The total contribution is
fwãdS= 1. (7.4.21)
Sometimes the field is more naturally given in non-cartesian coordinates.
Consider thus the field of a point charge q at the origin:
(7.4.22) where er is a unit vector in the radial direction and eo = 8.85419 ã 10-12 Coulombs2 / (Newton ã m2 ) is a constant called the permittivity of free space.
Let us find the surface integral over a sphere of radius R also centered at the origin. Consider a tiny patch of size dS on the sphere. The vector correspond- ing to it is erdS and the flux through it is E ã dS. The total flux out of the sphere is
iEãdS = i s 4 1reo q R2 er ã erdS (7.4.23)
q 1 dS (7.4.24)
= 41l'eo 8 R2
q (7.4.25)
= eo
where we have used the fact that the area of the sphere is 41l' R2 •
Problem 7.4.4. Show that the surface integral of V = iyz + jxz + kxy over a unit cube in the first octant, with one corner at the origin, is zero.
7 .5. Scalar Field and the Gradient
Let ¢> be a scalar field in three dimensions, say the temperature. The firSt question one asks is how it varies from point to point, i.e., its rate of change.
The change in¢> from one point (x,y,z) to a nearby point (x+dx,y+dy,z+
dz) is given to first order by
81/> 81/> 8,P
dt/> = -dx 8x + -dy 8y + -dz. 8z (7.5.1) Compare this to Eqn. (7.1.12) and notice that it looks like a dot product of two vectors. We can see that in the present case
dt/> = -dx ox ot/> + -dy 81/> 8y + -dz ot/> 8z =VI/>ã dr where (7.5.2) VI/> = .81/> Jot/> kot/>
1 8x + 8y + oz' and (7.5.3)
dr = idx + jdy + kdz (7.5.4)
is the displacement vector. One refers to V r/J as the gradient of 1/> or more often as "grad t/>", where "grad" rhymes with "sad". At each point in space V 1/> defines a vector field built from the various derivatives of the scalar field.
For example if
(7.5.5) then
VI/>= l(2xy + yz) + j(x2 + 1 + xz) + k(xy). (7.5.6)
The gradient contains a lot of information. We just saw that it tells us how much q, changes (to first order) when we move in any direction: we simply take the dot product of the gradient with the displacement vector. Next, if we write the dot product in the equivalent form as
dt/> = IVt~>lldrl cos9 (7.5.7) we see that for a given magnitude of displacement, ldrl, we get the maxi- mum change if the displacement is parallel to the vector V 1/>. Thus V 1/> points in the direction of greatest increase of 1/> and IVr/JI gives the greatest rate of change.
Let us pause to understand the significance of this result. Suppose you were in a temperature field T(x, y) in two dimensions and wanted to find the greatest rate of change. Since the rate of change is generally dependent on di- rection, you may think that you would have to move in every possible direc- tion, compute and tabulate the rates of change, and choose from your list the winner. Not so! You simply measure the rates of change in two perpendicular directions and make a vector out of those directional derivatives. This vector, the gradient, gives the magnitude and direction of the greatest rate of change.
To obtain the rate of change in any other direction, you take the projection of the gradient in that direction: as per Eqn. (7.5.7),
dljJ
-dr = IVI/Jicos8, (7.5.8)
where 8 is the angle between the gradient and the direction of interest.
In particular, if we move perpendicular to V 1/J, 1/J does not change to first order. If we draw a surface on which 1/J is constant, called a level surface, V 1/J
will be perpendicular to the surface at each point. Let us consider a simple function which measures the square of the distance from the origin:
(7.5.9) Thus the surfaces of constant 1/J are just spheres centered at the origin. For example the surface ¢ = 25 is a sphere of radius 5. Clearly
VI/J = 2xi + 2yj + 2zk (7.5.10)
points in the radial direction (since its components are proportional to that of the position vector) which is indeed the direction of atest increase. The rate of change in this direction is IVI/JI = 2 x2 + y2 + z2 = 2r, which is correct given that 1/J = r2 • Finally the gradient vector is clearly perpendicular to the (spherical) surface of constant ¢ or r.
Thus suppose you find yourself in a temperature field 1/J == T. If you feel too hot where you are, you must quickly compute the gradient at that point and move opposite to it; if you are too cold you must move along the gradi- ent, and if T is just right and you feel like moving around, you should move perpendicular to the gradient. Since the gradient changes in general from point to point, you must repeat all this after the first step. Here is an example. Let the temperature be given by
with VT = -2xi + 2yj. (7.5.11)
The temperature and its gradient are depicted in Fig. 7. 7.
For another example of the gradient, let us revisit the Lagrange multiplier method. When f ( x, y, z) is to be extremized with respect to the constraint
y
X
Figure 7.7. The plot of the temperature T = tP-x2 : darker regions are at higher tempera- ture. A few gradient vectors (not exactly to scale) are shown.
g ( x, y, z) = 0, we write the condition as 8f =), 8g
8x 8x (7.5.12)
and similarly for the other two coordinates. These can clearly be written as one vector equation:
Vf = )..Vg. (7.5.13)
This means that at the extremum, the normals to the surfaces g = 0 and f = c are parallel, or that the surfaces are in tangential contact. We can understand this as follows: if V f had a component along the constant g surface (g = 0)
we could move along the surface and vary f to first order. All this will be clearer if you go back and look at Fig. 3.1 in Chapter 3. As we move along the constraint circle of radius 2, we cross contours of fixed f. This means that by moving a little further we change the value of f. When we get to the extremum, the contour is tangent to the circle and there is no first order change in f .