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SECTION 4.3 FLUX AND DIVERGENCE 65

4.3. FLUX AND D IVERGENCE

From here on, we will almost always think of vector elds as representing ows of mate-rial. In this section, we seek to quantify how this ow is moving. The rst quantity westudy, outward ux, measures the net ow out of a region in the plane. The second quan-tity we study, divergence, measures ux per unit of area. In the next section, we establishthe relationship between these two quantities (Greens Theorem).

OUTWARD FLUXSuppose that we have a vector eld F which describes the velocity of moving air in a roomwith a lit replace and a glass of ice water. Near the replace, the air is being warmed, so itis expanding, which means that F near the replace is pointing out. Near the ice water, theopposite is happening, so F near the ice water is pointing in. In terms of ux, this meansthat the outward ux of this vector eld on a small region around the replace is positive,while in a small region around the ice water, it is negative.

Flux is the rate of ow through a unit area during a unit time. Various sorts of uxappear in physics, for example:

Newtons law of viscosity describes the rate of transfer of momentum across an area,

Fouriers law of conduction descibes the rate of heat ow across an area,

Ficks law of diffusion describes the rate of movement of molecules across an area.

From our point of view, given a 2 -dimensional vector eld F and a region R in the plane,the outward ux of F over R is the amount of material owing out of the region R . Sup-pose that the boundary of R is parameterized by the smooth curve r t . Then to computethe ux of F over R , we take the line integral of the amount of material owing out of theregion at each point along r t .

Consider the following region, R , whoseboundary is parameterized by the vector func-tion r t .

r t

r t0

n t0r t0

R

We want to calculate the amount of material leaving the region R at each point r t0 onthe boundary. At this point, the vector r t0 is tangent to the boundary of the region. There

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66 CHAPTER 4 VECTOR FIELDS IN TWO DIMENSIONS

are two unit vectors normal to r t0 . Let n t0 denote the unit vector which is normal tor t0 and points away from the region, which we refer to as the outward-pointing unit normalvector.

The ow at the point r t0 is given by the vector eld, F r t0 . We are only interestedin the amount of this ow which is leaving the region, i.e., the amount of this ow that isparallel to n t0 . Therefore, theamount of this ow at the point r t0 is the scalar projectionof F r t0 onto n t0 . Since we chose n t0 to be a unit vector, this is F r t0 n t0 . Theline integral (with respect to arc length) of this quantity gives us the total ow out of theregion,

Outward uxC

F n ds,

where C is the boundary of the region R , and n is the outward-pointing unit normal vectorto C .

In order to use this formula, we need to compute n

. First, we make a rule: we alwayschoose r t to parameterize the boundary of R in the counterclockwise direction (also knownas the positive orientation). Note that if r t parameterizes the boundary of a region incounterclockwise orientation, then if you walked along the path given by r t , the regionwould always lie to your left, while n would always lie to your right (so our choice is insome sense another righthand rule). This choice is arbitrary, but it is consistently used.Now, given that r t goes counterclockwise, n can be found by rotating r t to the right by90 . If r t x t , y t , then r t x t , y t , and

n t y t , x t

y t , x t .

Our formula for ux is with respect to arc length, so as usual,

ds x t , y t dt,

but notice that the speed quantity in this formula, x t , y t , is exactly the length weare dividing by in the equation for n t . Supposing, again as usual, that

F x, y P x, y i Q x, y j,

we can rewrite the formula for outward ux as

Outward ux C P x ,y i Q x, y j y t , x t

y t , x t x t , y t dt

C P x, y y t Q x, y x t dt.

We make one nal simplication in this formula: we distribute the dt to both terms of theintegrand and replace x t dt with dx and y t dt with dy .

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Outward Flux. Suppose that C is a piecewise smooth, simple,closed curve enclosing a region R in the plane and oriented in thecounterclockwise direction. Let F P i Q j be a vector eld andsuppose that P and Q have continuous rst order partial deriva-tives. Then the outward ux of F on R is

C F n ds

C P dy Q dx,

where n is the outward-pointing unit normal vector.

Example 1. Compute the outward ux of F xy 2x i x2 j on the disc of radius acentered at the origin.

Solution. The boundary of the given region is the circle of radius a , which we parameter-ize as

r t a cos t, a sin t , 0 t 2 .

We then have the following four quantities

P xy 2x a 2 cos t sin t 2a cos t,Q x 2 a 2 cos 2 t,dx a sin tdt,dy a cos tdt.

Substituting these into the integral gives us

Outward uxC

P dy Q dx,

2

0a 2 cos t sin t 2a cos t a cos t a 2 cos 2 t a sin t dt

2

0

2 a 3 cos 2 t sin t dt2

0

2 a 2 cos 2 t dt

The rst of these integrals can be done with a u-substitution; it is 0 . For the second integral,we use the identity cos 2 t 1 cos 2 t 2 , and this integral contributes 2 a 2 . Therefore theoutward ux of this vector eld over the disc of radius r centered at the origin is 2 a 2 .

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x, y x x, y

x x, y yx, y yn j

n j

n i

n i

We want to compute the ux of F over this rectangle. We know from earlier in thissection that the ux is given by the line integral C F n ds , where C is the boundaryof the rectangle and n is the outward-pointing unit normal vector. We dont have enoughinformation to compute this ux exactly, but we will be able to approximate it well enoughto get the correct limit.

Suppose, as usual, that

F x, y P x, y i Q x, y j.

Because C isnt smooth, we need to divide it into smooth pieces: the top, the bottom, theleft, and the right. For ow out of the top side of this rectangle, we can approximate F byF x, y y P x, y y i Q x, y y j. Our outward-pointing unit normal vectorfor this side is n j , and we have

Flow out of top sideTop

F j ds x

0Q x t, y y dt Q x, y y x.

Thefour other sides are similar, although the outward-pointing unit normalvectorschange.For example, for the bottom side, the outward-pointing unit normal vector is n j , so

Flow out of bottom sideBottom

F j ds x

0Q x t, y dt Q x, y x.

Adding these together, we have

Flow out of top and bottom sides Q x, y y Q x, y x.

Now recall the denition of partial derivatives,

Qy

lim y 0

Q x,y y Q x, y y

.

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70 CHAPTER 4 VECTOR FIELDS IN TWO DIMENSIONS

Therefore, assuming that Q y is continuous, the total ow out of these two sides can beapproximated with this partial derivative:

Flow out of top and bottom sides Qy

x y.

A similar computation for the left and right sides shows that their ow is approximated by a partial derivative of P :

Flow out of left and right sides P

x x y.

The total outward uxfor this rectangle is thereforeapproximately P

xQy

x y.

The area of the rectangle is x y, so if we divide by this we get ux per unit area. Taking

the limit as x

and y

approach 0

corrects for the inaccurate approximations we usedabove and gives the following formula.

Divergence. The divergence of the vector eld F P i Q j is thefunction

div F P

xQy

F .

Note that the expression F on the right of this denition is just a shorter way towrite the exact same formula; if we dene x, y , then

F x

,y

P, Q P

xQy

.

When used in this way, the symbol is referred to as the Del operator. While the dotproduct notation is convenient, note that is not really a vector (for example, what isF ?). Bending rules such as this is known as abuse of notation. (Another example isthe determinant form of the cross product.)

With this formula in hand, we return to our running example.

Example 3. Compute div F for the vector eld F xy 2x i x2 j.

Solution. By our formula for divergence,

div Fx

xy 2xy

x2 y 2 .

This shows that div F 0 , 0 2 , as we computed in Example 2, but this time with a lot lesswork.

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Because matter can neither be created nor destroyed, for vector elds that representows of material, the ux over a region really should be the double integral of divergence.In the next section, we prove this and more. To conclude this section, we return to ourrunning example to check this intuition.

Example 4. Show that the double integral of the divergence of the eld F xy 2x i x2 jover the disc of radius a centered at the origin is 2 a 2 .

Solution. For this vector eld, div F y 2 , so

Integral of divergenceR

y 2 dA,

where R is the disc of radius a centered at the origin. It is easiest to evaluate this doubleintegral in polar coordinates, where y 2 r sin 2 :

Integral of divergence2

0

a

0r sin 2 r d r d ,

2

0

r 3 sin 3

r 2r a

r 0d ,

2

0

a 3 sin 3

a2 d ,

a 3 cos 3

a2 2

0,

2 a 2 .

Note that this is the same quantity that we got in Example 1 when we integrated the uxof F around its boundary.

While we have a heuristic argument that ux should equal the (double) integral of divergence, this rule is based on physical intuition. Could it be that vector elds whicharise in nature behave more nicely than arbitrary vector elds 1, or does our equation forux equal the integral of our equation for divergence even when the vector elds do notarise in nature? This is our topic for the next section.

1In fact, in the 1800s, the fundamental forces of nature were believed to satisfy Laplaces equation,2 f x 2

2 f y 2

0

in the two-dimensional case, which would make them very special indeed. However, Einsteins general theoryof relativity, published in 1915, showed that nature is more complicated than this.

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ANSWERS TO SELECTED EXERCISES, SECTION 4 .31. 2

3. 2 sin

xsin

y5. 0

7.2

0

cos 2 t sin2

t dt 2 .

9.2

0

3 cos2

t sin t sin3

t dt 0 .

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4 2 Flux and Div