SURFACE INTEGRALS AND FLUX INTEGRALS

PORAMATE (TOM) PRANAYANUNTANA

Imagine water flowing through a fishing net stretched across a stream. Suppose we wantto measure the flow rate of water through the net, that is, the volume of fluid that passesthrough the surface per unit time. (See Figure 1.) This flow rate is called the flux of thefluid through the surface. We can also compute the flux of vector fields, such as electric andmagnetic fields, where no flow is actually taking place.

Figure 1. Flux measures rate of flow through a surface.

The Flux of a Constant Vector Field Through a Flat Surface If ~v is the velocityvector of a constant fluid flow and ~AS is the area vector of a flat surface S, then the totalflow through the surface in units of volume per unit time is called the flux of ~v through thesurface S and is given by

Flux =(~v � ~AS

).(1)

See Figure 2. Suppose when t = 0 seconds, the front part of the fluid was at the bottom ofthe skewed box, and when t = 1 second, that front part of the fluid has moved to the top ofthe skewed box. Therefore from t = 0 seconds to t = 1 second, the volume of the fluid thathas flowed through S in one unit time (one second) equals the volume of the skewed box inFigure 2. That is

Flux = flow rate = total amount of fluid that has flowed through S in 1 second(2)

=∥∥∥ ~AS

∥∥∥︸ ︷︷ ︸Base Area

· ‖~v‖ cos θ︸ ︷︷ ︸height

=(~v � ~AS

)Date: June 24, 2015.

Surface Integrals and Flux Integrals Poramate (Tom) Pranayanuntana

Figure 2. Flux of ~v through a flat surface with area vector ~AS is the volumeof this skewed box.

Figure 3. Surface S di-vided into small, almost flatpieces, showing a typical ori-entation vector n̂ and areavector ∆ ~AS

Figure 4. Flux of a vectorfield through a curved sur-face S.

The Flux Integral To calculate the flux of a vector field ~F which is not necessarily con-stant through a curved, oriented surface S, we divide the surface into a patchwork of small,approximately flat pieces (like a wire-frame representation of the surface) as shown in Figure3. Suppose a particular patch has area ∆AS. We pick an orientation vector n̂S at a point onthe patch and define the area vector of the patch, ∆ ~AS, as ∆ ~AS = n̂S∆AS. If the patches aresmall enough, we can assume that ~F is approximately constant on each piece. (See Figure4.) Then we know that

Flux through patch ≈(~F � ∆ ~AS

),

so, adding the fluxes through all the small pieces, we have

Flux through whole surface ≈∑(

~F � ∆ ~AS

).

As each patch becomes smaller and∥∥∥∆ ~AS

∥∥∥→ 0, the approximation gets better and we get

Flux through S = lim‖∆ ~AS‖→0

∑(~F � ∆ ~AS

).

June 24, 2015 Page 2 of 4

Surface Integrals and Flux Integrals Poramate (Tom) Pranayanuntana

Thus, provided the limit exists, we define the following:

The flux integral of the vector field ~F through the oriented surface S is∫S

(~F � d ~AS

)= lim‖∆ ~AS‖→0

∑(~F � ∆ ~AS

). (3)

If S is a closed surface oriented outward, we describe the flux through S as the flux

out of S, and it is denoted by

∮S

(~F � d ~AS

)to emphasize that S is a closed surface.

Flux Integrals Over Parameterized Surfaces We now consider how to compute theflux of a smooth vector field ~F through a smooth oriented surface, S, parameterized by

~r = ~r(s, t) = ~f(s, t), for (s, t) in some region T of the parameter space. We write

S : ~r = ~r(s, t) = ~f(s, t), (s, t) ∈ T.We consider a parameter rectangle on the surface S corresponding to a rectangular regionwith sides ∆s and ∆t in the parameter region, T . (See Figure 5.)

Figure 5. Parameter rectangle on the surface S corresponding to a smallrectangular region in the parameter region, T , in the parameter space.

June 24, 2015 Page 3 of 4

Surface Integrals and Flux Integrals Poramate (Tom) Pranayanuntana

If ∆s and ∆t are small, the area vector, ∆ ~AS, of the patch is approximately the area vectorof the parallelogram defined by the vectors

~r(s+ ∆s, t)− ~r(s, t)︸ ︷︷ ︸secant vector displaced from one point

to another point on surface S : ~r = ~f

corresponding to moving from (s, t)

to (s + ∆s, t) on parameter region T

≈ ∂~r

∂s∆s︸ ︷︷ ︸

tangent vector∂~r

∂son tangent plane:

~r = ~L, multiplied by the run ∆s

, and

~r(s, t+ ∆t)− ~r(s, t)︸ ︷︷ ︸secant vector displaced from one point

to another point on surface S : ~r = ~f

corresponding to moving from (s, t)

to (s, t + ∆t) on parameter region T

≈ ∂~r

∂t∆t︸ ︷︷ ︸

tangent vector∂~r

∂ton tangent plane:

~r = ~L, multiplied by the run ∆t

.

Thus

∆ ~AS ≈∂~r

∂s∆s× ∂~r

∂t∆t =

(∂~r

∂s× ∂~r

∂t

)∆s∆t.

From the reasoning above, we assume that the vector ~rs × ~rt is never zero and points in thedirection of the unit normal orientation vector n̂S. If the vector ~rs×~rt points in the oppositedirection of n̂S, we reverse the order of the cross-product. Replacing ∆ ~AS, ∆s, and ∆t byd ~AS, ds, and dt, we write

d ~AS =∂~r

∂sds× ∂~r

∂tdt =

(∂~r

∂s× ∂~r

∂t

)dsdt.

The Flux of a Vector Field through a Parameterized Surface The flux

of a smooth vector field ~F through a smooth oriented surface S parameterized

by ~r = ~r(s, t) = ~f(s, t), where (s, t) varies in a parameter region T , is given by∫S:~r(s,t),(s,t)∈T

(~F � d ~AS

)=

∫T

(~F (~r(s, t)) � (~rs × ~rt)

)dsdt︸︷︷︸dAT

. (4)

We choose the parameterization so that ~rs × ~rt is never zero and points in thedirection of n̂S everywhere.

June 24, 2015 Page 4 of 4