What do we mean by *Flux*? In the following we look at its
meaning. Recall that we said that

Let's consider an example to try and give this a slightly more precise
physical meaning. Suppose that we have a vector field,
**F** = -*y* **i** + *x* **j**,
which is the velocity field of a fluid with density 1. This means
that **F** has units of velocity, say m/s, and the density is in
mass/volume, say kg/m^{3}. Now, if we
ask for "the amount of fluid passing through the surface," we could
mean "the mass passing through per time unit," that is, mass/time, say,
kg/s.

So, how do we get this? We get mass from density times volume. The
volume will be the velocity times the surface area of the surface, which
gives the volume of fluid coming through the surface per second. To see
this, suppose that the velocity field were constant: say,
**F** = 2 **j**, a
constant flow in the j direction. Then think about a 1x1 square on the
*xz*-plane: in 1 second the fluid particles that are on the 1x1 square will
have moved 2m to the right, forming the outer edge of the volume of fluid
that has come through the surface. This is shown in the figure to the
right. The volume that has passed through the surface is therefore
2x1x1 = 2 m^{3}.

So, the volume of fluid passing through the surface is (velocity
perpendicular to the surface)(surface area). Thus the flux through a
little patch of surface area, *dS*, is
.
Here, **N** is a unit normal, which we've seen we can find in a
couple of different ways. If the surface is given parametrically as
**r**(*u*,*v*), then
.
Therefore the differential mass/second passing through the small patch
of surface given by *dS* is
.

Of course, for the case in figure 1 it's even easier than that:
the normal vector
is **N** = **j**, and because *dS* is just an area in
the *xz*-plane, *dS* = *dx* *dz*.

This gives the *surface integral* of the *vector field*
over the surface. We define
so that we can write the flux as
the first expression in the definition in the first equation on the
page. This is directly analagous to what we did with line integrals,
saying that

**Different Ways of Representing Surfaces**

How we approach finding **N** largely depends on how the surface
*S* is defined. We have two common ways this is done:

- As a function:
*z*=*f*(*x*,*y*). In this case we can find a normal by re-writing the surface as the zero*level surface*of the function*G*(*x*,*y*,*z*) =*z*-*f*(*x*,*y*) and remembering that the gradient of a level curve or surface is a vector perpendicular to the curve or surface: thus**N**= <-*f*, -_{x}*f*, 1> / sqrt(1 +_{y}*f*_{x}^{2}+*f*_{y}^{2}), and we've derived the expression for*dS*= sqrt(1 +*f*_{x}^{2}+*f*_{y}^{2})*dx**dy*, so*d***S**= <-*f*, -_{x}*f*, 1>_{y}*dx**dy*. - Parametrically:
**r**(*u*,*v*) =*x*(*u*,*v*)**i**+*y*(*u*,*v*)**j**+*z*(*u*,*v*)**k**. In this case, we use our parametric derivation of**N**and*dS*, and find*d***S**= (**r**_{u}x**r**_{v})*du**dv*.

the meaning of flux

Last modified: Fri Apr 9 15:10:29 EDT 2004

Comments to:glarose@umich.edu

©2004 Gavin LaRose, UM Math Dept.