The Flux Integral   

Let's suppose S is an oriented surface and that n denotes its unit normal. If S is placed within a vector field F = á M,N,P ñ , then we can imagine that the flow of the vectorfield is ''through'' the surface (but that the surface does not impede the flow, like light ''flowing'' through a clear sheet of plastic).

LiveGraphics3d Applet

Since F( x,y,z) is the velocity of the flow at each point point, the velocity of the flow in the direction of n is

projn( F) =  F·n n·n   n = ( F·n)   n

And since n is a unit vector, the speed of the flow in the direction of n is simply F·n, and it follows that during a short time Dt, the displacement in the direction of n is given by the product of the speed and Dt.  The volume of the flow through a patch with area DS_ij over a short period of time Dt will thus be practically the same as the volume of a small parallelpiped with height ( F·n)dt and base DS_ij. Thus, the volume of the flow through the patch over a short time period Dt is given by DV_ij = ( F·n) DtDS_ij, implying that the rate of change of the volume of that small box is

DVij Dt = ( F·n)   DSij

If DV denotes the total volume of the flow through the surface during a short time Dt, then DV/Dt is obtained by adding up the volumes of the individual patches and then letting the areas of the patches approach 0:

DV Dt = lim h®0  n å j = 1  m å k = 1    ( F·n)   DSij

(1)

The rate of change of the total volume of the flow through a surface is known as the Flux of the vectorfield through the surface. Equation (1) shows us that the flux is a surface integral of the form

Flux =  ó õ ó õ S    F·n  dS

(2)

where dS is the surface area differential. That is, the flux is the rate at which the vector field's flow passes through the surface.

For example, suppose S is the patch on the unit sphere corresponding to f in [ a, b] and q in [c, d] .  The normal to the unit sphere is the same as the position vector, which means that n = á x,y,z ñ at the point ( x,y,z) on the unit sphere. Since dS = sin( f)dfdq (see example 1), the flux integral through a patch on the unit sphere is

Flux = ó õ d c  ó õ b a  F· áx,y,z ñ   sin( f) dfdq

(3)

Also, x = sin( f) cos( q) , y = sin( f) sin( q) , and z = cos( f) on the unit sphere.      

EXAMPLE 2    Compute the flux of the vectorfield F = á y,-x,z ñ through the unit sphere in spherical coordinates.

LiveGraphics3d Applet

Solution: Equation (3) with q in [0,2p] and f in [ 0,p] implies that

Flux = ó õ 2p 0  ó õ p 0  á y,-x,z ñ· á x,y,z ñ   sin( f) dfdq = ó õ 2p 0&