Flow Across a Surface (Flux)

Consider a vector field v(x,y,z)=v1(x,y,z) i + v2(x,y,z) j + v3(x,y,z) k.

Think of a small (infinitesimal) piece of surface. The flow of v across the small piece of surface is in the direction of the unit normal to the surface and is v times the small piece of surface area (the bigger the piece of surface, the more flow!). Adding these infinitesimal pieces up (surface integrating!) yields flow across the surface:

Surface Integral of v.n dA

Recall that the normal vector at a point on a level curve was the gradient vector at that point.

The gradient of a "level surface" will give us the normal now.

Consider g(x,y,z)=z-f(x,y). g=c gives "level surfaces". All we care about is g=0.

The gradient will still give the normal:

The gradient of g(x,y,z) is
-Dxf i -Dyf j + 1 k
or
{-Dxf,-Dyf,1}.

The unit normal is given by
{-Dxf,-Dyf,1}/|| {-Dxf,-Dyf,1} ||

Example

Find the flux for the vector field v(x,y,z)=xi+yj+zk out of the surface
z=1-x^2-y^2, z>=0.

In[50]:=

  v[x_,y_,z_]:={x,y,z}
  f[x_,y_]:=1-x^2-y^2
  g[x_,y_,z_]:=z-(1-x^2-y^2)

In[51]:=

  grad3[g][x,y,z]

Out[51]=

  {2 x, 2 y, 1}

In[52]:=

  n=grad3[g][x,y,z]/norm3[grad3[g][x,y,z]]

Out[52]=

            2 x                    2 y
  {---------------------, ---------------------, 
               2      2               2      2
   Sqrt[1 + 4 x  + 4 y ]  Sqrt[1 + 4 x  + 4 y ]
   
              1
    ---------------------}
                2      2
    Sqrt[1 + 4 x  + 4 y ]

In[53]:=

  Simplify[v[x,y,f[x,y]].n]

Out[53]=

            2    2
       1 + x  + y
  ---------------------
              2      2
  Sqrt[1 + 4 x  + 4 y ]

In[54]:=

  Dx=D[f[x,y],x]
  Dy=D[f[x,y],y]

Out[54]=

  -2 x

Out[55]=

  -2 y

In[56]:=

  dA=Sqrt[1+Dx^2+Dy^2]

Out[56]=

              2      2
  Sqrt[1 + 4 x  + 4 y ]

In[57]:=

  Simplify[(v[x,y,f[x,y]].n)*dA]

Out[57]=

       2    2
  1 + x  + y

In[58]:=

  Integrate[1+x^2+y^2,{x,-1,1},
    {y,-Sqrt[1-x^2],Sqrt[1-x^2]}]

Out[58]=

  3 Pi
  ----
   2

Something you might have noticed while doing the example was the cancellation of the dA with ||normal vector||.

The unit normal is given by
{-Dxf,-Dyf,1}/|| {-Dxf,-Dyf,1} ||,
while dA is
dA=Sqrt[1+Dxf^2+Dyf^2] dydx
Notice that dA=|| {-Dxf,-Dyf,1} || dydx

So, surface integral v.n dA=double integral v.{-Dxf,-Dyf,1} dydx
or equivalently:

The surface integral v.n dA = double integral (-v1Dxf-v2Dyf+v3) dydx

Up to Some Multivariable Calculus Ideas Part II