Example 1

In[44]:=

  Clear[F,S,g,uN,v1,v2,v3]
  F[x_,y_,z_]:={x^2,y x,z x}
  v1={1,0,0};
  v2={0,2,0};
  v3={0,0,3};

To get a normal vector we need two vectors on the surface. They are easy enough to get here: v2-v1 and v3-v1.

In[45]:=

  vp=vectorPlot[{v1,v1},{v2-v1,v3-v1}];

This is probably a more common view.

In[46]:=

  Show[vp,AxesLabel->{"x","y","z"}, ViewPoint->{3,2,1}];

In[47]:=

  uN=Cross[v2-v1,v3-v1]/norm[Cross[v2-v1,v3-v1]]

Out[47]=

   6  3  2
  {-, -, -}
   7  7  7

We need the surface z=S(x,y).

Recall that uN={6/7, 3/7, 2/7}.
So,the plane is 6/7(x-1)+3/7 y +2/7 z=0
OR
z=3-3x-3/2 y

In[48]:=

  S[x_,y_]:=3-3 x-3/2 y
  g[x_,y_,z_]:=S[x,y]-z

This is only if you want to see a picture of the surface!

This is only if you want to see ...

We now need the region to integrate. Projecting onto the x y plane we get the region enclosed by x=0, y=0, and the line y=-2(x-1):

In[52]:=

  dA=norm[grad3[g][x,y,z]]

Out[52]=

  7
  -
  2

In[53]:=

  integrand=(F[x,y,S[x,y]].uN)dA

Out[53]=

                           3 y
        2   2 x (3 - 3 x - ---)
     6 x                    2     3 x y
  7 (---- + ------------------- + -----)
      7              7              7
  --------------------------------------
                    2

In[54]:=

  Simplify[integrand]

Out[54]=

  3 x

In[55]:=

  Plot[-2(x-1),{x,0,1},PlotStyle->{RGBColor[1,0,0]},AxesLabel->{"x","y"}];

In[56]:=

  Integrate[integrand,{x,0,1},{y,0,-2(x-1)}]

Out[56]=

  1

Up to Surface Integrals