Example
In[76]:=
Clear[v,S,g,n]
v[x_,y_,z_]:={x,y,z}
S[x_,y_]:=Sqrt[4-x^2-y^2]
g[x_,y_,z_]:=x^2+y^2+z^2-4
In[77]:=
p1=Plot3D[S[x,y],{x,-2,2},{y,-2,2},DisplayFunction->Identity];
p2=Plot3D[1,{x,-2,2},{y,-2,2},DisplayFunction->Identity];
p3=Plot3D[2,{x,-2,2},{y,-2,2},DisplayFunction->Identity];
Show[p1,p2,p3,DisplayFunction->$DisplayFunction,ViewPoint->{0,1,0}];

In[78]:=
grad3[g][x,y,z]
Out[78]=
{2 x, 2 y, 2 z}
In[79]:=
n=grad3[g][x,y,z]/norm[grad3[g][x,y,z]]
Out[79]=
2 x 2 y
{------------------------, ------------------------,
2 2 2 2 2 2
Sqrt[4 x + 4 y + 4 z ] Sqrt[4 x + 4 y + 4 z ]
2 z
------------------------}
2 2 2
Sqrt[4 x + 4 y + 4 z ]
In[80]:=
Simplify[n]
Out[80]=
x y
{------------------, ------------------,
2 2 2 2 2 2
Sqrt[x + y + z ] Sqrt[x + y + z ]
z
------------------}
2 2 2
Sqrt[x + y + z ]
In[81]:=
Simplify[v[x,y,S[x,y]].n]
Out[81]=
2 2 2 2
x + y + Sqrt[4 - x - y ] z
-----------------------------
2 2 2
Sqrt[x + y + z ]
In[82]:=
Dx=D[S[x,y],x]
Dy=D[S[x,y],y]
Out[82]=
x
-(-----------------)
2 2
Sqrt[4 - x - y ]
Out[83]=
y
-(-----------------)
2 2
Sqrt[4 - x - y ]
In[84]:=
dS=Sqrt[1+Dx^2+Dy^2]
Out[84]=
2 2
x y
Sqrt[1 + ----------- + -----------]
2 2 2 2
4 - x - y 4 - x - y
In[85]:=
Simplify[(v[x,y,S[x,y]].n)*dS]
Out[85]=
1 2 2 2 2
2 Sqrt[-----------] (x + y + Sqrt[4 - x - y ] z)
2 2
4 - x - y
---------------------------------------------------
2 2 2
Sqrt[x + y + z ]
Before we go any further, let's replace z with S(x,y).
In[86]:=
integrand=(v[x,y,f[x,y]].n)*dS/.z->S[x,y];
Simplify[integrand]
Out[86]=
2 2
1 x y
2 Sqrt[-----------] (-- + -- +
2 2 2 2
4 - x - y
2 2 2 2
Sqrt[4 - x - y ] (-9 + x + y )
--------------------------------)
2
This is what we must integrate, but over what region? The plane z=2 intersects the sphere at the "top" (0,0,2), and z=1 intersects the sphere in the circle x^2+y^2=3. Projecting this over the x,y plane gives the region, x^2+y^2<=3
In[87]:=
Integrate[integrand,{x,-Sqrt[3],Sqrt[3]},
{y,-Sqrt[3-x^2],Sqrt[3-x^2]}]
Out[87]=
-115 Pi
-------
6
Note: If you were doing this by hand, polar coordinates would be the way to go.