Surface Integrals   (continued)

 5 Note:  Domain D shown in the previous figure is the “area” projected on to the relevant plane.  In this case it is projected on to the x-y plane since the surface, S, was described by   z =  g(x,y).    If, on the other hand,  the surface was described by  y  =  g(x,z), then D would be the “area” projected on to the x-z plane, etc.   You have at least two options in evaluating surface integrals.   1.      You could use a transformation     dS  =  |  ru x rv  |  dA                           IS  =  ∫   ∫  f(x,y,z)  dS  =   ∫   ∫  f(x,y,z)  |  ru x rv  |  dA                                    S                              D   2.      You could attempt direct evaluation of the surface integral                          IS  =  ∫   ∫  f(x,y,z)  dS   When the surface, S, is described as follows:                     z  = g(x,y)              (recall let   u  =  x  and  v  =  y)            the position vector to an arbitrary point on S  is      r  =  xi  +  yj  +  g(x,y)k      Then one can use x and y rather than u and v as the parameters describing the surface, S.  So the normal vector to  S is   N(u,v)  =  N(x,y)  =   rx  x  ry     which yields the following 3 x 3 determinant.  Note:  N  is not a unit vector.                                    i             j           k                      i            j           k                N(x,y)  =    det     ∂x/∂x    ∂y/∂x    ∂g/∂x   =    det    1           0      ∂g/∂x                                     ∂x/∂y    ∂y/∂y    ∂g/∂y                  0           1      ∂g/∂y     N(x,y)  =  - ∂g/∂x i   -  ∂g/∂y j  + 1  k     and     | N(x,y) |  = √[1 + (∂g/∂x)2 + (∂g/∂x)2]     Thus      dS  =  √[1 + (∂g/∂x)2 + (∂g/∂x)2] dx dy        and                  IS  =  ∫ ∫ f(x, y, z) dS  =  ∫ ∫  f(x, y, g(x,y)) √[1 + (∂g/∂x)2 + (∂g/∂x)2] dx dy                        S                            D