Given a surface in the explicit form z=z(x, y ),define the outward normal to the
surface φ(x, y, z ) = z−z(x, y ) = 0 by
eˆn=grad φ
|grad φ|=−∂z∂x ˆe 1 −∂y∂z ˆe 2 +ˆe 3
√
1 + ( ∂z∂x )^2 + (∂z∂y )^2. (7 .93)
From equations (7.92) and (7.93), obtain the vector element of surface area
dS=ˆendS =(
−∂z
∂xˆe 1 −∂z
∂yˆe 2 +eˆ 3)
dx dy.Taking the dot product of both sides of the above equation with the unit vector ˆe 3
gives
|eˆ 3 ·ˆen|dS =dx dy or dS = dx dy
|ˆe 3 ·ˆen|=dx dy
cos γ(7 .94)with an absolute value placed upon the dot product to ensure that the surface area
is positive (i.e., recall that there are two normals to the surface which differ in sign).
In equation (7.94), the element of surface area has been expressed in terms of its
projection onto the xy plane. The angle γ=γ(x, y )is the angle between the outward
normal to the surface and the unit vector ˆe 3 .This representation of the element of
surface area is valid provided that cos γ= 0; That is, it is assumed that the surface
is such that the normal to the surface is nowhere parallel to the xy plane.
We have previously shown that for surfaces which have a normal parallel to the
xy plane, the element of surface area can be projected onto either of the planes x= 0
or y= 0.If the surface element is projected onto the plane x= 0 , then the element
of surface area takes the form
dS = dy dz
|ˆe 1 ·ˆen|(7 .95)and if projected onto the plane y= 0 it has the form
dS = dx dz
|ˆe 2 ·ˆen|. (7 .96)
If the element of surface area dS is projected onto the z = 0 plane, the total
surface area is then
S=∫∫RdS =∫∫Rdx dy
|ˆe 3 ·ˆen|,