The special case where a surface is in the form z = f
(x,y)
,
(x,y)
∈ U, yields a simple formula
which is used most often in this situation. You write the surface parametrically in the form
f
(x,y)
=
(x,y,f (x,y))
^{T} such that
(x,y)
∈ U. Then
( ) ( )
| 1 | | 0 |
fx = ( 0 ) ,fy = ( 1 )
fx fy
and
∘ ----------
|fx × fy| = 1+ f2y + fx2
so the area element is
∘ ----2----2
1+ fy + fxdxdy.
When the surface of interest comes in this simple form, people generally use this area element directly
rather than worrying about a parametrization and taking cross products.
In the case where the surface is of the form x = f
(y,z)
for
(y,z)
∈ U, the area element is
obtained similarly and is
∘ ----2---2-
1+ fy + fz
dydz. I think you can guess what the area element is if
y = f
(x,z)
.
There is also a simple geometric description of these area elements. Consider the surface z = f
(x,y)
.
This is a level surface of the function of three variables z − f
(x,y)
. In fact the surface is simply
z −f
(x,y)
= 0. Now consider the gradient of this function of three variables. The gradient is perpendicular
to the surface and the third component is positive in this case. This gradient is
(− fx,− fy,1)
and so the unit upward normal is just
√1+f12x+f2y-
(− fx,− fy,1)
. Now consider the following
picture.
PICT
In this picture, you are looking at a chunk of area on the surface seen on edge and so it seems
reasonable to expect to have dxdy = dV cosθ. But it is easy to find cosθ from the picture and the
properties of the dot product.
n ⋅k 1
cosθ = |n||k-| = ∘---2----2.
1+ fx + fy
Therefore, dA =
∘ ----------
1 + fx2+ f2y
dxdy as claimed.
Example 16.1.9Let z =
∘ -------
x2 + y2
where
(x,y)
∈ U for U =
{ 2 2 }
(x,y) : x + y ≤ 4
Find∫_{S}hdS where h
(x,y,z)
= x + z and S is the surface described as
( ∘ ------)
x,y, x2 + y2
for
(x,y)
∈ U.
Here you can see directly the angle in the above picture is
π
4
and so dV =
√ -
2
dxdy. If you do not see
this or if it is unclear, simply compute
Admittedly, the set C gets added in twice but this does not matter because its 2 dimensional volume equals
zero and therefore, the integrals over this set will also be zero.
I have been purposely vague about precise mathematical conditions necessary for the above procedures.
This is because the precise mathematical conditions which are usually cited are very technical and
at the same time far too restrictive. The most general conditions under which these sorts of
procedures are valid include things like Lipschitz functions defined on very general sets. These are
functions satisfying a Lipschitz condition of the form
|f (x)− f (y)|
≤ K
|x− y |
. For example,
y =
|x|
is Lipschitz continuous. This function does not have a derivative at every point. So it is
with Lipschitz functions. However, it turns out these functions have derivatives at enough
points to push everything through but this requires considerations involving the Lebesgue
integral. Lipschitz functions are also not the most general kind of function for which the above is
valid.