Consider the boundary of some three dimensional region such that a function f is defined on this boundary.
Imagine taking the value of this function at a point, multiplying this value by the area of an infinitesimal
chunk of area located at this point and then adding these up. This is just the notion of the
integral presented earlier only now there is a difference because this infinitesimal chunk of
area should be considered as two dimensional even though it is in three dimensions. However,
it is not really all that different from what was done earlier. It all depends on the following
fundamental definition which is just a review of the fact presented earlier that the area of a
parallelogram determined by two vectors in ℝ^{3} is the norm of the cross product of the two
vectors.
Definition 16.1.1Let u_{1},u_{2}be vectors in ℝ^{3}. The 2 dimensional parallelogram determined by thesevectors will be denoted by P
(u1,u2 )
and it is defined as
( )
{ ∑2 }
P (u1,u2) ≡ ( sjuj : sj ∈ [0,1]) .
j=1
Then the area of this parallelogram is
area P (u1,u2) ≡ |u1 × u2|.
Suppose then that x = f
(u)
where u ∈ U, a subset of ℝ^{2} and x is a point in V , a subset of 3
dimensional space. Thus, letting the Cartesian coordinates of x be given by x =
(x1,x2,x3)
^{T}, each x_{i}
being a function of u, an infinitesimal rectangle located at u_{0} corresponds to an infinitesimal parallelogram
located at f
(u0)
which is determined by the 2 vectors
{ }
∂f(∂uu0)dui
i
_{i=1}^{2}, each of which is tangent to the
surface defined by x = f
(u)
. (No sum on the repeated index.)
PICT
From Definition (16.1.1), the two dimensional volume of this infinitesimal parallelepiped located at
f
It might help to think of a lizard. The infinitesimal parallelepiped is like a very small scale on a lizard.
This is the essence of the idea. To define the area of the lizard sum up areas of individual
scales^{1} .
If the scales are small enough, their sum would serve as a good approximation to the area of the
lizard.
PICT
This motivates the following fundamental procedure which I hope is extremely familiar from the earlier
material.
Procedure 16.1.2Suppose U is a subset of ℝ^{2}and suppose f : U → f
(U )
⊆ ℝ^{3}is a one to one and C^{1}function. Then if h : f
(U)
→ ℝ, define the 2 dimensional surface integral∫_{f(U)
}h
(x)
dA according to thefollowing formula.
∫ ∫
h (x)dA ≡ h (f (u ))|fu1 (u )× fu2 (u)|du1du2.
f(U) U
Definition 16.1.3It is customary to write
|fu (u)× fu (u)|
1 2
=
∂(x1,x2,x3)
∂(u1,u2)
because this newnotation generalizes to far more general situations for which the cross product is not defined. Forexample, one can consider three dimensional surfaces in ℝ^{8}.
Example 16.1.4Find the area of the region labeled A in the following picture. The two circles areof radius 1, one has center
(0,0)
and the other has center
(1,0)
.
PICT
The circles bounding these disks are x^{2} + y^{2} = 1 and
(x− 1)
^{2} + y^{2} = x^{2} + y^{2}− 2x + 1 = 1. Therefore,
in polar coordinates these are of the form r = 1 and r = 2cosθ.
The set A corresponds to the set U, in the
(θ,r)
plane determined by θ ∈
[− π3, π3]
and for each value
of θ in this interval, r goes from 1 up to 2cosθ. Therefore, the area of this region is of the
form,
which can be obtained by using the trig. substitution, 2x = tanθ on the inside integral.
Note this all depends on being able to write the surface in the form, x = f
(u)
for u ∈ U ⊆ ℝ^{p}. Surfaces
obtained in this form are called parametrically defined surfaces. These are best but sometimes you have
some other description of a surface and in these cases things can get pretty intractable. For example, you
might have a level surface of the form 3x^{2} + 4y^{4} + z^{6} = 10. In this case, you could solve for z using methods
of algebra. Thus z =
∘610-−-3x2-− 4y4
and a parametric description of part of this level surface is
( 6∘ ------2----4)
x,y, 10 − 3x − 4y
for
(x,y)
∈ U where U =
{ 2 4 }
(x,y) : 3x + 4y ≤ 10
. But what if the level surface
was something like
( 2 ( 2 )) z
sin x + ln 7 +y sinx +sin(zx)e = 11sin(xyz)?
I really do not see how to use methods of algebra to solve for some variable in terms of the others. It isn’t
even clear to me whether there are any points
(x,y,z)
∈ ℝ^{3} satisfying this particular relation. However, if a
point satisfying this relation can be identified, the implicit function theorem from advanced calculus can
usually be used to assert one of the variables is a function of the others, proving the existence of a
parametrization at least locally. The problem is, this theorem does not give the answer in terms of known
functions so this is not much help. Finding a parametric description of a surface is a hard problem and
there are no easy answers. This is a good example which illustrates the gulf between theory and
practice.
Example 16.1.6Let U =
[0,12]
×
[0,2π]
and let f : U → ℝ^{3}be given by f
(t,s)
≡
(2 cost +coss,2sint+ sin s,t)
^{T}. Find a double integral for the surface area. A graph of this surfaceis drawn below.
PICT
Then
( ) ( )
| − 2 sint | | − sins |
ft = ( 2cost ) ,fs = ( coss )
1 0
and
( )
− coss
ft × fs = |( − sin s |)
− 2sin tcoss + 2costsins
If you really needed to find the number this equals, how would you go about finding it? This is an
interesting question and there is no single right answer. You should think about this. Here is an example for
which you will be able to find the integrals.
Example 16.1.7Let U =
[0,2π]
×
[0,2π]
and for
(t,s)
∈ U, let
T
f (t,s) = (2cost+ costcoss,− 2sint− sin tcoss,sins) .
Find the area of f
(U )
. This is the surface of a donut shown below. The fancy name for this shape is atorus.