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 26.1.1Let u_{1},u_{2}be vectors in ℝ^{3}. The 2 dimensionalparallelogram determined by these vectors will be denoted by P
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 26.1.2Suppose 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 surfaceintegral∫_{f(U)
}h
(x)
dA according to the following formula.
∫ ∫
h (x)dA ≡ h (f (u))|fu1 (u)× fu2 (u )|du1du2.
f(U) U
Definition 26.1.3It is customary to write
|fu1 (u)× fu2 (u)|
=
∂(∂x1(,u1x2,,ux2)3)
because this new notation generalizes to far more general situations for which thecross product is not defined. For example, one can consider three dimensionalsurfaces in ℝ^{8}.
Example 26.1.4Find the area of the region labeled A in the following picture.The two circles are of 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 ∈
[ π π]
− 3,3
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 =
∘ -------------
6 10− 3x2 − 4y4
and a parametric description
of part of this level surface is
( )
x,y, 6∘10-−-3x2 −-4y4
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 = 11 sin(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 26.1.6Let U =
[0,12]
×
[0,2π]
and let f : U → ℝ^{3}be given byf
(t,s)
≡
(2cost+ coss,2 sint+ sins,t)
^{T}. Find a double integral for the surfacearea. A graph of this surface is drawn below.
PICT
Then
( ) ( )
− 2sint − sins
ft = ( 2cost ) ,fs = ( coss )
1 0
and
( )
− coss
ft × fs = ( − sin s )
− 2sin tcos s+ 2costsin s
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 26.1.7Let 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 namefor this shape is a torus.
PICT
To find its area,
( − 2sin t− sintcoss ) ( − costsins )
f= ( − 2 cost− cos tcoss ) ,f = ( sintsins )
t 0 s coss
and so
|ft × fs|
=
(coss + 2)
so the area element is
(coss+ 2)
dsdt and the area
is
∫ ∫
2π 2π 2
0 0 (coss+ 2)dsdt = 8π
Example 26.1.8Let U =
[0,2π]
×
[0,2π]
and for
(t,s)
∈ U, let
T
f (t,s) = (2cost+ costcoss,− 2sint− sin tcoss,sins) .
Find∫_{f}
(U)
hdV where h
(x,y,z)
= x^{2}.
Everything is the same as the preceding example except this time it is an integral of
a function. The area element is