3Integration in ℝ2 and ℝ2
IA Vector Calculus
3.1 Integrals over subsets of R
2
Definition
(Surface integral)
.
Let
D ⊆ R
2
. Let
r
= (
x, y
) be in Cartesian
coordinates. We can approximate
D
by
N
disjoint subsets of simple shapes, e.g.
triangles, parallelograms. These shapes are labelled by I and have areas δA
i
.
x
y
D
To integrate a function
f
over
D
, we would like to take the sum
P
f
(
r
i
)
δA
i
,
and take the limit as
δA
i
→
0. But we need a condition stronger than simply
δA
i
→
0. We won’t want the areas to grow into arbitrarily long yet thin strips
whose area decreases to 0. So we say that we find an
`
such that each area can
be contained in a disc of diameter `.
Then we take the limit as
` →
0,
N → ∞
, and the union of the pieces tends
to D. For a function f(r), we define the surface integral as
Z
D
f(r) dA = lim
`→0
X
I
f(r
i
)δA
i
.
where
r
i
is some point within each subset
A
i
. The integral exists if the limit
is well-defined (i.e. the same regardless of what
A
i
and
r
i
we choose before we
take the limit) and exists.
If we take f = 1, then the surface integral is the area of D.
On the other hand, if we put
z
=
f
(
x, y
) and plot out the surface
z
=
f
(
x, y
),
then the area integral is the volume under the surface.
The definition allows us to take the
δA
i
to be any weird shape we want.
However, the sensible thing is clearly to take A
i
to be rectangles.
We choose the small sets in the definition to be rectangles, each of size
δA
I
=
δxδy
. We sum over subsets in a narrow horizontal strip of height
δy
with
y
and
δy
held constant. Take the limit as
δx →
0. We get a contribution
δy
R
x
y
f(y, x) dx with range x
y
∈ {x : (x, y) ∈ D}.
x
y
δy
y
x
y
Y
D
We sum over all such strips and take δy → 0, giving
Proposition.
Z
D
f(x, y) dA =
Z
Y
Z
x
y
f(x, y) dx
!
dy.
with x
y
ranging over {x : (x, y) ∈ D}.
Note that the range of the inner integral is given by a set
x
y
. This can be an
interval, or many disconnected intervals, x
y
= [a
1
, b
1
] ∪ [a
2
, b
2
]. In this case,
Z
x
y
f(x) dx =
Z
b
1
a
1
f(x) dx +
Z
b
2
a
2
f(x) dx.
This is useful if we want to integrate over a concave area and we have disconnected
vertical strips.
x
y
We could also do it the other way round, integrating over
y
first, and come up
with the result
Z
D
f(x, y) dA =
Z
X
Z
y
x
f(x, y) dy
dx.
Theorem
(Fubini’s theorem)
.
If
f
is a continuous function and
D
is a compact
(i.e. closed and bounded) subset of R
2
, then
ZZ
f dx dy =
ZZ
f dy dx.
While we have rather strict conditions for this theorem, it actually holds in many
more cases, but those situations have to be checked manually.
Definition (Area element). The area element is dA.
Proposition. dA = dx dy in Cartesian coordinates.
Example.
We integrate over the triangle bounded by (0
,
0)
,
(2
,
0) and (0
,
1).
We want to integrate the function f(x, y) = x
2
y over the area. So
Z
D
f(xy) dA =
Z
1
0
Z
2−2y
0
x
2
y dx
dy
=
Z
1
0
y
x
3
3
2−2y
0
dy
=
8
3
Z
1
0
y(1 − y)
3
dy
=
2
15
We can integrate it the other way round:
Z
D
x
2
y dA =
Z
2
0
Z
1−x/2
0
x
2
y dy dx
=
Z
2
0
x
2
1
2
y
2
1−x/2
0
dx
=
Z
2
0
x
2
2
1 −
x
2
2
dx
=
2
15
Since it doesn’t matter whether we integrate
x
first or
y
first, if we find it
difficult to integrate one way, we can try doing it the other way and see if it is
easier.
While this integral is tedious in general, there is a special case where it is
substantially easier.
Definition
(Separable function)
.
A function
f
(
x, y
) is separable if it can be
written as f (x, y) = h(y)g(x).
Proposition.
Take separable
f
(
x, y
) =
h
(
y
)
g
(
x
) and
D
be a rectangle
{
(
x, y
) :
a ≤ x ≤ b, c ≤ y ≤ d}. Then
Z
D
f(x, y) dx dy =
Z
b
a
g(x) dx
!
Z
d
c
h(y) dy
!