7Integral theorems
IA Vector Calculus
7.2 Relating and proving integral theorems
We will first show the following two equivalences:
– Stokes’ theorem ⇔ Green’s theorem
– 2D divergence theorem ⇔ Greens’ theorem
Then we prove the 2D version of divergence theorem directly to show that all of
the above hold. A sketch of the proof of the 3D version of divergence theorem
will be provided, because it is simply a generalization of the 2D version, except
that the extra dimension makes the notation tedious and difficult to follow.
Proposition. Stokes’ theorem ⇒ Green’s theorem
Proof.
Stokes’ theorem talks about 3D surfaces and Green’s theorem is about
2D regions. So given a region
A
on the (
x, y
) plane, we pretend that there is a
third dimension and apply Stokes’ theorem to derive Green’s theorem.
Let
A
be a region in the (
x, y
) plane with boundary
C
=
∂A
, parametrised
by arc length, (x(s), y(s), 0). Then the tangent to C is
t =
dx
ds
,
dy
ds
, 0
.
Given any P(x, y) and Q(x, y), we can consider the vector field
F = (P, Q, 0),
So
∇ × F =
0, 0,
∂Q
∂x
−
∂P
∂y
.
Then the left hand side of Stokes is
Z
C
F · dr =
Z
C
F · t ds =
Z
C
P dx + Q dy,
and the right hand side is
Z
A
(∇ × F) ·
ˆ
k dA =
Z
A
∂Q
∂x
−
∂P
∂y
dA.
Proposition. Green’s theorem ⇒ Stokes’ theorem.
Proof.
Green’s theorem describes a 2D region, while Stokes’ theorem describes
a 3D surface
r
(
u, v
). Hence to use Green’s to derive Stokes’ we need find some
2D thing to act on. The natural choice is the parameter space, u, v.
Consider a parametrised surface
S
=
r
(
u, v
) corresponding to the region
A
in
the
u, v
plane. Write the boundary as
∂A
= (
u
(
t
)
, v
(
t
)). Then
∂S
=
r
(
u
(
t
)
, v
(
t
)).
We want to prove
Z
∂S
F · dr =
Z
S
(∇ × F) · dS
given
Z
∂A
F
u
du + F
v
dv =
Z
A
∂F
v
∂u
−
∂F
u
∂v
dA.
Doing some pattern-matching, we want
F · dr = F
u
du + F
v
dv
for some F
u
and F
v
.
By the chain rule, we know that
dr =
∂r
∂u
du +
∂r
∂v
dv.
So we choose
F
u
= F ·
∂r
∂u
, F
v
= F ·
∂r
∂v
.
This choice matches the left hand sides of the two equations.
To match the right, recall that
(∇ × F) · dS = (∇ × F) ·
∂r
∂u
×
∂r
∂v
du dv.
Therefore, for the right hand sides to match, we want
∂F
v
∂u
−
∂F
u
∂v
= (∇ × F) ·
∂r
∂u
×
∂r
∂v
. (∗)
Fortunately, this is true. Unfortunately, the proof involves complicated suffix
notation and summation convention:
∂F
v
∂u
=
∂
∂u
F ·
∂r
∂v
=
∂
∂u
F
i
∂x
i
∂v
=
∂F
i
∂x
j
∂x
j
∂u
∂x
i
∂v
+ F
i
∂x
i
∂u∂v
.
Similarly,
∂F
u
∂u
=
∂
∂u
F ·
∂r
∂u
=
∂
∂u
F
j
∂x
j
∂u
=
∂F
j
∂x
i
∂x
i
∂v
∂x
j
∂u
+ F
i
∂x
i
∂u∂v
.
So
∂F
v
∂u
−
∂F
u
∂v
=
∂x
j
∂u
∂x
i
∂v
∂F
i
∂x
j
−
∂F
j
∂x
i
.
This is the left hand side of (∗).
The right hand side of (∗) is
(∇ × F) ·
∂r
∂u
×
∂r
∂v
= ε
ijk
∂F
j
∂x
i
ε
kpq
∂x
p
∂u
∂x
q
∂v
= (δ
ip
δ
jq
− δ
iq
δ
jp
)
∂F
j
∂x
i
∂x
p
∂u
∂x
q
∂v
=
∂F
j
∂x
i
−
∂F
i
∂x
j
∂x
i
∂u
∂x
j
∂v
.
So they match. Therefore, given our choice of
F
u
and
F
v
, Green’s theorem
translates to Stokes’ theorem.
Proposition. Greens theorem ⇔ 2D divergence theorem.
Proof. The 2D divergence theorem states that
Z
A
(∇ · G) dA =
Z
∂A
G · n ds.
with an outward normal n.
Write G as (Q, −P ). Then
∇ · G =
∂Q
∂x
−
∂P
∂y
.
Around the curve
r
(
s
) = (
x
(
s
)
, y
(
s
)),
t
(
s
) = (
x
0
(
s
)
, y
0
(
s
)). Then the normal,
being tangent to
t
, is
n
(
s
) = (
y
0
(
s
)
, −x
0
(
s
)) (check that it points outwards!). So
G · n = P
dx
ds
+ Q
dy
ds
.
Then we can expand out the integrals to obtain
Z
C
G · n ds =
Z
C
P dx + Q dy,
and
Z
A
(∇ · G) dA =
Z
A
∂Q
∂x
−
∂P
∂y
dA.
Now 2D version of Gauss’ theorem says the two LHS are the equal, and Green’s
theorem says the two RHS are equal. So the result follows.
Proposition. 2D divergence theorem.
Z
A
(∇ · G) dA =
Z
C=∂A
G · n ds.
Proof.
For the sake of simplicity, we assume that
G
only has a vertical component,
noting that the same proof works for purely horizontal
G
, and an arbitrary
G
is
just a linear combination of the two.
Furthermore, we assume that
A
is a simple, convex shape. A more complicated
shape can be cut into smaller simple regions, and we can apply the simple case
to each of the small regions.
Suppose G = G(x, y)
ˆ
j. Then
∇ · G =
∂G
∂y
.
Then
Z
A
∇ · G dA =
Z
X
Z
Y
x
∂G
∂y
dy
dx.
Now we divide
A
into an upper and lower part, with boundaries
C
+
=
y
+
(
x
)
and
C
−
=
y
−
(
x
) respectively. Since I cannot draw,
A
will be pictured as a circle,
but the proof is valid for any simple convex shape.
C
+
C
−
dy
Y
x
x
y
We see that the boundary of
Y
x
at any specific
x
is given by
y
−
(
x
) and
y
+
(
x
).
Hence by the Fundamental theorem of Calculus,
Z
Y
x
∂G
∂y
dy =
Z
y
+
(x)
y
−
(x)
∂G
∂y
dy = G(x, y
+
(x)) − G(x, y
−
(x)).
To compute the full area integral, we want to integrate over all
x
. However, the
divergence theorem talks in terms of d
s
, not d
x
. So we need to find some way
to relate d
s
and d
x
. If we move a distance
δs
, the change in
x
is
δs cos θ
, where
θ
is the angle between the tangent and the horizontal. But
θ
is also the angle
between the normal and the vertical. So cos θ = n ·
ˆ
j. Therefore dx =
ˆ
j · n ds.
In particular, G dx = G
ˆ
j · n ds = G · n ds, since G = G
ˆ
j.
However, at
C
−
,
n
points downwards, so
n ·
ˆ
j
happens to be negative. So,
actually, at C
−
, dx = −G · n ds.
Therefore, our full integral is
Z
A
∇ · G dA =
Z
X
Z
y
x
∂G
∂y
dY
dx
=
Z
X
G(x, y
+
(x)) − G(x, y
−
(x)) dx
=
Z
C
+
G · n ds +
Z
C
−
G · n ds
=
Z
C
G · n ds.
To prove the 3D version, we again consider
F
=
F
(
x, y, z
)
ˆ
k
, a purely vertical
vector field. Then
Z
V
∇ · F dV =
Z
D
Z
Z
xy
∂F
∂z
dz
!
dA.
Again, split
S
=
∂V
into the top and bottom parts
S
+
and
S
−
(ie the parts
with
ˆ
k · n ≥
0 and
ˆ
k · n <
0), and parametrize by
z
+
(
x, y
) and
z
−
(
x, y
). Then
the integral becomes
Z
V
∇ · F dV =
Z
D
(F (x, y, z
+
) − F (x, y, z
−
)) dA =
Z
S
F · n dS.