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Vector Calculus Theorems
Disclaimer: These lecture notes are not meant to replace the course textbook. The
content may be incomplete. Some topics may be unclear. These notes are only meant to
be a study aid and a supplement to your own notes. Please report any inaccuracies to the
professor.
Gauss’ Theorem (Divergence Theorem)
Consider a surface S with volume V. If we divide it in half into two volumes V 1
and V
2
with surface areas S 1
and S
2
, we can write:
1 2
S S S
Φ = ⋅ d = ⋅ d + ⋅ d
E A E A E A
v v v
since the electric flux through the boundary D between the two volumes is equal and
opposite (flux out of V 1
goes into V
2
Now let’s continue this process of dividing the original volume into a great number of
infinitesimal volumes, each cubic in shape:
i
i
S S
Φ = ⋅ d = ⋅ d
E A E A
v v
Consider now one of these small cubic volumes. Consider one corner of this cube at
position
0 0 0
x , y , z. The length of each side is Δ x , Δ y ,Δ z , and each face is perpendicular
to one of the coordinate axes.
S
1
V
1
D
S
2
V
2
We are interested in computing the flux passing through this small volume. The flux
through the top and bottom faces will only depend on
y
E since E ⋅ d A = 0 for the other 4
faces. Since the cube is infinitesimal, we can do a Taylor expansion of the field about
0 0 0
x , y , z and find the y component of the field at the center of the bottom face
x z
x y z
0 0 0 0 0 0
bottom
0 0 0 , , , ,
y y
y y x y z x y z
E E
x z
E E x y z
x z
Similarly for the top face , ,
x z
x y y z
we have:
0 0 0 0 0 0 0 0 0
top
0 0 0 , , , , , ,
y y y
y y x y z x y z x y z
E E E
x z
E E x y z y
x z y
So the net flux between top and bottom is:
top bottom
y y
Φ = Δ Δ x zE − Δ Δ x zE
top-bottom
The negative sign arises because the electric field points into one surface (chosen to be
the bottom) and out of the other (top). Thus,
0 0 0
x , y , z
x
y
z
Δ y
Δ z
Δ x
E
Differential form of Gauss’ Law
So how does Gauss’ Theorem relate to what we have learned in electromagnetism?
Consider Gauss’ Law, which relates the electric flux through a closed surface to the net
enclosed charge:
enc
0
S
q
d
E A
v
We can relate the enclosed charge to the volume integral of the differential charge density
per unit volume, ρ, in the enclosed volume:
enc
V
q = ρ dV
And by Gauss’ Theorem, we can relate the surface integral of E ⋅ d A to the volume
integral of ∇ ⋅ E , giving us:
0
V V
dV ρ dV
E
Now there is nothing special about what volume V we choose, so the above expression
must hold for all volumes. That means that the integrands must be equal:
0
∇ ⋅ E =
This is the differential form of Gauss’ Law. It holds for every point in space. When
combined with further differential laws of electromagnetism (see next section), we can
derive a differential equation for electromagnetic waves.
For example, consider a constant electric field:
0
E = E x. It is easy to see that the
divergence of E will be zero, so the charge density ρ=0 everywhere. Thus, the total
enclosed charge in any volume is zero, and by the integral form of Gauss’ Law the total
flux through the surface of that volume must be zero.
On the other hand, if
0
E = E x x + E ' y , then
0 0 0
x
E
E E
x
E
Note: If one integrates the obtained charge density within an certain enclosed volume,
you will get exactly the same amount of enclosed charge as if you used the integral form
of Gauss’ Law (try it, it really works!)
Stokes Theorem
Consider the line integral of a vector function around a closed curve C :
C
Γ = ⋅ d
F s
v
This integral is called the “circulation”. We use the right-hand rule to define the direction
of the area vector (perpendicular to the surface) with respect to the integration direction
(counter-clockwise in this case).
Now suppose we subdivide this surface into two regions, and calculate the circulation of
each around closed curves C 1
and C
2
1 2
1 2
C C
Γ = ⋅ d Γ = ⋅ d
F s F s
v v
It should be clear that in both cases, line segment AB is traversed in opposite directions,
so the contribution to Γ 1
is equal and opposite to Γ 2
Thus:
1 2
We can continue subdividing the surface into N subregions and we’ll get:
1
N
i
i =
Let’s let these regions become infinitesimal is size, and calculate the circulation for just
one infinitesimal area aligned in the x-y plane.
C
1
A
B
C
2
d s
F
C
A
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0
,
, ,
0 0 0 0
, , ,
y y x
x y
x y
x y x y
y x x
x y
x y x y x y
y x
F F
x F y
x F x y y F x y x
x y x
F
x F F y
x F x y y y F x y
x y y
F
F
x y
x y
If we define the infinitesimal area vector in the +z direction based on the RH rule:
y
x
z z
F
F
A
x y
Now suppose we did this for a loop in the y-z plane instead. Then xÆy and yÆz in the
above formula and we get:
y z
x x
F
F
A
y z
and if we did this for the z-x plane we get:
x z
y y
F
F
A
z x
So summarizing this with vector notation:
where we have introduced a curl operator:
curl
y y z z x x
F F
F F F F
y z x z x y
ΔΓ = Δ ⋅ ∇ ×
∇ × ≡ = − − − + −
A F
F F x y z
Thus our infinitesimal line integral for the circulation is:
i
i i
C
ΔΓ = ⋅ d = Δ ⋅ ∇ ×
F s A F
v
Now, coming back to the total circulation which is the sum of all the individual
contributions:
1 1
N N
i i
i = i =
Γ = ΔΓ = Δ ⋅ ∇ ×
A F
which when we take the infinitesimal limit gives us:
C S
Γ = ⋅ d = d ⋅ ∇ ×
F s A F
v
This is Stoke’s Theorem. It transforms a closed line integral of a vector function into a
surface integral of the curl of that function.
Why is this useful? Well…
Differential form of Faraday’s and Ampere’s Laws
Recall the integral form of Faraday’s Law of Induction:
B
C S S
d
dt
d d d
t t
B
E s B A A
v
Now let’s use the newly derived Stoke’s Theorem to transform the left side of the
equation involving the electric field:
C S
S S
d d
d d
t
⋅ = ∇ × ⋅
⇒ ∇ × ⋅ = − ⋅
E s E A
B
E A A
v
Now both and left sides of Faraday’s Law involves a surface integral over surface S. This
should be true for any arbitrary surface S, so the integrands must be equal:
t
∇ × = −
B
E
This is the differential form of Faraday’s Law! It holds for every point in space.
Now recall the integral form of Ampere’s Law:
0 enc 0
C S
⋅ d = μ i = μ ⋅ d
B s j A
v
where we have introduced the current density j , whose integral across the surface S gives
us the total current passing through it
Now Maxwell noted that to complete the symmetry between magnetic and electric fields,
there should be an additional term added to Ampere’s Law equivalent to Faraday’s Law
where a changing electric field induces a magnetic field (rather than vice versa):
