4Surfaces and surface integrals
IA Vector Calculus
4.3 Surface integral of vector fields
Just computing the area of a surface would be boring. Suppose we have a surface
S
parametrized by
r
(
u, v
), where (
u, v
) takes values in
D
. We would like to ask
how much “stuff” is passing through
S
, where the flow of stuff is given by a
vector field F(r).
We might attempt to use the integral
Z
D
|F| dS.
However, this doesn’t work. For example, if all the flow is tangential to the
surface, then nothing is really passing through the surface, but
|F|
is non-zero,
so we get a non-zero integral. Instead, what we should do is to consider the
component of F that is normal to the surface S, i.e. parallel to its normal.
Definition
(Surface integral)
.
The surface integral or
flux
of a vector field
F
(
r
)
over S is defined by
Z
S
F(r) · dS =
Z
S
F(r) · n dS =
Z
D
F(r(u, v)) ·
∂r
∂u
×
∂r
∂v
du dv.
Intuitively, this is the total amount of
F
passing through
S
. For example, if
F
is the electric field, the flux is the amount of electric field passing through a
surface.
For a given orientation, the integral
R
F·
d
S
is independent of the parametriza-
tion. Changing orientation is equivalent to changing the sign of
n
, which is in
turn equivalent to changing the order of
u
and
v
in the definition of
S
, which is
also equivalent to changing the sign of the flux integral.
Example. Consider a sphere of radius a, r(θ, ϕ). Then
∂r
∂θ
= ae
θ
,
∂r
∂ϕ
= a sin θe
ϕ
.
The vector area element is
dS = a
2
sin θe
r
dθ dϕ,
taking the outward normal n = e
r
= r/a.
Suppose we want to calculate the fluid flux through the surface. The velocity
field
u
(
r
) of a fluid gives the motion of a small volume of fluid
r
. Assume that
u
depends smoothly on
r
(and
t
). For any small area
δS
, on a surface
S
, the
volume of fluid crossing it in time δt is u · δS δt.
δS
u δt
n
So the amount of flow of u over at time δt through S is
δt
Z
S
u · dS.
So
R
S
u · dS is the rate of volume crossing S.
For example, let
u
= (
−x,
0
, z
) and
S
be the section of a sphere of radius
a
with 0 ≤ ϕ ≤ 2π and 0 ≤ θ ≤ α. Then
dS = a
2
sin θn dϕ dθ,
with
n =
r
a
=
1
a
(x, y, z).
So
n · u =
1
a
(−x
2
+ z
2
) = a(−sin
2
θ cos
2
ϕ + cos
2
θ).
Therefore
Z
S
u · dS =
Z
α
0
Z
2π
0
a
3
sin θ[(cos
2
θ − 1) cos
2
ϕ + cos
2
θ] dϕ dθ
=
Z
α
0
a
3
sin θ[π(cos
2
θ − 1) + 2π cos
2
θ] dθ
=
Z
α
0
a
3
π(3 cos
3
θ − 1) sin θ dθ
= πa
3
[cosθ − cos
3
θ]
α
0
= πa
3
cos α sin
2
α.
What happens when we change parametrization? Let r(u, v) and r(˜u, ˜v) be
two regular parametrizations for the surface. By the chain rule,
∂r
∂u
=
∂r
∂˜u
∂˜u
∂u
+
∂r
∂˜v
∂˜v
∂u
∂r
∂v
=
∂r
∂˜u
∂˜u
∂v
+
∂r
∂˜v
∂˜v
∂v
So
∂r
∂u
×
∂r
∂v
=
∂(˜u, ˜v)
∂(u, v)
∂r
∂˜u
×
∂r
∂˜v
where
∂(˜u,˜v)
∂(u,v)
is the Jacobian.
Since
d˜u d˜v =
∂(˜u, ˜v)
∂(u, v)
du dv,
We recover the formula
dS =
∂r
∂u
×
∂r
∂v
du dv =
∂r
∂˜u
×
∂r
∂˜v
d˜u d˜v.
Similarly, we have
dS =
∂r
∂u
×
∂r
∂v
du dv =
∂r
∂˜u
×
∂r
∂˜v
d˜u d˜v.
provided (u, v) and (˜u, ˜v) have the same orientation.