4Surfaces and surface integrals
IA Vector Calculus
4.2 Parametrized surfaces and area
However, specifying a surface by an equation
f
(
r
) =
c
is often not too helpful.
What we would like is to put some coordinate system onto the surface, so that
we can label each point by a pair of numbers (
u, v
), just like how we label points
in the x, y-plane by (x, y). We write r(u, v) for the point labelled by (u, v).
Example. Let S be part of a sphere of radius a with 0 ≤ θ ≤ α.
α
We can then label the points on the spheres by the angles θ, ϕ, with
r(θ, ϕ) = (a cos ϕ sin θ, a sin θ sin ϕ, a cos θ) = ae
r
.
We restrict the values of
θ, ϕ
by 0
≤ θ ≤ α
, 0
≤ ϕ ≤
2
π
, so that each point is
only covered once.
Note that to specify a surface, in addition to the function
r
, we also have
to specify what values of (
u, v
) we are allowed to take. This corresponds to a
region
D
of allowed values of
u
and
v
. When we do integrals with these surfaces,
these will become the bounds of integration.
When we have such a parametrization
r
, we would want to make sure this
indeed gives us a two-dimensional surface. For example, the following two
parametrizations would both be bad:
r(u, v) = u, r(u, v) = u + v.
The idea is that
r
has to depend on both
u
and
v
, and in “different ways”.
More precisely, when we vary the coordinates (
u, v
), the point
r
will change
accordingly. By the chain rule, this is given by
δr =
∂r
∂u
δu +
∂r
∂v
δv + o(δu, δv).
Then
∂r
δu
and
∂r
∂v
are tangent vectors to curves on
S
with
v
and
u
constant
respectively. What we want is for them to point in different directions.
Definition
(Regular parametrization)
.
A parametrization is regular if for all
u, v,
∂r
∂u
×
∂r
∂v
6= 0,
i.e. there are always two independent tangent directions.
The parametrizations we use will all be regular.
Given a surface, how could we, say, find its area? We can use our parametriza-
tion. Suppose points on the surface are given by
r
(
u, v
) for (
u, v
)
∈ D
. If we
want to find the area of D itself, we would simply integrate
Z
D
du dv.
However, we are just using
u
and
v
as arbitrary labels for points in the surface,
and one unit of area in
D
does not correspond to one unit of area in
S
. Instead,
suppose we produce a small rectangle in
D
by changing
u
and
v
by small
δu, δv
.
In
D
, this corresponds to a rectangle with vertices (
u, v
)
,
(
u
+
δu, v
)
,
(
u, v
+
δv
)
,
(
u
+
δu, v
+
δv
), and spans an area
δuδv
. In the surface
S
, these small
changes
δu, δv
correspond to changes
∂r
∂u
δu
and
∂r
∂v
δv
, and these span a vector
area of
δS =
∂r
∂u
×
∂r
∂v
δuδv = n δS.
Note that the order of u, v gives the choice of the sign of the unit normal.
The actual area is then given by
δS =
∂r
∂u
×
∂r
∂v
δu δv.
Making these into differentials instead of deltas, we have
Proposition. The vector area element is
dS =
∂r
∂u
×
∂r
∂v
du dv.
The scalar area element is
dS =
∂r
∂u
×
∂r
∂v
du dv.
By summing and taking limits, the area of S is
Z
S
dS =
Z
D
∂r
∂u
×
∂r
∂v
du dv.
Example. Consider again the part of the sphere of radius a with 0 ≤ θ ≤ α.
α
Then we have
r(θ, ϕ) = (a cos ϕ sin θ, a sin θ sin ϕ, a cos θ) = ae
r
.
So we find
∂r
∂θ
= ae
θ
.
Similarly, we have
∂r
∂ϕ
= a sin θe
ϕ
.
Then
∂r
∂θ
×
∂r
∂ϕ
= a
2
sin θ e
r
.
So
dS = a
2
sin θ dθ dϕ.
Our bounds are 0 ≤ θ ≤ α, 0 ≤ ϕ ≤ 2π.
Then the area is
Z
2π
0
Z
α
0
a
2
sin θ dθ dϕ = 2πa
2
(1 −cos α).