5Smooth embedded surfaces (in ℝ3)
IB Geometry
5.3 Surfaces of revolution
So far, we do not have many examples of surfaces. We now describe a nice way
of obtaining surfaces — we obtain a surface
S
by rotating a plane curve
η
around
a line
`
. We may wlog assume that coordinates a chosen so that
`
is the
z
-axis,
and η lies in the x − z plane.
More precisely, we let η : (a, b) → R
3
, and write
η(u) = (f(u), 0, g(u)).
Note that it is possible that a = −∞ and/or b = ∞.
We require
kη
0
(
u
)
k
= 1 for all
u
. This is sometimes known as parametrization
by arclength. We also require
f
(
u
)
>
0 for all
u >
0, or else things won’t make
sense.
Finally, we require that
η
is a homeomorphism to its image. This is more
than requiring η to be injective. This is to eliminate things like
Then S is the image of the following map:
σ(u, v) = (f(u) cos v, f(u) sin v, g(u))
for
a < u < b
and 0
≤ v ≤
2
π
. This is not exactly a parametrization, since it is
not injective (
v
= 0 and
v
= 2
π
give the same points). To rectify this, for each
α ∈ R, we define
σ
α
: (a, b) × (α, α + 2π) → S,
given by the same formula, and this is a homeomorphism onto the image. The
proof of this is left as an exercise for the reader.
Assuming this, we now show that this is indeed a parametrization. It is
evidently smooth, since
f
and
g
both are. To show this is a parametrization,
we need to show that the partial derivatives are linearly independent. We can
compute the partial derivatives and show that they are non-zero. We have
σ
u
= (f
0
cos v, f
0
sin v, g
0
)
σ
v
= (−f sin v, f cos v, 0).
We then compute the cross product as
σ
u
× σ
v
= (−fg
0
cos v, −fg
0
sin v, ff
0
).
So we have
kσ
u
× σ
v
k = f
2
(g
02
+ f
02
) = f
2
6= 0.
Thus every σ
α
is a valid parametrization, and S is a valid embedded surface.
More generally, we can allow
S
to be covered by several families of parametriza-
tions of type
σ
α
, i.e. we can consider more than one curve or more than one axis
of rotation. This allows us to obtain, say,
S
2
or the embedded torus (in the old
sense, we cannot view
S
2
as a surface of revolution in the obvious way, since we
will be missing the poles).
Definition (Parallels). On a surface of revolution, parallels are curves of the
form
γ(t) = σ(u
0
, t) for fixed u
0
.
Meridians are curves of the form
γ(t) = σ(t, v
0
) for fixed v
0
.
These are generalizations of the notions of longitude and latitude (in some
order) on Earth.
In a general surface of revolution, we can compute the first fundamental form
with respect to σ as
E = kσ
u
k
2
= f
02
+ g
02
= 1,
F = σ
u
· σ
v
= 0
G = kσ
v
k
2
= f
2
.
So its first fundamental form is also of the simple form, like the geodesic polar
coordinates.
Putting these explicit expressions into the geodesic formula, we find that the
geodesic equations are
¨u = f
df
du
˙v
2
d
dt
(f
2
˙v) = 0.
Proposition. We assume k˙γk = 1, i.e. ˙u
2
+ f
2
(u) ˙v
2
= 1.
(i) Every unit speed meridians is a geodesic.
(ii) A (unit speed) parallel will be a geodesic if and only if
df
du
(u
0
) = 0,
i.e. u
0
is a critical point for f.
Proof.
(i)
In a meridian,
v
=
v
0
is constant. So the second equation holds. Also, we
know k˙γk = |˙u| = 1. So ¨u = 0. So the first geodesic equation is satisfied.
(ii) Since o = o
u
, we know f(u
0
)
2
˙v
2
= 1. So
˙v = ±
1
f(u
0
)
.
So the second equation holds. Since
˙v
and
f
are non-zero, the first equation
is satisfied if and only if
df
du
= 0.