Orienting surfaces

Given some curve C from the point p to point q, we can parametrize C by a function c(t), for a ≤ t ≤ b. In this case, we know that c(a) = p and c(b) = q. We could say that the curve is oriented from p to q. At any point c(t) along the curve, you could tell its orientation by its unit tangent vector

T =

We could also parametrize the curve backward, going from q to p, by a function d(s), for a ≤ s ≤ b, with d(a) = q and d(b) = p. We could say that the curve parametrized by d has the opposite orientation as the curve parametrized by c. You could tell this by looking at the unit tangent vector at the point d(s):

T =

If c(t) = d(t), the two unit tangent vectors would point in the opposite directions, as shown below.

We don’t usually use this language when talking about curves. The concept of parametrizing a curve in opposite directions is simple enough to understand without having to talk about orientation and unit tangent vectors. But this is the language we will use with parametrized surfaces. So if you can understand orientation of curves as discussed above, it will help you understand the orientation of surfaces, which is a little trickier.

It turns out that, like curves, we can parametrize a surface in two different orientations. The orientation of a curve is given by the unit tangent vector n; the orientation of a surface is given by the unit normal vector n. Unless we are dealing with an unusual surface, a surface has two sides. We can pick the normal vector to point out one side of the surface, or we can pick the normal vector to point out the other side of the surface. Our choice of normal vector specifies the orientation of the surface. We call the side of the surface with the normal vector the positive side of the surface.

As an example, consider the sphere of radius R centered at the origin. Recalling our spherical coordinates, we could paramatrize the sphere using

Φ(θ,ϕ) = (R sin ϕ cos θ,R sin ϕ sin θ,R cos ϕ)

for 0 ≤ θ ≤ 2π and 0 ≤ ϕ ≤ π. Remember that a formula for the unit normal vector is

n =

∂Φ-(s,t) × ∂Φ-(s,t)
∥-∂s---------∂t------∥
∥ ∂Φ ∂Φ ∥
∥∥ ---(s,t) × ---(s,t)∥∥
∂s ∂t

only in this case, we have θ and ϕ rather than s and t.

Let’s find an unit normal vector of the sphere.

(θ,ϕ)	= (-R sin ϕ sin θ,R sin ϕ cos θ, 0)
(θ,ϕ)	= (R cos ϕ cos θ,R cos ϕ sin θ,-R sin ϕ)
(θ,ϕ) ×(θ,ϕ)	=
	= -R² sin ²ϕ cos θi - R² sin ²ϕ sin θj
	- R² sin ϕ cos ϕ(sin ²θ + cos ²θ)k
	= -R² sin ²ϕ cos θi - R² sin ²ϕ sin θj
	- R² sin ϕ cos ϕk
	= -R² sin ϕ
	= R² sin ϕ
	= R² sin ϕ
	= R² sin ϕ
	= R² sin ϕ

Our unit normal vector is

n	=
	= - sin ϕ cos θi - sin ϕ sin θj - cos ϕk
	= (- sin ϕ cos θ,- sin ϕ sin θ,- cos ϕ)

This is the normal vector for the point on the surface given by Φ(θ,ϕ) = (R sin ϕ cos θ,R sin ϕ sin θ,R cos ϕ). Note that n points in the opposite direction as Φ(θ,ϕ) (as n is -1∕R times Φ(θ,ϕ)). Consequently, n points back toward the origin, i.e., points inside the sphere. For example, if θ = 0 and ϕ = π∕2, then Φ(0,π∕2) = (R, 0, 0) and n = (-1, 0, 0). The vector n for any value of θ or ϕ is shown in the figure below. Notice that n always points toward the inside of the sphere. In this case, the inside of the sphere is the positive side of the surface.

What if we changed parameterization so that ϕ was the first variable?

Φ(ϕ,θ) = (R sin ϕ cos θ,R sin ϕ sin θ,R cos ϕ)

Then the normal would be ∂ Φ
----
∂ϕ × ∂Φ
----
∂θ . How does this compare to ∂Φ
----
∂ θ × ∂ Φ
----
∂ ϕ ? Recalling the properties of the cross product, we can conclude these vectors are opposites: ∂Φ--
∂ϕ × ∂-Φ-
∂θ = - ∂Φ--
∂θ × ∂Φ--
∂ ϕ . In this case, unit normal is