Orientation of Surface | Sahithyan's Notes

An orientation of a surface is a choice of which side is the positive side, given by a continuous unit normal field $\hat{n}$ over the whole surface.

Orientable Surface

A surface is orientable iff a continuous choice of unit normal vector can be made over the entire surface.

Two-sided surfaces are orientable. One-sided surfaces (such as Möbius strip) are non-orientable.

A smooth orientable surface has exactly 2 orientations, $\hat{n}$ and $-\hat{n}$ . The orientation fixes the sign of a flux integral over the surface.

If $\sigma$ is smooth and orientable with unit normal $\hat{n}$ at each point, the direction of $\hat{n}$ can be chosen so that $\hat{n} = \hat{n}(x,y,z)$ varies continuously over the surface. Such a choice of unit normal field is an orientation of $\sigma$ .

For a closed surface, the positive orientation is the outward orientation.

Positive Orientation of a Boundary Curve

Let $\sigma$ be an oriented surface with unit normal field $\hat{n}$ , and let $C$ be the boundary curve of $\sigma$ . The positive orientation of $C$ is the direction determined by the chosen orientation of $\sigma$ .

Equivalently, walking along $C$ with the head pointing in the direction of $\hat{n}$ keeps the surface on the left.

Orientation of a Smooth Parametric Surface

Let $\sigma$ be a smooth parametric surface given by $\boldsymbol{r}(u,v) = x(u,v)\hat{i} + y(u,v)\hat{j} + z(u,v)\hat{k}$ . If

\boldsymbol{N}(u,v) = \frac{\partial \boldsymbol{r}}{\partial u} \times \frac{\partial \boldsymbol{r}}{\partial v} \ne \boldsymbol{0}

then

\hat{n} = \frac{\boldsymbol{N}(u,v)}{\lVert \boldsymbol{N}(u,v) \rVert} = \frac{\dfrac{\partial \boldsymbol{r}}{\partial u} \times \dfrac{\partial \boldsymbol{r}}{\partial v}}{\left\lVert \dfrac{\partial \boldsymbol{r}}{\partial u} \times \dfrac{\partial \boldsymbol{r}}{\partial v} \right\rVert}

defines one orientation of $\sigma$ . The opposite orientation is $-\hat{n}$ .

Orientation of Non-Parametric Surfaces

Surfaces of the form $z = g(x,y)$ , $y = g(x,z)$ , $x = g(y,z)$ can be parametrized using their independent variables as parameters:

$\boldsymbol{r}(x,y) = x\hat{i} + y\hat{j} + g(x,y)\hat{k}$ , for $z = g(x,y)$
$\boldsymbol{r}(x,z) = x\hat{i} + g(x,z)\hat{j} + z\hat{k}$ , for $y = g(x,z)$
$\boldsymbol{r}(y,z) = g(y,z)\hat{i} + y\hat{j} + z\hat{k}$ , for $x = g(y,z)$

Each can be written as a level surface $G(x,y,z) = 0$ by moving all terms to one side. The vector

\hat{n} = \frac{\nabla G}{\lVert \nabla G \rVert}

is a unit normal vector to the surface, whenever $\nabla G \ne \boldsymbol{0}$ .

Examples

Cylinder

Find the orientation of the cylinder $x^2 + y^2 = 1$ , $0 \le z \le 2$ , given by $\boldsymbol{r}(\theta, z) = \cos\theta\,\hat{i} + \sin\theta\,\hat{j} + z\hat{k}$ , and state which way it points.

Solution Reveal Hide

The partial derivatives are

\frac{\partial \boldsymbol{r}}{\partial \theta} = -\sin\theta\,\hat{i} + \cos\theta\,\hat{j}, \qquad \frac{\partial \boldsymbol{r}}{\partial z} = \hat{k}

\boldsymbol{N} = \frac{\partial \boldsymbol{r}}{\partial \theta} \times \frac{\partial \boldsymbol{r}}{\partial z} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{vmatrix} = \cos\theta\,\hat{i} + \sin\theta\,\hat{j}

$\lVert \boldsymbol{N} \rVert = 1 \ne 0$ , so

\hat{n} = \cos\theta\,\hat{i} + \sin\theta\,\hat{j}

At $(\cos\theta, \sin\theta, z)$ , $\hat{n}$ equals the position vector’s $x$ and $y$ parts, so it points away from the $z$ -axis. This is the outward orientation. The inward orientation is $-\hat{n}$ .

Paraboloid

Find the orientation of the paraboloid $z = x^2 + y^2$ that points upward.

Solution Reveal Hide

Write the surface as $G(x,y,z) = z - x^2 - y^2 = 0$ .

\nabla G = -2x\,\hat{i} - 2y\,\hat{j} + \hat{k}, \qquad \lVert \nabla G \rVert = \sqrt{4x^2 + 4y^2 + 1}

$\nabla G \ne \boldsymbol{0}$ everywhere since its $\hat{k}$ component is $1$ .

\hat{n} = \frac{-2x\,\hat{i} - 2y\,\hat{j} + \hat{k}}{\sqrt{4x^2 + 4y^2 + 1}}

The $\hat{k}$ component is positive, so $\hat{n}$ points upward. The downward orientation is $-\hat{n}$ .

Sphere

Find the outward orientation of the sphere $x^2 + y^2 + z^2 = a^2$ , where $a > 0$ .

Solution Reveal Hide

Write the surface as $G(x,y,z) = x^2 + y^2 + z^2 - a^2 = 0$ .

\nabla G = 2x\,\hat{i} + 2y\,\hat{j} + 2z\,\hat{k}, \qquad \lVert \nabla G \rVert = 2\sqrt{x^2 + y^2 + z^2} = 2a

\hat{n} = \frac{x\,\hat{i} + y\,\hat{j} + z\,\hat{k}}{a}

$\hat{n}$ is the position vector scaled to unit length, so it points away from the origin. The inward orientation is $-\hat{n}$ .