# Homework Help: What does it mean that a curve or a surface is oriented?

1. Apr 28, 2007

### kasse

What does it mean that a curve or a surface is oriented?

2. Apr 28, 2007

### HallsofIvy

Integrating a function of a single variable "from a to b" will give the negative of the integral of the same function "from b to a". Those are two different "orientations" of the same line. Of course, in one dimension, we can just say "from a to b" or "from b to a". In higher dimensions, it's a little more complicated. Integrating around a closed path in two dimensions we might specify "integrating in the counterclockwise direction" (often the "default" choice) or "integrating in the clockwise direction.

Take Stokes' Theorem, for example,
$$\int_S\int \left[\left(\nabla x \vec{v}\right)\cdot\vec{n}\right]d\sigma= \int_C \vec{v}\cdot d\vec{r}$$
where C is the (necessarily closed) boundary of S.

The right hand side will differ in sign depending on the direction you go around C: a choice of "orientation". Similarly, we can take $\vec{n}$, the unit normal to surface point in either of two directions. The vector field $\vec{v}$ will have positive dot product with one of those, negative dot product with the other. Again, that is a choice of "orientation".

Typically, problems in calculus books about integrating a vector field over a surface specify "oriented by upward normals", or "downward normals", or perhaps "rightward normals" if that is simpler. Problems involving a close surface like a sphere might say "oriented by outward normals" or "oriented by inward normals". Again, those are different choices for orientation. Every curve or surface has two possible "orientations". For a curve, it is "this direction" or "that direction". For a surface it is the direction in which the unit normals point.

Getting back to Stokes' Theorem, in order to make the right and left sides equal, we must choose "corresponding" orientations. The standard "correspondence" is this. Imagine walking around the boundary standing "up" so that your head is pointing the same way as the unit normals (in the given orientation of the surface). If your left side is toward the surface (i.e. inward), then you are walking around the boundary in the 'corresponding' orientation for the boundary.

For a specific example, let S be the surface z= 4- x2- y2, the paraboloid with vertex (0,0,4), for z non-negative. The boundary of that surface is the circle x2+ y2= 4 in the xy-plane. We can specifically find a normal to the surface by writing it as F(x,y,z)= z+ x2+ y2= 4 and taking the gradient of F: $2x\vec{i}+ 2y\vec{j}+ \vec{k}$. Since the z-component of that is positive (+1), that is the "upward" orientation. Specifically, at (2, 0, 0), which is on the boundary, that is $4\vec{i}+ \vec{k}$. If I "stand on" the boundary, with my head pointed in that direction and my left side toward the surface (the vector $-\vec{i}+ 4\ve{k}$ is perpendicular to the normal vector and can represent my "left arm"- it has positive z-component so is pointing in the direction of the surface above the xy-plane) I would be facing in the counter clockwise direction of the circle. That is the "corresponding" orientation for the boundary. On the other hand, if I take the normal vector in the opposite direction, by multiplying by -1, $+\vec{i}- 4\vec{k}$, I would have the same surface "oriented by downward normals". Standing with my head downward (sounds uncomfortable doesn't it!), but still with my left side (now the normal $4\vec{i}+ \vec{k}$ "upward", my left side is not rightward (x component is +4) so I am facing clockwise on the circle. Now clockwise orientation of the boundary curve "corresponds" to "downward" orientation of the surface.