6.6 Surface integrals  (Page 9/27)

Orientation of a surface

Recall that when we defined a scalar line integral, we did not need to worry about an orientation of the curve of integration. The same was true for scalar surface integrals: we did not need to worry about an “orientation” of the surface of integration.

On the other hand, when we defined vector line integrals, the curve of integration needed an orientation. That is, we needed the notion of an oriented curve to define a vector line integral without ambiguity. Similarly, when we define a surface integral of a vector field, we need the notion of an oriented surface. An oriented surface is given an “upward” or “downward” orientation or, in the case of surfaces such as a sphere or cylinder, an “outward” or “inward” orientation.

Let S be a smooth surface. For any point $\left(x,y,z\right)$ on S, we can identify two unit normal vectors $\text{N}$ and $\text{−}\text{N}.$ If it is possible to choose a unit normal vector N at every point $\left(x,y,z\right)$ on S so that N varies continuously over S , then S is “ orientable .” Such a choice of unit normal vector at each point gives the orientation of a surface     S . If you think of the normal field as describing water flow, then the side of the surface that water flows toward is the “negative” side and the side of the surface at which the water flows away is the “positive” side. Informally, a choice of orientation gives S an “outer” side and an “inner” side (or an “upward” side and a “downward” side), just as a choice of orientation of a curve gives the curve “forward” and “backward” directions.

Closed surfaces such as spheres are orientable: if we choose the outward normal vector at each point on the surface of the sphere, then the unit normal vectors vary continuously. This is called the positive orientation of the closed surface ( [link] ). We also could choose the inward normal vector at each point to give an “inward” orientation, which is the negative orientation of the surface.

A portion of the graph of any smooth function $z=f(x,y)$ is also orientable. If we choose the unit normal vector that points “above” the surface at each point, then the unit normal vectors vary continuously over the surface. We could also choose the unit normal vector that points “below” the surface at each point. To get such an orientation, we parameterize the graph of $f$ in the standard way: $\text{r}(x,y)=⟨x,y,f(x,y)⟩,$ where x and y vary over the domain of $f.$ Then, ${\text{t}}_x=⟨1,0,f_x⟩$ and ${\text{t}}_y=⟨0,1,f_y⟩,$ and therefore the cross product ${\text{t}}_x\times {\text{t}}_y$ (which is normal to the surface at any point on the surface) is $⟨\text{−}{f}_x,\text{−}{f}_y,1⟩.$ Since the z component of this vector is one, the corresponding unit normal vector points “upward,” and the upward side of the surface is chosen to be the “positive” side.