Stoke’s theorem as just presented needs no apology. However, it is helpful in applications to have some additional geometric insight.
To begin with, suppose the surface S of interest is a parallelogram in ℝ3 determined by the two vectors a,b. Thus S = R
the point p being a corner of the parallelogram S. Then orient ∂S consistent with the counter clockwise orientation on ∂Q. Thus, following this orientation on S you go from p to p + a to p + a + b to p + b to p. Then Stoke’s theorem implies that with this orientation on ∂S,
Now recall a,b,a × b forms a right hand system.
Thus, if you were walking around ∂S in the direction of the orientation with your left hand over the surface S, the normal vector a × b would be pointing in the direction of your head.
More generally, if S is a surface which is not necessarily a parallelogram but is instead as described in Theorem 28.3.4, you could consider a small rectangle Q contained in U and orient the boundary of R
As explained above, this is true of the tangent parallelogram, and by continuity of Rv,Ru, the normals to the surface R
You can imagine filling U with such non-overlapping regions Qi. Then orienting ∂R(Qi) consistent with the counter clockwise orientation on Qi, and adding the resulting line integrals, the line integrals over the common sides cancel as indicated in the following picture and the result is the line integral over ∂S.
Thus there is a simple relation between the field of normal vectors on S and the orientation of ∂S. It is simply this. If you walk along ∂S in the direction mandated by the orientation, with your left hand over the surface, the nearby normal vectors in Stoke’s theorem will point roughly in the direction of your head.
This also illustrates that you can define an orientation for ∂S by specifying a field of unit normal vectors for the surface, which varies continuously over the surface, and require that the motion over the boundary of the surface is such that your head points roughly in the direction of nearby normal vectors as you walk along the boundary with your left hand over S. The existence of such a continuous field of normal vectors is what constitutes an orientable surface.