12.2. STOKES THEOREM FROM GREEN’S THEOREM 291

12.2.1 The Normal and the OrientationStoke’s theorem as just presented needs no apology. However, it is helpful in applicationsto have some additional geometric insight.

To begin with, suppose the surface S of interest is a parallelogram in R3 determined bythe two vectors a,b. Thus S = R(Q) where Q = [0,1]× [0,1] is the unit square and for(u,v) ∈ Q,

R(u,v)≡ ua+ vb+p,

the point p being a corner of the parallelogram S. Then orient ∂S consistent with thecounter clockwise orientation on ∂Q. Thus, following this orientation on S you go from pto p+a to p+a+b to p+b to p. Then Stoke’s theorem implies that with this orientationon ∂S, ∫

∂SF ·dR=

∫S

∇×F ·nds

where n=Ru ×Rv/ |Ru×Rv| = a×b/ |a×b| . Now recall a,b,a×b forms a righthand system.

aba×b

p+a

p+a+b

S

p

Thus, if you were walking around ∂S in the direction of the orientation with your lefthand over the surface S, the normal vector a×b would be pointing in the direction of yourhead.

More generally, if S is a surface which is not necessarily a parallelogram but is insteadas described in Theorem 12.2.4, you could consider a small rectangle Q contained in Uand orient the boundary of R(Q) consistent with the counter clockwise orientation on ∂Q.Then if Q is small enough, as you walk around ∂R(Q) in the direction of the describedorientation with your left hand over R(Q), your head points roughly in the direction ofRu×Rv.

Q

u0

∆v

∆u

Rv(u0)∆v

Ru(u0)∆u

R(Q)

As explained above, this is true of the tangent parallelogram, and by continuity ofRv,Ru, the normals to the surface R(Q)Ru×Rv (u) for u ∈ Q will still point roughly inthe same direction as your head if you walk in the indicated direction over ∂R(Q), meaningthe angle between the vector from your feet to your head and the vector Ru×Rv (u) is lessthan π/2.

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