490 CHAPTER 18. DIFFERENTIAL FORMS
Then
dω = Pydy∧dx+Pzdz∧dx+Qxdx∧dy+Qzdz∧dy+Rxdx∧dz+Rydy∧dz
Now let r : [0,1]2→ R3,r (s,u) = (x(s,u) ,y(s,u) ,z(s,u)). Then letting the various func-tions be defined at r (s,u) ,∫
rdω =
∫[0,1]2
(Py det
(ys yuxs xu
)+Pz det
(zs zuxs xu
)+Qx det
(xs xuys yu
)
+Qz det(
zs zuys yu
)+Rx det
(xs xuzs zu
)+Ry det
(ys yuzs zu
))dsdu
This equals ∫[0,1]2
(Ry−Qz)(yszu− yuzs)+(Pz−Rx)(xuzs− xszu)
+(Qx−Py)(xsyu− xuys)dsdu
By the definition of surface area on S≡ r([0,1]2
), see Definition 14.2.1 the area increment
on the surface r([0,1]2
)is√|yszu− yuzs|2 + |xuzs− xszu|2 + |xsyu− xuys|2dsdu
= det(Dr (u,s)∗Dr (s,u)
)1/2 dsdu
also the vector (Ry−Qz)i+(Pz−Rx) j+ (Qx−Py)k is the curl of the vector field F ≡Pi+Qj+Rj. Thus in more familiar calculus notation, the above integral is of the form∫
S≡r([0,1]2)∇×F ·pdS where p is a vector which is in the direction of
(yszu− yuzs)i+(xuzs− xszu)j+(xsyu− xuys)k.
It happens that this vector p is perpendicular to the surface S at every point where rs×ru ̸=0,which is seen to occur in the above formula. Recall that this follows from noting that, as(hopefully) discussed in beginning calculus,you have rs ·rs×ru = ru ·rs×ru = 0 and thisis sufficient to claim, based on geometric reasoning that it is perpendicular to the surface.Thus
∫r dω is one side of the usual Stoke’s theorem from calculus. You could generalize
to chains of rectangles as well.Next consider what happens with
∫∂r ω . This is just like it was with Green’s theorem
but with more terms. To save on space, P(x(1,u) ,y(1,u) ,z(1,u)) is denoted as P(1,u)with similar considerations for Q and R. Then this results in∫ 1
0
(P(1,u)
∂x(1,u)∂u
+Q(1,u)∂y(1,u)
∂u+R(1,u)
dz(1,u)du
)du
−∫ 1
0
(P(0,u)
∂x(0,u)∂u
+Q(0,u)∂y(0,u)
∂u+R(1,u)
dz(1,u)du
)du
+∫ 1
0P(s,0)
∂x(s,0)∂ s
+Q(s,0)∂y(s,0)
∂ s+R(s,0)
dz(s,0)ds
ds
−∫ 1
0P(s,1)
∂x(s,1)∂ s
+Q(s,1)∂y(s,1)
∂ s+R(s,1)
dz(s,1)ds
ds