18.8. EXAMPLES OF STOKE’S THEOREM 489

Recall that terms like dx∧dx are zero because they result in a determinant which equals 0.Then let r be smooth and map [a,b]× [c,d] to R2.

r (s,u)≡ (x(s,u) ,y(s,u))T .

Then∫r dω ≡∫ b

a

∫ d

cPy (x(s,u) ,y(s,u))

∣∣∣∣ ys yuxs xu

∣∣∣∣+Qx (x(s,u) ,y(s,u))∣∣∣∣ xs xu

ys yu

∣∣∣∣dsdu

=∫ b

a

∫ d

c[Qx (x(s,u) ,y(s,u))−Py (x(s,u) ,y(s,u))] (xsyu− xuys)dsdu

Let U = r ([a,b]× [c,d]) . Then by the change of variables theorem, this equals∫U(Qx (x,y)−Py (x,y))sgn(xsyu− xuys)dxdy

Next consider∫

∂r ω . For r (s,u) = (x(s,u) ,y(s,u))T , this equals∫ d

cP(x(1,u) ,y(1,u))

∂x(1,u)∂u

+Q(x(1,u) ,y(1,u))∂y(1,u)

∂udu

−∫ d

cP(x(0,u) ,y(0,u))

∂x(0,u)∂u

+Q(x(0,u) ,y(0,u))∂y(0,u)

∂udu

+∫ b

aP(x(s,0) ,y(s,0))

∂x(s,0)∂ s

+Q(x(s,0) ,y(s,0))∂y(s,0)

∂ sds

−∫ b

aP(x(s,1) ,y(s,1))

∂x(s,1)∂ s

+Q(x(s,1) ,y(s,1))∂y(s,1)

∂ sds

This is computing a line integral by summing the contributions around the edges of [a,b]×[c,d]. It is

∫C Pdx+Qdy where the orientation on this curve C comes from the counter

clockwise orientation on the boundary of [a,b]× [c,d] as shown in the picture.

a b

c

d

Thus if xsyu− xuys > 0, this is just Green’s theorem from calculus. Thus this gives aproof of this important theorem provided the region U is the C1 image of a rectangle. Onecould generalize to consider chains of rectangles which, as mentioned above yields fairlygeneral surfaces in R2. One could also include Lipschitz maps for even more generality.However, this falls short of the best results for Green’s theorem which involve a rectifiablesimple closed curve with U its interior. Rectifiable only requires the curve to have finitelength.

18.8.4 Stoke’s Theorem from CalculusNext let k = 2 and p = 3. Thus dω is a 2 form so ω is a 1 form. Say

ω = P(x,y,z)dx+Q(x,y,z)dy+R(x,y,z)dz