18.3. STOKES THEOREM 481

can yield a pointy place in the curve resulting from the given parametrization so it wouldnot deserve to be called a smooth curve. It is the same here. Since Dr is allowed tovanish, one can have r ([a,b]) many different kinds of sets. However, if you insist that Drbe invertible, the points on the box would be preserved by an application of the implicitfunction theorem.

The above could be improved by using the above to approximate functions which arenot C2 with functions which are, obtained by mollifying, and then passing to a limit to getmore general situations. In particular, consider r a function in C1 ([a,b] ;Rp). Then byLemma 16.3.1 there is a sequence of functions {rn} each C2 which converges uniformlyto r and such that Drn converges uniformly to Dr on [a,b]. Then Stoke’s theorem holdswith r replaced with rn and so, passing to a limit as n→ ∞ one obtains Stoke’s theoremfor r. This yields the following corollary.

Corollary 18.3.4 Let ω = ∑I α I (x)dxi1 ∧·· ·∧dxik−1 be an k−1 form, each α I beingC1 ([a,b]). Let r : [a,b]→ Rp, p≥ k be in C1 ([a,b] ;Rp). Then

∫∂r ω =

∫r dω .

Proof: From the above,∫

∂rnω =

∫rn

dω where rn is C2. Both terms involve integralsover [a,b] or [a,b]l and the convergence is uniform, so one can pass to a limit as n→ ∞

retaining the same formula with rn replaced with r. ■Suppose you have two boxes [a,b] and [c,d] and these intersect on a common face, say

the lth face. Thus cl = bl . Then in the above description for the boundary integral on thecommon face, the two contributions cancel because you have the same thing except one has|blal and the other has |dl

bl. Therefore, you would get Stokes theorem for the box consisting

of these two pasted together, the boundary integrals consisting of the sum of the boundaryintegrals of the remaining faces. Continuing this way, consider a chain of these boxes suchthat each box intersects another along a complete face. Then you could do the above foreach pair of boxes in the chain and note that the boundary integrals will cancel along thecommon faces. Thus, in place of a single box you could have a much more complicatedshape and the boundary integral would take place over exactly those faces which do nothave intersection with faces of other boxes whereas

∫r dω would take place over the union

of the boxes since the boundaries are sets of measure zero relative to mk. The followingpicture illustrates what is meant.

Thus, adding over the boxes yields a parameter domain which looks like the followingfor∫r dω .

By now, it should be clear that fairly general regions can be included. Also, we onlyneed to have aI continuous on the face of two of these intersecting boxes. This is an analogof a piece-wise smooth curve.