18.8. EXAMPLES OF STOKE’S THEOREM 491

As before, this is a line integral of the form∫

C Pdx+Qdy+Rdz. where C is an orientedcurve bounding the surface S. This orientation will determine the direction of the vector pabove.

18.8.5 The Divergence TheoremIn this case, we have a parameter domain [a,b]⊆ Rp and the differential form is

ω =p

∑r=1

αr (x)(−1)r−1 dx1∧·· ·∧dx̂r ∧·· ·∧dxp

where dx̂r with the hat means that dxr is omitted. The reason for the (−1)r−1 is to makeminus signs disappear in dω . This led to the divergence theorem

±∫r([a,b])

p

∑j=1

∂α j

∂x j(x)dH k =

∫r(∂ [a,b])

p

∑j=1

α j (x)N j (x)dH k−1 (x)

where the + sign holds if and only if∂(x1,··· ,xp)∂(u1,··· ,up)

≥ 0. We needed to have r is one to one off

a set Ŝ where r(Ŝ)

has H p measure zero and rbl ,ral are also one to one or at least one toone off a set S where rbl (S) ,ral (S) have H p−1 measure zero. It remains to consider thevector N which has jth component N j. I want to argue that this vector is a.e. normal tor (∂ [a,b]) and that sometimes it is an outer normal. Recall that for x= rbl (ul) , N j (x) =

(−1) j+l ∂(x1,··· ,x̂ j ,··· ,xp)∂(u1,··· ,ûl ··· ,up)

1J∗(x)

. Thus for a point on the boundary, for i ̸= l,

xui ·N =p

∑j=1

x j,uiNj (x) =

p

∑j=1

x j,ui (−1)l+ j ∂ (x1, · · · , x̂ j, · · · ,xp)

∂ (u1, · · · , ûl · · · ,up)

1J∗ (x)

=1

J∗ (x)det(xu1 · · · xui · · · xui · · · xup

)= 0

since it is a determinant with two equal columns. This involved expanding along the lth

column which was filled by xui . Also, expanding along this column,

xul ·N =1

J∗ (x)det(Dr) = sgn(det(Dr))

It follows that N is perpendicular to r (∂ [a,b]) and that the angle between xul and N is

no more than 90 degrees if∂(x1,··· ,xp)∂(u1,··· ,up)

≥ 0. Now on the face where ul = bl , xul points away

from this face and so N points in roughly the same direction and is an “outer” normal.Similar considerations apply when ul = al but here the −xul points away and we use −Nbecause of the subtraction in the boundary integrals. This yields the following.

Theorem 18.8.1 Let r be Lipschitz and one to one off S where r (S) has H p mea-sure zero and suppose a similar condition holds for ral and rbl . Let α j be C1

ω =p

∑j=1

α j (x)(−1) j−1 dx1∧·· ·∧dx̂ j ∧·· ·∧dxp, α(x) = (α1 (x) , ...,α p (x))