29.8. THE REYNOLDS TRANSPORT FORMULA 1031

where n = (n1, · · · ,nn) is the H p−1 measurable unit vector of Lemma 29.7.8. Also, if F isa vector field such that each component is in C1

c (Rp) , then∫U

∇ ·F(x)dmp =∫

∂UF ·ndH p−1. (29.7.39)

Proof: To obtain 29.7.38 apply Lemma 29.7.8 to w = ek. Then to obtain 29.7.39 fromthis, ∫

U∇ ·F(x)dmp

=p

∑j=1

∫U

Fj, jdmp =p

∑j=1

∫∂U

Fjn jdH p−1

=∫

∂U

p

∑j=1

Fjn jdH p−1 =∫

∂UF ·ndH p−1.

What is the geometric significance of the vector, n? Recall that in the part of the bound-ary contained in Qi, this vector points in the same direction as the vector

Ni (x1, · · · ,xp−1,gi (x1, · · · ,xp−1))

given by

(−gi,1 (x1, · · · ,xp−1) ,−gi,2 (x1, · · · ,xp−1) , · · · ,−gi,p−1 (x1, · · · ,xp−1) ,1) (29.7.40)

in the case where k = p. This vector is the gradient of the function,

xp−gi (x1, · · · ,xp−1)

and so is perpendicular to the level surface given by

xp−gi (x1, · · · ,xp−1) = 0

in the case where gi is C1. It also points away from U so the vector n is the unit outernormal. The other cases work similarly.

29.8 The Reynolds Transport FormulaNext is an interesting version of the chain rule for Lipschitz maps. The proof of this theoremis based on the following lemma.

Lemma 29.8.1 If h : Rn→ Rn is Lipschitz, then if h(x) = 0 for all x ∈ A, then

det(Dh(x)) = 0 a.e.x ∈ A

Proof: By the area formula, 0 =∫{0} #(y)dy =

∫A |det(Dh(x))|dx, and so it follows

that det(Dh(x)) = 0 a.e.