14.3. DIVERGENCE THEOREM 401

where

mp−1

(k−1

∏j=1

(ai

j,bij)×

p

∏j=k+1

(ai

j,bij)\Ai

)= 0.

and we assume there is a constant C such that for all i and j,∣∣∣ ∂gi

∂x j

∣∣∣ ≤C off Ai and that ineach variable, g can be recovered from integrating an appropriate partial derivative. Thatis, each gk is absolutely continuous in each variable.

To illustrate the above here is a picture.

U ∩Qi

Qi U ∩QiQi

Recall from calculus that if z−g(x̂) = 0 then to get a normal vector to the level surface,it will be ± the gradient.

Lemma 14.3.2 Let α1, · · · ,α p be real numbers and let A(α1, · · · ,α p) be the matrixwhich has 1+α2

i in the iith slot and α iα j in the i jth slot when i ̸= j. Then detA = 1+∑

pi=1 α2

i .

Proof of the claim: The matrix, A(α1, · · · ,α p) is of the form

A(α1, · · · ,α p) =

1+α2

1 α1α2 · · · α1α pα1α2 1+α2

2 α2α p...

. . ....

α1α p α2α p · · · 1+α2p

Now consider the product of a matrix and its transpose, BT B below.

1 0 · · · 0 α10 1 0 α2...

. . ....

0 1 α p−α1 −α2 · · · −α p 1



1 0 · · · 0 −α10 1 0 −α2...

. . ....

0 1 −α pα1 α2 · · · α p 1

 (14.10)

This product equals a matrix of the form(

A(α1, · · · ,α p) 00 1+∑

pi=1 α2

i

). Therefore,(

1+∑pi=1 α2

i)

det(A(α1, · · · ,α p)) = det(B)2 = det(BT)2. However, using row operations,

detBT = det

1 0 · · · 0 α10 1 0 α2...

. . ....

0 1 α p0 0 · · · 0 1+∑

pi=1 α2

i

= 1+p

∑i=1

α2i

and therefore, (1+

p

∑i=1

α2i

)det(A(α1, · · · ,α p)) =

(1+

p

∑i=1

α2i

)2