29.3. THE DIVERGENCE THEOREM 573

z = ψ(x,y)

z = φ(x,y)

x

z

y

nz =1

(1+ψ2x+ψ2

y)1/2

nz =−1

(1+φ2x+φ2

y)1/2

nz = 0

Of course, many three dimensional sets are cylindrical in each of the coordinate direc-tions. For example, a ball or a rectangle or a tetrahedron are all cylindrical in each direction.The following lemma allows the exchange of the volume integral of a partial derivative foran area integral in which the derivative is replaced with multiplication by an appropriatecomponent of the unit exterior normal.

Lemma 29.3.2 Suppose V is cylindrical in the z direction and that φ and ψ are thefunctions in the above definition. Assume φ and ψ are C1 functions and suppose F is a C1

function defined on V . Also, let n= (nx,ny,nz) be the unit exterior normal to ∂V . Then∫V

∂F∂ z

(x,y,z) dV =∫

∂VFnz dA.

Proof: From the fundamental theorem of calculus,∫V

∂F∂ z

(x,y,z) dV =∫

D

∫ψ(x,y)

φ(x,y)

∂F∂ z

(x,y,z) dzdxdy (29.3)

=∫

D[F (x,y,ψ (x,y))−F (x,y,φ (x,y))] dxdy

Now the unit exterior normal on the top of V , the surface (x,y,ψ (x,y)) is

1√ψ2

x +ψ2y +1

(−ψx,−ψy,1

).

This follows from the observation that the top surface is the level surface z−ψ (x,y) = 0and so the gradient of this function of three variables is perpendicular to the level surface.It points in the correct direction because the z component is positive. Therefore, on the topsurface

nz =1√

ψ2x +ψ2

y +1

Similarly, the unit normal to the surface on the bottom is

1√φ

2x +φ

2y +1

(φ x,φ y,−1

)