406 CHAPTER 22. CALCULUS OF VECTOR FIELDS

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

1√φ

2x +φ

2y +1

(φ x,φ y,−1

)and so on the bottom surface,

nz =−1√

φ2x +φ

2y +1

Note that here the z component is negative because since it is the outer normal it must pointdown. On the lateral surface, the one where (x,y) ∈ ∂D and z ∈ [φ (x,y) ,ψ (x,y)], nz = 0.

The area element on the top surface is dA=√

ψ2x +ψ2

y +1dxdy while the area element

on the bottom surface is√

φ2x +φ

2y +1dxdy. Therefore, the last expression in (22.3) is of

the form,

∫D

F (x,y,ψ (x,y))

nz︷ ︸︸ ︷1√

ψ2x +ψ2

y +1

dA︷ ︸︸ ︷√ψ2

x +ψ2y +1dxdy+

∫D

F (x,y,φ (x,y))

nz︷ ︸︸ ︷ −1√φ

2x +φ

2y +1

dA︷ ︸︸ ︷√

φ2x +φ

2y +1dxdy

+∫

Lateral surfaceFnz dA,

the last term equaling zero because on the lateral surface, nz = 0. Therefore, this reducesto∫

∂V Fnz dA as claimed. ■The following corollary is entirely similar to the above.

Corollary 22.3.3 If V is cylindrical in the y direction, then∫V

∂F∂y

dV =∫

∂VFny dA

and if V is cylindrical in the x direction, then∫V

∂F∂x

dV =∫

∂VFnx dA

With this corollary, here is a proof of the divergence theorem.

Theorem 22.3.4 Let V be cylindrical in each of the coordinate directions and let F be aC1 vector field defined on V . Then∫

V∇ ·F dV =

∫∂V

F ·ndA.