Page 121 - Vector Analysis

§4.3 The Surface Integrals 117

The area of the triangle with vertices ψ(u0, v0), ψ(u0 + h, v0), ψ(u0, v0 + k) is

A1 = 1 ››(ψ(u0 + h, v0) ´ ) ˆ ( + k) ´ ψ(u0, v0))››R3 .
2 ψ(u0, v0) ψ(u0, v0

By the mean value theorem, for each component j P t1, 2, 3u, we have

ψj(u0 + h, v0) ´ ψj(u0, v0) = ψ,1 (u0 + θ1jh, v0)h ,
ψj(u0, v0 + k) ´ ψj(u0, v0) = ψ,2 (u0, v0 + θ2jk)k

for some θij P (0, 1); thus if ψ is of class C 1,

ψ(u0 + h, v0) ´ ψ(u0, v0) = ψ,1 (u0, v0)h + E1(u0, v0; h)h ,
ψ(u0, v0 + k) ´ ψ(u0, v0) = ψ,2 (u0, v0)k + E2(u0, v0; k)k ,

where E1 and E2 are bounded vector-valued functions satisfying that lim E1(u0, v0; h) = 0
hÑ0

and lim E2(u0, v0; k) = 0. Therefore,
kÑ0

( )( )
ψ(u0 + h, v0) ´ ψ(u0, v0) ˆ ψ(u0, v0 + k) ´ ψ(u0, v0)
lim hk ´ ψ,1 (u0, v0) ˆ ψ,2 (u0, v0) = 0 .

(h,k)Ñ(0,0)

Since ?g = }ψ,1 ˆψ,2 }R3, we have

A1 = 1 ag(u0, v0)hk + f1(u0, v0; h, k)hk
2

for some function f1 which converges to 0 as (h, k) Ñ (0, 0) and is bounded since ∇ψ

is bounded. Similarly, the area of the triangle with vertices ψ(u0 + h, v0), ψ(u0, v0 + k),

ψ(u0 + h, v0 + k) is

A2 = 1 ag(u0, v0 )hk + f2(u0, v0; h, k)hk .
2

Taking (4.10) into account, we find that

the surface area of ( u0 + h] ˆ [v0, v0 + ) = ag(u0, v0)hk + f (u0, v0; h, k)hk (4.11)
ψ [u0, k]

for some bounded function f (¨, ¨; ¨, ¨) which converges to 0 as the last two variables h, k
approach 0.

Now consider the surface area of ψ([a, a + L] ˆ [b, b + W ]). Let ε ą 0 be given. Choose
N ą 0 such that

ˇˇf (u, v; h, k)ˇˇ ă ε @0 ă h ă L , 0 ă k ă W and (u, v) P [a, a + L] ˆ [b, b + W ] ,
2LW N N