Page 90 - Vector Analysis

86 CHAPTER 3. Multiple Integrals

Then B S ˆ [m, M ] Ď Ť ∆ ˆ [m, M ], and as above

∆PP,∆XB S‰H

ż

1B Sˆ[m,M](z) dz ď ÿ νn(∆) ˆ (M ´ m) ď (M ´ m)U (1S, P) ă ε .

B Sˆ[m,M ] ∆PP,∆XB S‰H

Therefore, B S ˆ [m, M ] has volume zero; thus we establish that f is Riemann integrable

over A.

Next we prove (3.5). Note that A Ď S ˆ [m, M ]; thus Theorem 3.20 and the Fubini

Theorem imply that

żż f A(x, y) d(x, y) = ż (żM )
f (x, y) d(x, y) = f A(x, y) dy dx

A Sˆ[m,M ] Sm

ż (żM )
= f A(x, y) dy dx .

Sm

Noting that [m, M ] has a boundary of volume zero in R, and for each x P S, f A(x, ¨) is

continuous except perhaps at y = φ1(x) and y = φ2(x), Theorem 3.23 implies that f A(x, ¨)

żM żM

is Riemann integrable over [m, M ] for each x P S; thus f A(x, y) dy = f A(x, y) dy

which further implies that mm

ż ż (żM )

f (x, y) d(x, y) = f A(x, y) dy dx . (3.6)

A Sm

For each fixed x P S, let Ax = ␣y P R ˇ φ1(x) ď y ď φ2(x)(. Then f A(x, y) = f (x, y)1Ax(y)
ˇ

for all (x, y) P S ˆ [m, M ] or equivalently, f A(x, ¨) = f (x, ¨)|Ax for all x P S; thus Proposition

3.14 (a) implies that

żM ż ż φ2(x)
f A(x, y) dy = f (x, y) dy =
f (x, y) dy @x P S. (3.7)
˝
m Ax φ1(x)

Combining (3.6) and (3.7), we conclude (3.5).

Example 3.27. Let A = ␣(x, y) P R2 ˇ 0 ď x ď 1, x ď y ď 1(, and f : A Ñ R be given by
ˇ

f (x, y) = xy. Then Corollary 3.26 implies that

ż ż 1(ż 1) ż 1 xy2 ˇy=1 ż 1 (x x3 ) 1 1 1
xy dy dx ˇ dx dx
f (x, y) dA = = = ´ = ´ = .
0 2 ˇy=x 02 2 48 8
A 0x

On the other hand, since A = ␣(x, y) P R2 ˇ 0 ď y ď 1, 0 ď x ď y(, we can also evaluate the
ˇ

integral of f over A by

ż ż 1(ż y ) ż 1 x2y ˇx=y ż 1 y3 1
xy dx dy ˇ dy
xy dA = = 0 2 ˇx=0 = 02 dy = 8 .

A 00