Page 91 - Vector Analysis

§3.4 The Fubini theorem 87

Example 3.28. Let A = ␣(x, y) P R2 ˇ 0 ď x ď ? ď y ď 1(, and f : A Ñ R be given by
ˇ 1, x

f (x, y) = ey3. Then Corollary 3.26 implies that

ż ż 1 (ż 1 )

f (x, y) dA = ? ey3dy dx .

A 0x

Since we do not know how to compute the inner integral, we look for another way of finding

the integral. Observing that A = ␣(x, y) P R2 ˇ 0 ď y ď 1, 0 ď x ď y2(, we have
ˇ

ż ż 1 (ż y2 ) ż 1 1 ˇy=1 e ´ 1
ey3dx dy ˇ
f (x, y) dA = = y2ey3 dy = ey3 = .
3 ˇy=0 3
A 00 0

Example 3.29. Let A Ď R3 be the set ␣(x1, x2, x3) P R3 ˇ x1 ě 0, x2 ě 0, x3 ě 0, and x1 +
ˇ

x2 + x3 ď 1(, and f : A Ñ R be given by f (x1, x2, x3) = (x1 + x2 + x3)2. Let S =

[0, 1] ˆ [0, 1] ˆ [0, 1], and f : R3 Ñ R be the extension of f by zero outside A. Then

Theorem 3.23 implies that f is Riemann integrable. Write xp1 = (x2, x3), xp2 = (x1, x3) and
xp3 = (x1, x2). Theorem 3.20 implies that

żż

f (x)dx = f (x)dx ,

AS

and Theorem 3.25 implies that

ż ż (ż )

f (x)dx = f (xp3, x3)dxp3 dx3 .

S [0,1] [0,1]ˆ[0,1]

Let Ax3 = ␣(x1, x2) P R2 ˇ x1 ě 0, x2 ě 0, x1 + x2 ď 1 ´ x3(. Then for each x3 P [0, 1],
ˇ

ż ż ż 1´x3 ( ż 1´x3´x2 )

f (xp3, x3)dxp3 = f (xp3, x3)dxp3 = f (x1, x2, x3)dx1 dx2 .

[0,1]ˆ[0,1] Ax3 0 0

Computing the iterated integral, we find that

ż ż 1 [ ż 1´x3 ( ż 1´x3´x2 )]
(x1 + x2 + x3)2dx1 dx2 dx3
f (x)dx =

A 00 0 ]
[ ż 1´x3 dx2 dx3
= ż1 (x1 + x2 + x3)3 ˇx1=1´x3´x2
0 3 ˇ
0
ˇx1=0

ż1 [ ż 1´x3 (1 (x2 + x3 )3 ) ]
3 3 dx2 dx3
= 0 0 ´

= ż1 (1 ´ x3 + x43 ) = 1 ´ 1 + 1 = 15 ´ 10 + 1 = 1 .
4 3 12 dx3 4 6 60 60 10
0