Chapter 4. Multiple Integrals
§1. Integrals on Rectangles
Let R be a rectangle in IR2 given by
R = [a1 , b1 ] × [a2 , b2 ] := {(x, y) ∈ IR2 : a1 ≤ x ≤ b1 , a2 ≤ y ≤ b2 }.
Let P1 be a partition of [a1 , b1 ]:
P1 : a1 = x0 < x1 < · · · < xm = b1 ,
and let P2 be a partition of [a2 , b2 ]:
P2 : a2 = y0 < y1 < · · · < yn = b2 .
For i ∈ {1, . . . , m} and j ∈ {1, . . . , n}, let Cij be the rectangle [xi−1 , xi ]×[yj−1 , yj ]. We will
refer to Cij as the (i, j) call. The area of Cij is ∆Aij = ∆xi ∆yj , where ∆xi := xi − xi−1
p
and ∆yj := yj − yj−1 . The diameter of Cij is d(Cij ) = (xi − xi−1 )2 + (yj − yj−1 )2 . Let
P = P1 × P2 be the collection Cij : i = 1, . . . , m, j = 1, . . . , n . Then P is a partition
of R.
Now let f be a bounded real-valued function on the rectangle R. Given a partition
P = {Cij : i = 1, . . . , m, j = 1, . . . , n of R, let
mij := inf{f (x, y) : (x, y) ∈ Cij } and Mij := sup{f (x, y) : (x, y) ∈ Cij }.
The upper sum U (f, P ) and the lower sum L(f, P ) for the function f and the partition
P are defined by
U (f, P ) :=
m X
n
X
Mij ∆Aij
and L(f, P ) :=
i=1 j=1
m X
n
X
i=1 j=1
The upper integral U (f ) of f over R is defined by
U (f ) := inf{U (f, P ) : P is a partition of R}
and the lower integral L(f ) of f over R is defined by
L(f ) := sup{L(f, P ) : P is a partition of R}.
1
mij ∆Aij .
If P 0 = P10 × P20 and P 00 = P100 × P200 are partitions of R, then L(f, P 0 ) ≤ U (f, P 00 ). In
order to prove this assertion, we let P1 be a common refinement of P10 and P100 , and let P2
be a common refinement of P20 and P200 . Then for P := P1 × P2 we have
L(f, P 0 ) ≤ L(f, P ) ≤ U (f, P ) ≤ U (f, P 00 ).
Consequently, L(f ) ≤ U (f ).
A bounded function f on R is said to be Riemann integrable if L(f ) = U (f ). In
this case, we write
ZZ
f (x, y) dA := L(f ) = U (f ).
R
Theorem 1.1. A bounded function f on a rectangle R is integrable if and only if for each
ε > 0 there exists a partition P of R such that
U (f, P ) − L(f, P ) < ε.
Proof. Suppose that f is integrable on R. For ε > 0, there exist
such that
ε
and U (f, P 00 ) < U (f ) +
L(f, P 0 ) > L(f ) −
2
Let P a common refinement of P 0 and P 00 . Then we have
L(f ) −
partitions P 0 and P 00
ε
.
2
ε
ε
< L(f, P 0 ) ≤ L(f, P ) ≤ U (f, P ) ≤ U (f, P 00 ) < U (f ) + .
2
2
Since L(f ) = U (f ), it follows that U (f, P ) − L(f, P ) < ε.
Conversely, suppose that for each ε > 0 there exists a partition P of R such that
U (f, P ) − L(f, P ) < ε. Then we have
U (f ) ≤ U (f, P ) = U (f, P ) − L(f, P ) + L(f, P ) < ε + L(f, P ) ≤ ε + L(f ).
Since ε > 0 is arbitrary, we conclude that U (f ) ≤ L(f ). Hence f is integrable.
Theorem 1.2. If f is a continuous function on a rectangle R, then it is integrable on R.
Proof. Consider ε > 0. Since f is uniformly continuous on R, there exists some δ > 0
such that
(x, y), (x0 , y 0 ) ∈ R
and
p
(x − x0 )2 + (y − y 0 )2 < δ
imply
|f (x, y) − f (x0 , y 0 )| <
ε
,
A
where A is the area of the rectangle R. Let P = {Cij : i = 1, . . . , m, j = 1, . . . , n}
be a partition of R such that d(Cij ) < δ for all i = 1, . . . , m and j = 1, . . . , n. Let
2
mij := inf{f (x, y) : (x, y) ∈ Cij } and Mij := sup{f (x, y) : (x, y) ∈ Cij }. Since f
attains its maximum and minimum on each closed cell Cij , we have Mij − mij < ε/A for
i = 1, . . . , m and j = 1, . . . , n. Consequently,
U (f, P ) − L(f, P ) =
m X
n
X
m
(Mij − mij )∆Aij <
i=1 j=1
n
ε XX
∆Aij = ε.
A i=1 j=1
By Theorem 1.1 we conclude that f is integrable.
The following elementary properties of the integral can be easily established.
Theorem 1.3. Let f and g be integrable functions on a rectangle R and let c be a real
number. Then
RR
RR
(1) cf is integrable on R and R (cf )(x, y) dA = c R f (x, y) dA;
RR
RR
RR
(2) f + g is integrable on R and R (f + g)(x, y) dA = R f (x, y) dA + R g(x, y) dA;
RR
RR
(3) if f (x, y) ≤ g(x, y) for all (x, y) ∈ R, then R f (x, y) dA ≤ R g(x, y) dA.
§2. Repeated Integrals
The following theorem demonstrates that the evaluation of the double integral of an
integrable function on a rectangle can be reduced to repeated integrals.
Theorem 2.1. Let f be a bounded, real-valued function that is integrable on a rectangle
R = [a1 , b1 ] × [a2 , b2 ]. Suppose that, for each y in [a2 , b2 ], the function f (x, y) is an
Rb
integrable function of x on [a1 , b1 ]. Then the function F (y) := a11 f (x, y) dx is an integrable
function of y on [a2 , b2 ] and
ZZ
Z
b2 Z b1
f (x, y) dA =
R
a2
f (x, y) dx dy.
a1
Proof. Let ε > 0 be given. Since the function f is integrable on R, there exists a partition
P of the rectangle R such that U (f, P ) − L(f, P ) < ε, by Theorem 1.1. Suppose that
P = {Cij : i = 1, . . . , m, j = 1, . . . , n}, where each Cij = [xi−1 , xi ] × [yj−1 , yj ] with
a1 = x0 < x1 < · · · < xm = b1 and a2 = y0 < y1 < · · · < yn = b2 . For i = 1, . . . , m and
j = 1, . . . , n, let mij := inf{f (x, y) : (x, y) ∈ Cij } and Mij := sup{f (x, y) : (x, y) ∈ Cij }.
Moreover, let kj := inf{F (y) : yj−1 ≤ y ≤ yj } and Kj := sup{F (y) : yj−1 ≤ y ≤ yj } for
j = 1, . . . , n. If yj−1 ≤ y ≤ yj , then mij ≤ f (x, y) ≤ Mij for x ∈ [xi−1 , xi ], i = 1, . . . , m.
Hence,
Z b1
m
m
X
X
mij ∆xi ≤ F (y) =
f (x, y) dx ≤
Mij ∆xi .
i=1
a1
3
i=1
It follows that
L(f, P ) =
n X
m
X
mij ∆xi ∆yj ≤
j=0 i=0
n
X
kj ∆yj ≤ L(F, [a2 , b2 ])
j=0
≤ U (F, [a2 , b2 ]) ≤
n
X
n X
m
X
Kj ∆yj ≤
j=0
Mij ∆xi ∆yj = U (f, P ).
j=0 i=0
Since U (f, P ) − L(f, P ) < ε, it follows that U (F, [a2 , b2 ]) − L(F, [a2 , b2 ]) < ε whenever
ε > 0. Therefore, U (F, [a2 , b2 ]) = L(F, [a2 , b2 ]). In other words, F is integrable on [a2 , b2 ].
Consequently,
Z
b2
L(f, P ) ≤
ZZ
F (y) dy ≤ U (f, P )
and L(f, P ) ≤
a2
f (x, y) dA ≤ U (f, P ).
R
Since U (f, P ) − L(f, P ) < ε, we conclude that
ZZ
Z
−ε <
b2
f (x, y) dA −
F (y) dy < ε
R
for any ε > 0. This shows that
RR
a2
f (x, y) dA =
R
R b2
F (y) dy.
a2
Of course, the symmetric version of the theorem holds. If f is integrable on the
rectangle R = [a1 , b1 ] × [a2 , b2 ], and if for each x in [a1 , b1 ], the function f (x, y) is an
Rb
integrable function of y on [a2 , b2 ]. Then the function G(x) := a22 f (x, y) dy is an integrable
function of x on [a1 , b1 ] and
ZZ
Z
b1 Z b2
f (x, y) dA =
R
f (x, y) dy dx.
a1
a2
Example. Let f (x, y) := xexy for (x, y) ∈ R = [0, 2] × [0, 1]. Since f is continuous on R,
we may evaluate the double integral by repeated integrals in either order. We have
ZZ
xy
xe
Z 2 Z
dA =
R
1
xy
xe
0
Z
=
Z
dy dx =
0
0
2
exy
1
0
dx
2
(ex − 1) dx = [ex − x]20 = (e2 − 2) − (1 − 0) = e2 − 3.
0
4
§3. Riemann Domains
Given a subset E of a metric space (X, ρ), we use E ◦ to denote the set of all interior
points of E. Then Bd(E) := E \ E ◦ is the set of all boundary points of E.
In what follows, by a closed rectangle we mean a set of the form [a1 , b1 ] × [a2 , b2 ],
where −∞ < a1 < b1 < ∞ and −∞ < a2 < b2 < ∞. The area of a closed rectangle
R = [a1 , b1 ] × [a2 , b2 ] is defined to be
A(R) := (b1 − a1 )(b2 − a2 ).
A subset E of IR2 is called a null set if for any ε > 0 there exists a finite collection
Pm
{R1 , R2 , . . . , Rm } of closed rectangles such that E ⊆ ∪m
i=1 Ri and
i=1 A(Ri ) < ε. In the
◦
above definition, the rectangles R1 , . . . , Rm could be so chosen that E ⊆ ∪m
i=1 Ri . Indeed,
0
} of
if E is a null set, then for any ε > 0, there exists a finite collection {R10 , R20 , . . . , Rm
P
m
m
0
0
closed rectangles such that E ⊆ ∪i=1 Ri and i=1 A(Ri ) < ε/2. For each rectangle Ri0
we can find a closed rectangle Ri such that Ri0 ⊂ Ri◦ and A(Ri ) < 2A(Ri0 ). Consequently,
Pm
Pm
◦
0
E ⊆ ∪m
i=1 Ri and
i=1 A(Ri ) < 2
i=1 A(Ri ) < ε.
The following properties of null sets can be deduced from the above definition at once.
(1) The empty set is a null set.
(2) If E is a null set, then so is E.
(3) If E is a null set and K ⊆ E, then K is a null set.
(4) If E and F are null sets, then E ∪ F is a null set.
A function g on a closed interval [a, b] is said to be a Lipschitz function if there
exists a real number M > 0 such that |g(s) − g(t)| ≤ M |s − t| whenever a ≤ s, t ≤ b.
Theorem 3.1. Let u1 and u2 be continuous functions on a closed interval [a, b]. If one of
u1 and u2 is a Lipschitz function, then K := {(u1 (t), u2 (t)) : a ≤ t ≤ b} is a null set.
Proof. Without loss of any generality we may assume that u1 is a Lipschitz function on
[a, b]. Thus, there exists a real number M > 0 such that |u1 (s)−u1 (t)| ≤ M |s−t| whenever
a ≤ s, t ≤ b. Let ε > 0 be given. Since u2 is uniformly continuous on [a, b], there exists
δ > 0 such that
s, t ∈ [a, b] and |s − t| < δ
|u2 (s) − u2 (t)| < η :=
imply
ε
.
4M (b − a)
Choose k in IN such that h := (b − a)/k < δ. Partition the interval [a, b] at the equally
spaced points tj = a + jh, j = 0, 1, . . . , k. Let
Rj := [u1 (tj ) − M h, u1 (tj ) + M h] × [u2 (tj ) − η, u2 (tj ) + η],
5
j = 1, . . . , k.
If tj−1 ≤ t ≤ tj for some j, then |t − tj | ≤ h < δ. Consequently,
|u1 (t) − u1 (tj )| ≤ M |t − tj | ≤ M h
and |u2 (t) − u2 (tj )| ≤ η.
Hence, (u1 (t), u2 (t)) ∈ Rj . This shows that K ⊆ ∪kj=1 Rj . Moreover, we have
k
X
A(Rj ) = k(2M h)(2η) < 4kM
j=1
b−a
ε
= ε.
k 4M (b − a)
Therefore, K is a null set.
A bounded set E in IR2 is called a Riemann domain if its boundary Bd(E) is a null
set. Clearly, a null set is a Riemann domain.
Theorem 3.2. If E and F are Riemann domains in IR2 , then E ∪ F , E ∩ F , and E \ F
are all Riemann domains.
Proof. It suffices to show that the following three relations hold for two subsets E and F
of a metric space:
Bd(E∪F ) ⊆ Bd(E)∪Bd(F ),
Bd(E∩F ) ⊆ Bd(E)∪Bd(F ),
Bd(E\F ) ⊆ Bd(E)∪Bd(F ).
First, since E ∪ F = E ∪ F and (E ∪ F )◦ ⊇ E ◦ ∪ F ◦ , we have
Bd(E ∪ F ) = E ∪ F \ (E ∪ F )◦ ⊆ (E ∪ F ) \ (E ◦ ∪ F ◦ ) = (E \ (E ◦ ∪ F ◦ )) ∪ (F \ (E ◦ ∪ F ◦ )).
But E \ (E ◦ ∪ F ◦ ) ⊆ E \ E ◦ = Bd(E) and F \ (E ◦ ∪ F ◦ ) ⊆ F \ F ◦ = Bd(F ). This shows
that Bd(E ∪ F ) ⊆ Bd(E) ∪ Bd(F ).
Second, since E ∩ F ⊆ E ∩ F and E ◦ ∩ F ◦ = (E ∩ F )◦ , we have
Bd(E ∩ F ) = E ∩ F \ (E ∩ F )◦ ⊆ (E ∩ F ) \ (E ◦ ∩ F ◦ ) = ((E ∩ F ) \ E ◦ ) ∪ ((E ∩ F ) \ F ◦ ).
But (E ∩ F ) \ E ◦ ⊆ E \ E ◦ = Bd(E) and (E ∩ F ) \ F ◦ ⊆ F \ F ◦ = Bd(F ). This shows
that Bd(E ∩ F ) ⊆ Bd(E) ∩ Bd(F ).
Third, we observe that
Bd(E \ F ) = E \ F \ (E \ F )◦ ⊆ E ⊆ (E \ E ◦ ) ∪ (E ◦ \ F ) ∪ (F \ F ◦ ) ∪ F ◦ .
Since E ◦ \ F is an open set and E ◦ \ F ⊆ E \ F , we have E ◦ \ F ⊆ (E \ F )◦ . Moreover,
since F ◦ is an open set and F ◦ ∩ (E \ F ) ⊆ F ∩ (E \ F ) = ∅, we have F ◦ ⊆ (E \ F )c . This
shows that (E ◦ \ F ) ∩ Bd(E \ F ) = ∅ and F ◦ ∩ Bd(E \ F ) = ∅. Therefore,
Bd(E \ F ) ⊆ (E \ E ◦ ) ∪ (F \ F ◦ ) = Bd(E) ∪ Bd(F ).
The proof of the theorem is complete.
6
§4. Integrals on General Domains
In this section we study integrals on general domains. First, we establish the following
theorem which is an extension of Theorem 1.2.
Theorem 4.1. Let f be a bounded, real-valued function defined on a closed rectangle R.
Let E denote the set of points in R where f is discontinuous. If E is a null set, then f is
integrable on R.
Proof. Since f is bounded, there exists a positive number M such that −M ≤ f (x, y) ≤ M
for all (x, y) ∈ R. Let ε > 0 be given. Since E is a null set, there exist closed rectangles
Ps
R1 , . . . , Rs such that E ⊆ ∪sr=1 Rr◦ and r=1 A(Rr ) < ε. Without loss of any generality
we may assume that Rr ⊆ R for r = 1, . . . , s. Let K := R \ ∪sr=1 Rr◦ . Then K is a bounded
closed set. Since f is continuous on K, there exists some δ > 0 such that
p
(x − x0 )2 + (y − y 0 )2 < δ imply |f (x, y) − f (x0 , y 0 )| < ε.
(x, y), (x0 , y 0 ) ∈ K and
Let P = {Cij : i = 1, . . . , m, j = 1, . . . , n} be a partition of R such that d(Cij ) < δ for all
i = 1, . . . , m and j = 1, . . . , n and that each Rr (r = 1, . . . , s) is the union of certain cells
Cij . Let mij := inf{f (x, y) : (x, y) ∈ Cij } and Mij := sup{f (x, y) : (x, y) ∈ Cij }. Then
we have
U (f, P ) − L(f, P ) =
n
m X
X
(Mij − mij )∆Aij =
i=1 j=1
X
(Mij − mij )∆Aij ,
(i,j)∈Γ
where Γ is the index set {(i, j) : i = 1, . . . , m, j = 1, . . . , n}. Let Γ1 be the set of all those
indices (i, j) for which Cij is a subset of some Rr . Then Γ = Γ1 ∪ Γ2 , where Γ2 := Γ \ Γ1 .
If (i, j) ∈ Γ2 , then Cij ∩ (∪sr=1 Rr◦ ) = ∅. In other words, Cij ⊆ K whenever (i, j) ∈ Γ2 .
Thus, Mij − mij < ε for (i, j) ∈ Γ2 . Consequently,
X
X
(Mij − mij )∆Aij < ε
∆Aij ≤ εA,
(i,j)∈Γ2
(i,j)∈Γ2
where A = A(R) is the area of the rectangle R. Furthermore, since −M ≤ f (x, y) ≤ M
for all (x, y) ∈ R, we have
X
X
(Mij − mij )∆Aij ≤ 2M
A(Cij ).
(i,j)∈Γ1
(i,j)∈Γ1
But ∪(i,j)∈Γ1 Cij ⊆ ∪sr=1 Rr . Hence,
X
A(Cij ) ≤
s
X
r=1
(i,j)∈Γ1
7
A(Rr ) < ε.
Combining the above estimates, we obtain
X
(Mij − mij )∆Aij =
(i,j)∈Γ
X
(Mij − mij )∆Aij +
(i,j)∈Γ1
X
(Mij − mij )∆Aij ≤ (A + 2M )ε.
(i,j)∈Γ2
Therefore, by Theorem 1.1 we conclude that f is integrable.
Let f be a bounded real-valued function defined on a bounded subset E of IR2 . Choose
a closed rectangle R such that R ⊇ E. Let f˜ be the function on R given by
f (x, y) for (x, y) ∈ E,
˜
f (x, y) :=
0
for (x, y) ∈ R \ E.
If f˜ is integrable on R, then we say that f is integrable on E and define
ZZ
ZZ
f (x, y) dA :=
f˜(x, y) dA.
E
R
Evidently, the above definition is independent of the choice of the rectangle R.
If E is a null set, then
ZZ
f (x, y) dA = 0.
E
To prove this assertion, we choose a closed rectangle R such that R◦ ⊃ E. Let f˜ be the
function on R defined by f˜(x, y) = f (x, y) for (x, y) ∈ E and f˜(x, y) = 0 for (x, y) ∈ R \ E.
Since R◦ \ E is an open set, f˜ is continuous on R \ E. Moreover, E is a null set because
E is a null set. Thus, the set of points in R where f˜ is discontinuous is a null set. In light
RR
of the proof of Theorem 4.1, f˜ is integrable on R and R f˜(x, y) dA = 0.
The following theorem is an extension of Theorem 1.3 to integrals on general domains.
Theorem 4.2. Let f and g be integrable functions on a bounded set E in IR2 and let c
be a real number. Then
RR
RR
(1) cf is integrable on E and E (cf )(x, y) dA = c E f (x, y) dA;
RR
RR
RR
(2) f + g is integrable on E and E (f + g)(x, y) dA = E f (x, y) dA + E g(x, y) dA;
RR
RR
(3) if f (x, y) ≤ g(x, y) for all (x, y) ∈ E, then E f (x, y) dA ≤ E g(x, y) dA.
The following theorem gives a useful property of integrals.
Theorem 4.3. Let f be a bounded function on E = E1 ∪ E2 , where E1 and E2 are
bounded sets in IR2 such that E1 ∩ E2 is a null set. If f is integrable on both E1 and E2 ,
then f is integrable on E and
ZZ
ZZ
ZZ
f (x, y) dA =
f (x, y) dA +
f (x, y) dA.
E
E1
E2
8
Proof. Choose a closed rectangle R such that R ⊃ E. Let g, g1 , g2 , g3 be the functions
on R defined as follows:
f (x, y) for (x, y) ∈ E,
0
for (x, y) ∈ R \ E,
f (x, y) for (x, y) ∈ E1 ,
0
for (x, y) ∈ R \ E1 ,
f (x, y) for (x, y) ∈ E2 ,
0
for (x, y) ∈ R \ E2 ,
−f (x, y)
0
g(x, y) :=
g1 (x, y) :=
g2 (x, y) :=
g3 (x, y) :=
for (x, y) ∈ E1 ∩ E2 ,
for (x, y) ∈ R \ (E1 ∩ E2 ).
Then g = g1 + g2 + g3 . By our assumption, g1 and g2 are integrable on R,
ZZ
ZZ
ZZ
g1 (x, y) dA =
R
f (x, y) dA
and
ZZ
g2 (x, y) dA =
E1
R
Moreover, since E1 ∩ E2 is a null set, g3 is integrable on R and
Therefore, g = g1 + g2 + g3 is integrable on R and
ZZ
ZZ
f (x, y) dA =
E
ZZ
g(x, y) dA =
R
RR
g (x, y) dA
R 3
= 0.
ZZ
g1 (x, y) dA +
R
f (x, y) dA.
E2
g2 (x, y) dA.
R
This established the desired result.
Theorem 4.4. Let E be a bounded set and G an open set in IR2 . Suppose that G ⊆ E
and E \ G is a null set. If f is a bounded function on E and if f is continuous on G, then
f is integrable on E.
Proof. Choose a closed rectangle R such that R◦ ⊃ E. Let f˜ be the function on R given
by f˜(x, y) := f (x, y) for (x, y) ∈ E and f˜(x, y) := 0 for (x, y) ∈ R \ E. Let K be the
set of those points in R where f˜ is discontinuous. By our assumption f is continuous on
the open set G. Moreover, f˜(x, y) = 0 for all (x, y) ∈ R◦ \ E. Hence, f˜ is continuous on
the open set R◦ \ E. Furthermore, f˜ is continuous on Bd(R). Therefore, K ⊆ E \ G. By
our assumption, E \ G is a null set. Consequently, K is a null set. By Theorem 4.1, f˜ is
integrable on R. In other words, f is integrable on E.
9
§5. Evaluation of Double Integrals
In this section we discuss how to reduce double integrals on Riemann domains to
repeated integrals.
A set E in IR2 is said to be y-simple if it can be represented as
E = (x, y) ∈ IR2 : a ≤ x ≤ b, φ1 (x) ≤ y ≤ φ2 (x) ,
where φ1 and φ2 are continuous functions on a bounded closed interval [a, b]. It is easily
seen that E is a Riemann domain. Let f be a bounded function on E. If f is continuous
on E ◦ , then f is integrable on E and
ZZ
Z b Z φ2 (x)
f (x, y) dA =
f (x, y) dy dx.
E
a
φ1 (x)
In order to prove this formula, we choose R = [a, b] × [c, d], where
c := inf{φ1 (x) : a ≤ x ≤ b} and d := sup{φ2 (x) : a ≤ x ≤ b}.
Let f˜ be the function on R defined by f˜(x, y) = f (x, y) for (x, y) ∈ E and f˜(x, y) = 0 for
x ∈ R \ E. By Theorem 2.1 we have
ZZ
Z b Z d
f˜(x, y) dA =
f˜(x, y) dy dx.
R
a
c
For each fixed x ∈ [a, b], f˜(x, y) is an integrable function of y on [c, d]. But f˜(x, y) = f (x, y)
for φ1 (x) ≤ y ≤ φ2 (x) and f˜(x, y) = 0 for c ≤ y < φ1 (x) or φ2 (x) < y ≤ d. Hence,
Z d
Z φ2 (x)
˜
f (x, y) dy =
f (x, y) dy.
c
φ1 (x)
Therefore we obtain
ZZ
ZZ
Z b Z
˜
f (x, y) dA =
f (x, y) dA =
E
R
a
φ2 (x)
f (x, y) dy dx.
φ1 (x)
A set E in IR2 is said to be x-simple if it can be represented as
E = (x, y) ∈ IR2 : c ≤ y ≤ d, ψ1 (y) ≤ x ≤ ψ2 (y) ,
where ψ1 and ψ2 are continuous functions on a bounded closed interval [c, d]. Let f be a
bounded function on E. If f is continuous on E ◦ , then f is integrable on E and
ZZ
Z d Z ψ2 (y)
f (x, y) dA =
f (x, y) dx dy.
E
c
ψ1 (y)
10
Example 1. Evaluate the double integral
ZZ
(2xy + y 2 ) dA,
E
where E is the triangle in IR2 with vertices (0, 0), (1, 0), and (1, 2).
Solution. The domain E can be described as
E = {(x, y) ∈ IR2 : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2x}.
Thus, the double integral is evaluated as
ZZ
Z 1 Z
2
2x
2
(2xy + y ) dA =
(2xy + y ) dy dx =
E
0
0
1
Z
20 3
x dx = x4
3
3
0
xy 2 +
0
h5
=
Z 1h
i1
=
0
y 3 iy=2x
dx
3 y=0
5
.
3
Example 2. Evaluate the repeated integral
1
Z
Z
0
1
2
yex dx dy.
y2
R 2
Solution. Note that the integral ex dx cannot be expressed in a closed form. We may
write this repeated integral as a double integral:
1
Z
Z
1
x2
ye
ZZ
dx dy =
y2
0
2
yex dA,
E
where E = {(x, y) ∈ IR2 : 0 ≤ y ≤ 1, y 2 ≤ x ≤ 1}. The domain E is also y-simple and can
be described as
√
E = {(x, y) ∈ IR2 : 0 ≤ x ≤ 1, 0 ≤ y ≤ x}.
We therefore have
Z
1
Z
1
x2
ye
y2
0
Z
1
=
0
Z
=
0
1
√
ZZ
dx dy =
2
yex dA
E
√
Z 1h 2
y x2 iy= x
ye dy dx =
e
dx
2
y=0
0
0
h 1 2 i1
x x2
1
e dx =
ex
= (e − 1).
2
4
4
0
Z
x
x2
Example 3. Let r1 and r2 be two real numbers such that 0 < r1 < r2 . Evaluate the
RR
double integral E xy dA, where E = {(x, y) ∈ IR2 : r12 ≤ x2 + y 2 ≤ r22 , y ≥ 0}.
11
Solution. We observe that E = E1 ∪ E2 ∪ E3 , where
q
E1 := (x, y) ∈ IR2 : −r2 ≤ x ≤ −r1 , 0 ≤ y ≤ r22 − x2 ,
q
q
2
2
2
E2 := (x, y) ∈ IR : −r1 ≤ x ≤ r1 , r1 − x ≤ y ≤ r22 − x2 ,
q
2
E3 := (x, y) ∈ IR : r1 ≤ x ≤ r2 , 0 ≤ y ≤ r22 − x2 .
By Theorem 4.3,
E2
E1
xy dA.
xy dA +
xy dA +
xy dA =
E
ZZ
ZZ
ZZ
ZZ
E3
Each of the domains E1 , E2 , and E3 is y-simple. We have
ZZ
Z −r1 Z √r22 −x2
Z −r1
−(r22 − r12 )2
1
xy dA =
xy dy dx =
x(r22 − x2 ) dx =
,
8
E1
−r2
0
−r2 2
ZZ
Z r1 Z √r22 −x2
Z r1
1
xy dA =
x(r22 − r12 ) dx = 0,
√ 2 2 xy dy dx =
2
E2
−r1
r1 −x
−r1
ZZ
Z r2 Z √r22 −x2
Z r2
(r2 − r12 )2
1
x(r22 − x2 ) dx = 2
.
xy dA =
xy dy dx =
8
E3
r1
0
r1 2
RR
RR
The value of E xy dS is the sum of these three numbers. Therefore E xy dA = 0.
§6. Area
The area of a Riemann domain E in IR2 is defined to be
ZZ
A(E) :=
1 dA.
E
This integral is well defined, since the constant function 1 on a Riemann domain is integrable. Clearly, if E is a null set, then A(E) = 0. Moreover, if E1 and E2 are Riemann
domains, then
A(E1 ∪ E2 ) = A(E1 ) + A(E2 ) − A(E1 ∩ E2 ).
Two sets E1 and E2 in IR2 are said to be nonoverlapping if (E1 ∩ E2 )◦ = ∅. If
E1 and E2 are two nonoverlapping Riemann domains, then Bd(E1 ∩ E2 ) is a null set and
(E1 ∩ E2 )◦ = ∅; hence, A(E1 ∩ E2 ) = 0. Consequently, A(E1 ∪ E2 ) = A(E1 ) + A(E2 ). More
generally, if E1 , . . . , Em are mutually nonoverlapping Riemann domains and E = ∪m
i=1 Ei ,
then
m
X
A(E) =
A(Ei ).
i=1
12
Theorem 6.1. Let E be a Riemann domain in IR2 . For any given ε > 0 there exist two
nonoverlapping closed sets G and H in IR2 such that
(1) Each of the sets G and H is the union of finitely many nonoverlapping squares in IR2 ,
(2) G ⊆ E ◦ ⊆ E ⊆ G ∪ H, and
(3) A(H) < ε.
Proof. Let ε > 0 be given. Since E is a Riemann domain, its boundary Bd(E) is a
null set. Hence, there exists a finite collection {R1 , R2 , . . . , Rm } of closed rectangles such
Pm
that Bd(E) ⊆ ∪m
k=1 Rk and
k=1 A(Rk ) < ε/4. Suppose that Rk = [ak , bk ] × [ck , dk ],
k = 1, . . . , m. Choose h > 0 such that h < min{(bk −ak )/2, (dk −ck )/2} for all k = 1, . . . , m.
Consider squares
Qij := [ih, (i + 1)h] × [jh, (j + 1)h],
(i, j) ∈ ZZ2 .
These squares are mutually nonoverlapping and their union is IR2 . For each k ∈ {1, . . . , m},
let Ik be the set of those indices (i, j) for which Qij ∩ Rk 6= ∅. Let Hk := ∪(i,j)∈Ik Qij . If
(x, y) ∈ Qij and Qij ∩ Rk 6= ∅, then ak − h ≤ x ≤ bk + h and ck − h ≤ y ≤ dk + h. This
shows that Hk ⊆ [ak − h, bk + h] × [ck − h, dk + h]. It follows that
A(Hk ) ≤ (bk − ak + 2h)(dk − ck + 2h) ≤ 2(bk − ak )2(dk − ck ) = 4A(Rk ).
Let H := ∪m
k=1 Hk . Then
A(H) ≤
m
X
A(Hk ) ≤ 4
k=1
m
X
A(Rk ) < ε.
k=1
Now consider those indices (i, j) ∈
/ I := ∪m
/ I, then
k=1 Ik . If (i, j) ∈
Qij ∩ Bd(E) ⊆ Qij ∩ (∪m
k=1 Rk ) = ∅.
Thus, Qij does not contain any boundary point of E. If Qij ∩ E ◦ 6= ∅, then Qij does not
contain any exterior point, for otherwise the line segment joining an interior point of E
and an exterior point of E must intersect the boundary of E. In other words, Qij ⊆ E ◦ .
Let G be the union of those squares Qij for which (i, j) ∈
/ I and Qij ∩ E ◦ 6= ∅. Then G
and H are nonoverlapping, G ⊆ E ◦ and E ⊆ G ∪ H, as desired.
For a vector v in IR2 , we use Tv to denote the mapping from IR2 to IR2 given by
Tv x = x + v, x ∈ IR2 . We call Tv the translation by v. The following theorem asserts
that the area is invariant under translation.
13
Theorem 6.2. Let E be a Riemann domain in IR2 , and let v ∈ IR2 . Then Tv (E) is a
Riemann domain and
A Tv (E) = A(E).
Proof. We observe that Tv is a one-to-one continuous mapping from IR2 onto IR2 and
its inverse mapping is T−v . Thus, Tv (E ◦ ) is the interior of Tv (E) and Tv (Bd(E)) is the
boundary of Tv (E). We write E + v for Tv (E). Let ε > 0 be given. By Theorem 6.1, there
exist two nonoverlapping closed sets G and H in IR2 such that each of the sets G and H
is the union of finitely many nonoverlapping squares in IR2 , G ⊆ E ◦ ⊆ E ⊆ G ∪ H, and
that A(H) < ε. If Q is a square, then Q + v is a square and A(Q + v) = A(Q). Since each
of the sets G and H is the union of finitely many nonoverlapping squares in IR2 , we have
A(G + v) = A(G) and A(H + v) = A(H) < ε. But Bd(E + v) = Bd(E) + v ⊆ H + v.
Consequently, Bd(E + v) is a null set and hence E + v is a Riemann domain. Moreover,
we have
A(G) ≤ A(E) ≤ A(G ∪ H) < A(G) + ε
and
A(G + v) ≤ A(E + v) ≤ A (G + v) ∪ (H + v) < A(G + v) + ε.
Since A(G) = A(G + v), we deduce that for any ε > 0,
−ε < A(E + v) − A(E) < ε.
Therefore, A(E + v) = A(E).
A linear mapping L on IR2 has the form
x
L 1
x2
We have
a11
=
a21
a
det L = 11
a21
a12
a22
x1
,
x2
x1
x2
∈ IR2 .
a12 = a11 a22 − a12 a21 .
a22 If det L 6= 0, then we say that L is nonsingular. Clearly, L is nonsingular if and only if
L is a bijective mapping on IR2 .
If there is a real number λ 6= 0 such that
x
L 1
x2
λ
=
0
0
1
x1
x2
x
or L 1
x2
14
1 0
=
0 λ
x1
,
x2
then L is called an elementary linear mapping of the first type. Let R = [a, b] × [c, d]
be a closed rectangle, If L(x1 , x2 ) = (λx1 , x2 ), then L(R) = [λa, λb] × [c, d] for λ > 0 or
L(R) = [λb, λa] × [c, d] for λ < 0. In both cases we obtain
A L(R) = |λ|(b − a)(d − c) = | det L|A(R).
This is also true if L(x1 , x2 ) = (x1 , λx2 ).
A mapping L on IR2 is called an elementary linear mapping of the second type, if
there is a real number µ such that
x1
1 µ
x1
x1
1 0
x1
L
=
or L
=
.
x2
0 1
x2
x2
µ 1
x2
If L(x1 , x2 ) = (x1 + µx2 , x2 ), we have
L(R) = {(x1 , x2 ) : c ≤ x2 ≤ d, a + µx2 ≤ x1 ≤ b + µx2 }.
This is a x-simple domain and its area is
A L(R) = (b − a)(d − c) = | det L|A(R).
The above relation is also valid if L(x1 , x2 ) = (x1 , x2 + µx1 ).
A mapping L on IR2 is called an elementary linear mapping of the third type if
L(x1 , x2 ) = (x2 , x1 ). In this case, L(R) = [c, d] × [a, b]. It follows that
A L(R) = (b − a)(d − c) = | det L|A(R).
To summarize, we have proved that for every elementary linear mapping L and every
rectangle R in IR2 ,
A L(R) = | det L|A(R).
Note that a nonsingular linear mapping can be represented as a composition of finitely
many elementary linear mappings.
Theorem 6.3. Let E be a Riemann domain in IR2 . If L is a linear mapping on IR2 , then
L(E) is a Riemann domain and
A L(E) = | det L|A(E).
Proof. If det L = 0, then L(E) is included in a line segment. Hence, L(E) is a null set
and A(L(E)) = 0 = | det L|A(E). In what follows, we assume that det L 6= 0, i.e., L is a
nonsingular linear mapping.
15
Suppose that L is an elementary linear mapping on IR2 . Since E is a Riemann domain,
by Theorem 6.1 there exist two nonoverlapping closed sets G and H in IR2 such that
each of the sets G and H is the union of finitely many nonoverlapping squares in IR2 ,
G ⊆ E ◦ ⊆ E ⊆ G ∪ H, and that A(H) < ε. If Q is a square, then A(L(Q)) = | det L|A(Q).
Since each of the sets G and H is the union of finitely many nonoverlapping squares
in IR2 , we have A(L(G)) = | det L|A(G) and A(L(H)) = | det L|A(H) < | det L|ε. But
Bd(L(E)) = L(Bd(E)) ⊆ L(H). Consequently, Bd(L(E)) is a null set and hence L(E) is
a Riemann domain. Moreover, we have
| det L|A(G) ≤ | det L|A(E) < | det L|A(G) + | det L|ε
and
| det L|A(G) = A(L(G)) ≤ A(L(E)) ≤ A L(G) ∪ L(H) < | det L|A(G) + | det L|ε.
We deduce that for any ε > 0,
−| det L|ε < A(L(E)) − | det L|A(E) < | det L|ε.
Therefore, A(L(E)) = | det L|A(E).
Finally, suppose that L is a nonsingular linear mapping on IR2 . Then L can be
represented as
L = Lk · · · L2 L1 ,
where L1 , L2 , . . . , Lk are elementary linear mappings on IR2 . Let Ej := Lj · · · L1 (E) for
j = 1, 2, . . . , k. An inductive argument shows that E1 , E2 , . . . , Ek are Riemann domains.
Moreover, by what has been proved
A(Ej ) = | det Lj |A(Ej−1 ),
j = 1, . . . , k,
where E0 := E. Consequently,
A(L(E)) = A(Ek ) = | det Lk | · · · | det L2 || det L1 |A(E) = | det L|A(E),
where we have used the fact det L = det Lk · · · det L2 det L1 to derive the last equality.
A mapping S from IR2 to IR2 is called an affine mapping if there exist a linear
mapping L on IR2 and a vector v in IR2 such that
Sx = Lx + v,
x ∈ IR2 .
If E is a Riemann domain in IR2 , then Theorems 6.2 and 6.3 tell us that S(E) is a Riemann
domain and
A S(E) = | det L|A(E).
16
§7. Smooth Mappings
In this section we investigate the action of a continuously differentiable mapping on
Riemann domains.
Let φ = (φ1 , φ2 ) be a mapping from an open set U in IR2 to IR2 . Suppose that the
partial derivatives D1 φ1 , D2 φ1 , D1 φ2 , and D2 φ2 exist and are continuous on U . The
Jacobian matrix of φ at x ∈ U is
D1 φ1 (x) D2 φ1 (x)
Dφ(x) :=
.
D1 φ2 (x) D2 φ2 (x)
If x, y ∈ U and kDφ(z)k ≤ M for all z in the line segment [x, y], then the mean value
theorem tells us that kφ(x) − φ(y)k ≤ M kx − yk.
Let ρ be the metric of the Euclidean plane IR2 . For a subset E of IR2 and x ∈ IR2 ,
define ρ(x, E) := inf{ρ(x, y) : y ∈ E}. For r > 0, let Br (E) := {x ∈ IR2 : ρ(x, E) < r}.
Then Br (E) is an open set and Br (E) = {x ∈ IR2 : ρ(x, E) ≤ r}. It can be easily verified
that
Br (E) ⊆ E ∪ Br (Bd(E)).
Moreover, if E is a line segment of length b, then
A(Br (E)) < (b + 2r)2r.
Theorem 7.1. Let φ be a continuously differentiable mapping from an open set U in IR2
to IR2 . If E is a Riemann domain in IR2 such that E ⊂ U , then φ(E) is a Riemann domain
and
ZZ
Jφ (x1 , x2 ) dA.
A φ(E) ≤
E
Proof. Choose r > 0 such that F := Br (E) ⊂ U . Let M := sup{kDφ(x)k : x ∈ F }.
Then M < ∞ because kDφk is continuous on the compact set F . For x, y ∈ F we have
kφ(x) − φ(y)k ≤ M kx − yk. Let ε > 0 be given. There exists δ > 0 such that
x, y ∈ F and kx − yk < δ
imply
kDφ(x) − Dφ(y)k < ε.
√
Let Q be a square of side length h such that 0 < h < δ/ 2 and Q ⊆ F . Choose an
arbitrary point a ∈ Q. Let Sa be the affine mapping on IR2 given by
Sa x := φ(a) + Dφ(a)(x − a),
17
x ∈ IR2 .
Then Sa (Q) is a parallelogram and its area A(Sa (Q)) = |Jφ (a)|A(Q), by Theorem 6.3.
Let ψ := φ − Sa . Then ψ(a) = 0 and Dψ(x) = Dφ(x) − Dφ(a). For x ∈ Q we have
√
kx − ak ≤ 2h < δ; hence,
kDψ(x)k = kDφ(x) − Dφ(a)k < ε.
It follows that
√
kφ(x) − Sa (x)k = kψ(x) − ψ(a)k ≤ εkx − ak < ε 2h
∀ x ∈ Q.
In other words, φ(x) ∈ Bε√2h (Sa (Q)) for all x ∈ Q. Consequently,
φ(Q) ⊆ Bε√2h (Sa (Q)) ⊆ Sa (Q) ∪ Bε√2h (Bd(Sa (Q))).
Note that Bd(Q) is the union of four line segments of length h. Hence, Sa (Bd(Q)) is the
union of four line segments of length ≤ M h. But Bd(Sa (Q)) = Sa (Bd(Q)). Therefore, the
√
√
area of the Riemann domain Bε√2h (Bd(Sa (Q))) is less than 4(M h + 2ε 2h)2ε 2h. We
thereby obtain
A(φ(Q)) ≤ |Jφ (a)|h2 + M 0 εh2 = |Jφ (a)| + M 0 ε A(Q),
(∗)
√
√
where M 0 := 8 2(M + 2 2ε). Let N := sup{|Jφ (x)| : x ∈ F }. Then N < ∞.
√
Let ε, δ, M , M 0 and N be given as above. Choose h such that 0 < h < min{r, δ}/ 2.
Consider squares of the form Cij := [ih, (i + 1)h] × [jh, (j + 1)h] for (i, j) ∈ ZZ2 . Let
Γ := {(i, j) ∈ ZZ2 : Cij ∩ E 6= ∅}.
In light of our choice of h, Cij ⊆ F whenever (i, j) ∈ Γ. Moreover, E ⊆ ∪(i,j)∈Γ Cij . It
follows that φ(E) ⊆ ∪(i,j)∈Γ φ(Cij ). For each (i, j) ∈ Γ, choose aij ∈ Cij such that
|Jφ (aij )| = inf{|Jφ (x)| : x ∈ Cij }.
By the estimate (∗) we have
A(φ(Cij )) ≤ |Jφ (aij )| + M 0 ε A(Cij ).
Let
Γ1 := {(i, j) ∈ ZZ2 : Cij ∩ Bd(E) 6= ∅}
18
and H := ∪(i,j)∈Γ1 Cij .
Since E is a Riemann domain, E \ E ◦ = Bd(E) is a null set. Therefore, h > 0 can be
chosen so small that A(H) < ε. Consequently,
X
X A(φ(Cij )) ≤
|Jφ (aij )| + M 0 ε A(Cij ) ≤ (N + M 0 ε)A(H) < ε(N + M 0 ε).
(i,j)∈Γ1
(i,j)∈Γ1
It follows that A(φ(Bd(E))) < ε(N + M 0 ε) whenever ε > 0. Hence, φ(Bd(E)) is a null set.
Let Γ2 := Γ \ Γ1 . For (i, j) ∈ Γ2 we have Cij ⊆ E ◦ . Furthermore,
X
X
X
A(φ(E)) ≤
A(φ(Cij )) =
A(φ(Cij )) +
A(φ(Cij )).
(i,j)∈Γ
(i,j)∈Γ1
(i,j)∈Γ2
The first sum was estimated above. For the second sum we have the following estimate:
X
X
X
A(φ(Cij )) ≤
|Jφ (aij )|A(Cij ) + M 0 ε
A(Cij ).
(i,j)∈Γ2
P
Note that
X
(i,j)∈Γ2
(i,j)∈Γ2
A(Cij ) ≤ A(E). Since |Jφ (aij )| ≤ |Jφ (x)| for all x ∈ Cij , we have
ZZ
X ZZ
|Jφ (aij )|A(Cij ) ≤
|Jφ (x1 , x2 )| dA ≤
|Jφ (x1 , x2 )| dA.
(i,j)∈Γ2
(i,j)∈Γ2
(i,j)∈Γ2
Cij
E
Combining the above estimates, we obtain
ZZ
A(φ(E)) ≤
|Jφ (x1 , x2 )| dA + M 0 εA(E) + ε(N + M 0 ε).
E
Since this estimate is valid whenever ε > 0, we conclude that
ZZ
A(φ(E)) ≤
|Jφ (x1 , x2 )| dA.
E
It remains to show that φ(E) is a Riemann domain. Let V := {x ∈ U : Jφ (x) 6= 0}
and K := {x ∈ E : Jφ (x) = 0}. Then V is an open set, while K is a closed set. Clearly,
E ◦ \ K is an open set contained in V . Since Jφ (x) 6= 0 for all x ∈ V , φ|V is an open
mapping, by the inverse mapping theorem. Thus, φ(E ◦ \ K) is an open set contained in
φ(E). It follows that φ(E ◦ \ K) ⊆ (φ(E))◦ . Moreover, since φ is a continuous mapping on
the compact set E, we have φ(E) = φ(E). Consequently,
Bd(φ(E)) = φ(E) \ (φ(E))◦ ⊆ φ(E) \ φ(E ◦ \ K) ⊆ φ(E \ E ◦ ) ∪ φ(K).
We have shown that φ(E \ E ◦ ) is a null set. To prove that φ(K) is a null set, we set
Γ3 := {(i, j) ∈ ZZ2 : Cij ∩ K 6= ∅}.
If (i, j) ∈ Γ3 , then Jφ (aij ) = 0. By the estimate (∗) we have A(φ(Cij )) ≤ M 0 εA(Cij ).
Hence,
X
X
A(φ(K)) ≤
A(φ(Cij )) ≤ M 0 ε
A(Cij ) ≤ M 0 εA(F ).
(i,j)∈Γ3
(i,j)∈Γ3
This shows that φ(K) is a null set. Consequently, Bd(φ(E)) is a null set.
19
§8. Change of Variables in Double Integrals
In this section we establish a general formula for change of variables in double integrals.
Theorem 8.1. Let U be an open set in IR2 , and let E be a closed Riemann domain such
that E ⊂ U . Suppose that φ is a continuously differentiable mapping from U to IR2 . If f
is a nonnegative continuous function on the domain φ(E), then
ZZ
ZZ
f φ(x1 , x2 ) Jφ (x1 , x2 ) dx1 dx2 .
f (u1 , u2 ) du1 du2 ≤
φ(E)
E
Proof. Let ε > 0 be given. Since f ◦ φ is continuous on the compact set E, there exists
δ > 0 such that
x, y ∈ E and kx − yk < δ
|f (φ(x)) − f (φ(y))| < ε.
imply
Partition the domain E into mutually disjoint Riemann domains E1 , E2 , . . . , En such that
E = ∪nj=1 Ej and the diameter of each domain Ej is less than δ. For j = 1, . . . , n, let
Mj := sup{f (φ(x)) : x ∈ Ej } and mj := inf{f (φ(x)) : x ∈ Ej }.
Since the diameter of each domain Ej is less than δ, we have Mj − mj ≤ ε for j = 1, . . . , n.
By our assumption, f is nonnegative. Moreover, φ(E) ⊆ ∪nj=1 φ(Ej ). Hence,
ZZ
f (u1 , u2 ) du1 du2 ≤
φ(E)
n ZZ
X
j=1
f (u1 , u2 ) du1 du2 ≤
φ(Ej )
n
X
Mj A(φ(Ej )).
j=1
By Theorem 7.1 we assert that
n
X
Mj A(φ(Ej )) ≤
j=1
n ZZ
X
j=1
Mj |Jφ (x1 , x2 )| dx1 dx2 .
Ej
Since Mj ≤ mj + ε ≤ f (φ(x1 , x2 )) + ε for all (x1 , x2 ) ∈ Ej , we obtain
ZZ
f (u1 , u2 ) du1 du2 ≤
φ(E)
n
X
ZZ
Mj A(φ(Ej )) ≤
f φ(x1 , x2 ) + ε Jφ (x1 , x2 ) dx1 dx2 .
E
j=1
The desired result follows after letting ε → 0+ .
Theorem 8.2. Suppose that φ is a one-to-one mapping from an open set U in IR2 onto
an open set V in IR2 . Let E be a closed Riemann domain such that E ⊂ U . If both φ and
20
φ−1 are continuously differentiable, and if f is a continuous function on the domain φ(E),
then
ZZ
ZZ
f (u1 , u2 ) du1 du2 =
f φ(x1 , x2 ) Jφ (x1 , x2 ) dx1 dx2 .
φ(E)
E
Proof. First, consider the case that f is nonnegative. By Theorem 8.1,
ZZ
ZZ
f φ(x1 , x2 ) Jφ (x1 , x2 ) dx1 dx2 .
f (u1 , u2 ) du1 du2 ≤
φ(E)
E
On the other hand, applying Theorem 8.1 to the mapping φ−1 and the function (f ◦ φ)|Jφ |
on E = φ−1 (φ(E)), we obtain
ZZ
φ−1 (φ(E))
ZZ
f φ(x1 , x2 ) Jφ (x1 , x2 ) dx1 dx2
f φ ◦ φ−1 (u1 , u2 ) Jφ (φ−1 (u1 , u2 ))Jφ−1 (u1 , u2 ) du1 du2 .
≤
φ(E)
But φ ◦ φ−1 is the identity mapping on V . By the chain rule we have
Jφ (φ−1 (u1 , u2 ))Jφ−1 (u1 , u2 ) = 1.
Consequently,
ZZ
f φ(x1 , x2 ) Jφ (x1 , x2 ) dx1 dx2 ≤
E
ZZ
f (u1 , u2 ) du1 du2 .
φ(E)
Thus, the change of variable formula as stated in the theorem is valid for the case that f
is nonnegative.
For the general case, we may write f = f + − f − , where f + := (|f | + f )/2 and
f − := (|f | − f )/2. Then both f + and f − are nonnegative continuous functions on φ(E).
The change of variable formula is valid for both f + and f − . Therefore, we conclude that
it is also valid for f .
For the special case that f = 1 on φ(E), Theorem 8.2 yields the following result:
ZZ
A(φ(E)) =
Jφ (x1 , x2 ) dx1 dx2 .
E
The following stronger version of Theorem 8.2 is often used in applications of change
of variables for double integrals.
21
Theorem 8.3. Suppose that φ is a one-to-one mapping from an open Riemann domain
U in IR2 onto an open Riemann domain V in IR2 . Let f be a bounded continuous function
on V . If both φ and φ−1 are continuously differentiable, and if |Jφ | is bounded on U , then
ZZ
ZZ
f φ(x1 , x2 ) Jφ (x1 , x2 ) dx1 dx2 .
f (u1 , u2 ) du1 du2 =
V
U
Proof. Let I1 and I2 denote the integral on the left and the right of the above equation respectively. By our assumption, there exists a positive real number M such that
|f (u1 , u2 )| ≤ M for all (u1 , u2 ) ∈ V and f φ(x1 , x2 ) Jφ (x1 , x2 ) ≤ M for all (x1 , x2 ) ∈ U .
Let ε > 0 be given. Since U and V are open Riemann domains, there exists a compact
Riemann domain K of U such that A(U \ K) < ε and A(V \ φ(K)) < ε. By Theorem 8.2
we obtain
ZZ
ZZ
f (u1 , u2 ) du1 du2 =
f φ(x1 , x2 ) Jφ (x1 , x2 ) dx1 dx2 .
φ(K)
K
On the other hand we have
Z Z
ZZ
I1 −
=
f
(u
,
u
)
du
du
1
2
1
2
V \φ(K)
φ(K)
f (u1 , u2 ) du1 du2 ≤ M A(V \ φ(K)) < M ε.
A similar argument shows that
ZZ
I2 −
f φ(x1 , x2 ) Jφ (x1 , x2 ) dx1 dx2 ≤ M A(U \ K) < M ε.
U
Consequently, |I1 − I2 | < 2M ε for any ε > 0. Therefore, I1 = I2 .
We are in a position to investigate double integrals in polar coordinates. Consider the
mapping φ : (r, θ) 7→ (x, y) from IR2 to IR2 given by
x = r cos θ,
y = r sin θ,
(r, θ) ∈ IR2 .
The Jcobian determinant of φ is
∂x
∂r
Jφ (r, θ) = ∂y
∂r
∂x
∂θ
∂y
∂θ
cos θ
=
sin θ
−r sin θ = r.
r cos θ Thus Jφ (r, θ) 6= 0 for r 6= 0. But φ is not one-to-one on IR2 . Let r1 , r2 , θ1 and θ2 be real
numbers such that 0 ≤ r1 < r2 and θ1 < θ2 ≤ θ1 + 2π. It is easily seen that φ is one-to-one
22
on the open domain U := {(r, θ) : r1 < r < r2 , θ1 < θ < θ2 }. Hence, if f is a bounded
continuous function on V := φ(U ), then by Theorem 8.3 we obtain
ZZ
Z θ2 Z r 2
f (r cos θ, r sin θ) r dr dθ.
f (x, y) dx dy =
V
θ1
r1
In particular, if r1 = 0, θ1 = 0, and θ2 = 2π, then V = V1 \ {(x, y) : 0 ≤ x ≤ 1, y = 0},
where V1 is the open disk {(x, y) ∈ IR2 : x2 + y 2 ≤ r22 }. Hence, for a bounded continuous
function f on the disk V1 we have
ZZ
ZZ
Z 2π Z r2
f (r cos θ, r sin θ) r dr dθ.
f (x, y) dx dy =
f (x, y) dx dy =
V1
V
0
0
Example 1. Evaluate the double integral
ZZ
2
2
e−x −y dx dy,
V
where V is the open disk {(x, y) : x2 + y 2 < b2 } with b > 0.
Solution. By using polar coordinates we obtain
ZZ
Z 2π Z b
b
2
2
2
f (x, y) dx dy =
e−r r dr dθ = 2π −e−r /2 0 = π(1 − e−b ).
V
0
0
Example 2. Evaluate the double integral
ZZ
xy(x2 + y 2 ) dx dy,
E
where E is the domain in the first quadrant bounded by the curves x2 −y 2 = 1, x2 −y 2 = 4,
xy = 1, and xy = 2.
Solution. Let ψ : (x, y) 7→ (u, v) be the mapping from IR2 to IR2 given by
u = x2 − y 2 ,
v = 2xy,
(x, y) ∈ IR2 .
The mapping ψ is not one-to-one on IR2 , but it is one-to-one on the first quadrant. Indeed,
√
we have x2 + y 2 = u2 + v 2 . Hence, if x > 0 and y > 0, then
qp
qp
2
2
x = ( u + v + u)/2 and y = ( u2 + v 2 − u)/2.
Let φ be the mapping (u, v) 7→ (x, y) as given above. Then φ : ψ(E) → E is the inverse of
the mapping ψ : E → ψ(E). Note that ψ(E) = {(u, v) : 1 ≤ u ≤ 4, 2 ≤ v ≤ 4}. We have
∂u ∂u ∂x ∂y 2x −2y = 4(x2 + y 2 ).
Jψ (x, y) = ∂v ∂v = 2y
2x
∂x
∂y
23
It follows that
Jφ (u, v) =
1
1
1
.
=
= √
2
2
Jψ (x, y)
4(x + y )
4 u2 + v 2
Therefore,
ZZ
2
ZZ
2
xy(x + y ) dx dy =
E
ψ(E)
1
1
vp 2
u + v2 √
du dv =
2
8
4 u2 + v 2
Z 4 Z
4
v dv du =
1
2
9
.
4
Example 3. For a > 0, find the area of the domain
Q := {(x, y) ∈ IR2 : x2/3 + y 2/3 ≤ a2/3 }.
Solution. Consider the mapping φ : (r, t) 7→ (x, y) given by
(r, t) ∈ IR2 .
y = r sin3 t,
x = r cos3 t,
Let E := {(r, t) ∈ IR2 : 0 ≤ r ≤ a, 0 ≤ t ≤ 2π}. It is easily seen that φ(E) = Q. The
Jacobian determinant of φ is
∂x ∂x cos3 t −3r cos2 t sin t ∂r
∂t
= 3r sin2 t cos2 t.
Jφ (r, t) = ∂y ∂y = 3
2
sin
t
3r
sin
t
cos
t
∂r
∂t
Therefore, the area of the domain Q is
ZZ
Z
2π Z a
|Jφ (r, t)| dr dt =
A(Q) = A(φ(E)) =
E
0
3r dr sin2 t cos2 t dt.
0
It follows that
3a2
A(Q) =
2
Z
0
2π
3a2
sin2 (2t)
dt =
4
8
Z
0
2π
2π
3a2 3
1 − cos 4t
dt =
t − sin 4t/4 0 = πa2 .
2
16
8
24