Chapter 4. Multiple Integrals 1. Integrals on Rectanglesrjia/Math414/Notes/chap4.pdfChapter 4. Multiple Integrals 1. Integrals on Rectangles Let Rbe a rectangle in IR2 given by R=

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 willrefer to Cij as the (i, j) call. The area of Cij is ∆Aij = ∆xi∆yj , where ∆xi := xi − xi−1and ∆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 partitionP are defined by

U(f, P ) :=m∑i=1

n∑j=1

Mij∆Aij and L(f, P ) :=m∑i=1

n∑j=1

mij∆Aij .

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

If P ′ = P ′1 × P ′2 and P ′′ = P ′′1 × P ′′2 are partitions of R, then L(f, P ′) ≤ U(f, P ′′). Inorder to prove this assertion, we let P1 be a common refinement of P ′1 and P

′′1 , and let P2

be a common refinement of P ′2 and P′′2 . Then for P := P1 × P2 we have

L(f, P ′) ≤ L(f, P ) ≤ U(f, P ) ≤ U(f, P ′′).

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 ∫∫R

f(x, y) dA := L(f) = U(f).

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 partitions P ′ and P ′′

such thatL(f, P ′) > L(f)− ε

2and U(f, P ′′) < U(f) +

ε

2.

Let P a common refinement of P ′ and P ′′. Then we have

L(f)− ε2< L(f, P ′) ≤ L(f, P ) ≤ U(f, P ) ≤ U(f, P ′′) < U(f) + ε

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 > 0such that

(x, y), (x′, y′) ∈ R and√

(x− x′)2 + (y − y′)2 < δ imply |f(x, y)− f(x′, y′)| < εA,

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 fattains its maximum and minimum on each closed cell Cij , we have Mij −mij < ε/A fori = 1, . . . ,m and j = 1, . . . , n. Consequently,

U(f, P )− L(f, P ) =m∑i=1

n∑j=1

(Mij −mij)∆Aij <ε

A

m∑i=1

n∑j=1

∆Aij = ε.

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

(1) cf is integrable on R and∫∫R(cf)(x, y) dA = c

∫∫Rf(x, y) dA;

(2) f + g is integrable on R and∫∫R(f + g)(x, y) dA =

∫∫Rf(x, y) dA+

∫∫Rg(x, y) dA;

(3) if f(x, y) ≤ g(x, y) for all (x, y) ∈ R, then∫∫Rf(x, y) dA ≤

∫∫Rg(x, y) dA.

§2. Repeated Integrals

The following theorem demonstrates that the evaluation of the double integral of anintegrable 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 anintegrable function of x on [a1, b1]. Then the function F (y) :=

∫ b1a1f(x, y) dx is an integrable

function of y on [a2, b2] and∫∫R

f(x, y) dA =∫ b2a2

[∫ b1a1

f(x, y) dx]dy.

Proof. Let ε > 0 be given. Since the function f is integrable on R, there exists a partitionP of the rectangle R such that U(f, P ) − L(f, P ) < ε, by Theorem 1.1. Suppose thatP = {Cij : i = 1, . . . ,m, j = 1, . . . , n}, where each Cij = [xi−1, xi] × [yj−1, yj ] witha1 = x0 < x1 < · · · < xm = b1 and a2 = y0 < y1 < · · · < yn = b2. For i = 1, . . . ,m andj = 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} forj = 1, . . . , n. If yj−1 ≤ y ≤ yj , then mij ≤ f(x, y) ≤ Mij for x ∈ [xi−1, xi], i = 1, . . . ,m.Hence,

m∑i=1

mij∆xi ≤ F (y) =∫ b1a1

f(x, y) dx ≤m∑i=1

Mij∆xi.

3

It follows that

L(f, P ) =n∑j=0

m∑i=0

mij∆xi∆yj ≤n∑j=0

kj∆yj ≤ L(F, [a2, b2])

≤ U(F, [a2, b2]) ≤n∑j=0

Kj∆yj ≤n∑j=0

m∑i=0

Mij∆xi∆yj = U(f, P ).

Since U(f, P ) − L(f, P ) < ε, it follows that U(F, [a2, b2]) − L(F, [a2, b2]) < ε wheneverε > 0. Therefore, U(F, [a2, b2]) = L(F, [a2, b2]). In other words, F is integrable on [a2, b2].Consequently,

L(f, P ) ≤∫ b2a2

F (y) dy ≤ U(f, P ) and L(f, P ) ≤∫∫

R

f(x, y) dA ≤ U(f, P ).

Since U(f, P )− L(f, P ) < ε, we conclude that

−ε <∫∫

R

f(x, y) dA−∫ b2a2

F (y) dy < ε

for any ε > 0. This shows that∫∫Rf(x, y) dA =

∫ b2a2F (y) dy.

Of course, the symmetric version of the theorem holds. If f is integrable on therectangle R = [a1, b1] × [a2, b2], and if for each x in [a1, b1], the function f(x, y) is anintegrable function of y on [a2, b2]. Then the functionG(x) :=

∫ b2a2f(x, y) dy is an integrable

function of x on [a1, b1] and

∫∫R

f(x, y) dA =∫ b1a1

[∫ b2a2

f(x, y) dy]dx.

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

∫∫R

xexy dA =∫ 2

0

[∫ 10

xexy dy

]dx =

∫ 20

[exy

]10dx

=∫ 2

0

(ex − 1) dx = [ex − x]20 = (e2 − 2)− (1− 0) = e2 − 3.

4

§3. Riemann Domains

Given a subset E of a metric space (X, ρ), we use E◦ to denote the set of all interiorpoints 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 rectangleR = [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{R1, R2, . . . , Rm} of closed rectangles such that E ⊆ ∪mi=1Ri and

∑mi=1A(Ri) < ε. In the

above definition, the rectangles R1, . . . , Rm could be so chosen that E ⊆ ∪mi=1R◦i . Indeed,if E is a null set, then for any ε > 0, there exists a finite collection {R′1, R′2, . . . , R′m} ofclosed rectangles such that E ⊆ ∪mi=1R′i and

∑mi=1A(R

′i) < ε/2. For each rectangle R

′i

we can find a closed rectangle Ri such that R′i ⊂ R◦i and A(Ri) < 2A(R′i). Consequently,E ⊆ ∪mi=1R◦i and

∑mi=1A(Ri) < 2

∑mi=1A(R

′i) < 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 thereexists 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 ofu1 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| whenevera ≤ s, t ≤ b. Let ε > 0 be given. Since u2 is uniformly continuous on [a, b], there existsδ > 0 such that

s, t ∈ [a, b] and |s− t| < δ imply |u2(s)− u2(t)| < η :=ε

4M(b− a).

Choose k in IN such that h := (b − a)/k < δ. Partition the interval [a, b] at the equallyspaced points tj = a+ jh, j = 0, 1, . . . , k. Let

Rj := [u1(tj)−Mh, u1(tj) +Mh]× [u2(tj)− η, u2(tj) + η], j = 1, . . . , k.

5

If tj−1 ≤ t ≤ tj for some j, then |t− tj | ≤ h < δ. Consequently,

|u1(t)− u1(tj)| ≤M |t− tj | ≤Mh and |u2(t)− u2(tj)| ≤ η.

Hence, (u1(t), u2(t)) ∈ Rj . This shows that K ⊆ ∪kj=1Rj . Moreover, we have

k∑j=1

A(Rj) = k(2Mh)(2η) < 4kMb− ak

ε

4M(b− a)= ε.

Therefore, K is a null set.

A bounded set E in IR2 is called a Riemann domain if its boundary Bd(E) is a nullset. 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 \ Fare all Riemann domains.

Proof. It suffices to show that the following three relations hold for two subsets E and Fof 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 showsthat 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 showsthat 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. Thisshows 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 followingtheorem 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 numberM such that−M ≤ f(x, y) ≤Mfor all (x, y) ∈ R. Let ε > 0 be given. Since E is a null set, there exist closed rectanglesR1, . . . , Rs such that E ⊆ ∪sr=1R◦r and

∑sr=1A(Rr) < ε. Without loss of any generality

we may assume that Rr ⊆ R for r = 1, . . . , s. Let K := R \∪sr=1R◦r . Then K is a boundedclosed set. Since f is continuous on K, there exists some δ > 0 such that

(x, y), (x′, y′) ∈ K and√

(x− x′)2 + (y − y′)2 < δ imply |f(x, y)− f(x′, y′)| < ε.

Let P = {Cij : i = 1, . . . ,m, j = 1, . . . , n} be a partition of R such that d(Cij) < δ for alli = 1, . . . ,m and j = 1, . . . , n and that each Rr (r = 1, . . . , s) is the union of certain cellsCij . Let mij := inf{f(x, y) : (x, y) ∈ Cij} and Mij := sup{f(x, y) : (x, y) ∈ Cij}. Thenwe have

U(f, P )− L(f, P ) =m∑i=1

n∑j=1

(Mij −mij)∆Aij =∑

(i,j)∈Γ

(Mij −mij)∆Aij ,

where Γ is the index set {(i, j) : i = 1, . . . ,m, j = 1, . . . , n}. Let Γ1 be the set of all thoseindices (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=1R◦r) = ∅. In other words, Cij ⊆ K whenever (i, j) ∈ Γ2.Thus, Mij −mij < ε for (i, j) ∈ Γ2. Consequently,∑

(i,j)∈Γ2

(Mij −mij)∆Aij < ε∑

(i,j)∈Γ2

∆Aij ≤ εA,

where A = A(R) is the area of the rectangle R. Furthermore, since −M ≤ f(x, y) ≤ Mfor all (x, y) ∈ R, we have∑

(i,j)∈Γ1

(Mij −mij)∆Aij ≤ 2M∑

(i,j)∈Γ1

A(Cij).

But ∪(i,j)∈Γ1Cij ⊆ ∪sr=1Rr. Hence,∑(i,j)∈Γ1

A(Cij) ≤s∑r=1

A(Rr) < ε.

7

Combining the above estimates, we obtain∑(i,j)∈Γ

(Mij−mij)∆Aij =∑

(i,j)∈Γ1

(Mij−mij)∆Aij +∑

(i,j)∈Γ2

(Mij−mij)∆Aij ≤ (A+2M)ε.

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. Choosea closed rectangle R such that R ⊇ E. Let f̃ be the function on R given by

f̃(x, y) :={f(x, y) for (x, y) ∈ E,0 for (x, y) ∈ R \ E.

If f̃ is integrable on R, then we say that f is integrable on E and define∫∫E

f(x, y) dA :=∫∫

R

f̃(x, y) dA.

Evidently, the above definition is independent of the choice of the rectangle R.If E is a null set, then ∫∫

E

f(x, y) dA = 0.

To prove this assertion, we choose a closed rectangle R such that R◦ ⊃ E. Let f̃ be thefunction on R defined by f̃(x, y) = f(x, y) for (x, y) ∈ E and f̃(x, y) = 0 for (x, y) ∈ R\E.Since R◦ \ E is an open set, f̃ is continuous on R \ E. Moreover, E is a null set becauseE is a null set. Thus, the set of points in R where f̃ is discontinuous is a null set. In lightof the proof of Theorem 4.1, f̃ is integrable on R and

∫∫Rf̃(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 cbe a real number. Then

(1) cf is integrable on E and∫∫E

(cf)(x, y) dA = c∫∫Ef(x, y) dA;

(2) f + g is integrable on E and∫∫E

(f + g)(x, y) dA =∫∫Ef(x, y) dA+

∫∫Eg(x, y) dA;

(3) if f(x, y) ≤ g(x, y) for all (x, y) ∈ E, then∫∫Ef(x, y) dA ≤

∫∫Eg(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 arebounded 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∫∫

E

f(x, y) dA =∫∫

E1

f(x, y) dA+∫∫

E2

f(x, y) dA.

8

Proof. Choose a closed rectangle R such that R ⊃ E. Let g, g1, g2, g3 be the functionson R defined as follows:

g(x, y) :={f(x, y) for (x, y) ∈ E,0 for (x, y) ∈ R \ E,

g1(x, y) :={f(x, y) for (x, y) ∈ E1,0 for (x, y) ∈ R \ E1,

g2(x, y) :={f(x, y) for (x, y) ∈ E2,0 for (x, y) ∈ R \ E2,

g3(x, y) :={−f(x, y) for (x, y) ∈ E1 ∩ E2,0 for (x, y) ∈ R \ (E1 ∩ E2).

Then g = g1 + g2 + g3. By our assumption, g1 and g2 are integrable on R,∫∫R

g1(x, y) dA =∫∫

E1

f(x, y) dA and∫∫

R

g2(x, y) dA =∫∫

E2

f(x, y) dA.

Moreover, since E1 ∩ E2 is a null set, g3 is integrable on R and∫∫Rg3(x, y) dA = 0.

Therefore, g = g1 + g2 + g3 is integrable on R and∫∫E

f(x, y) dA =∫∫

R

g(x, y) dA =∫∫

R

g1(x, y) dA+∫∫

R

g2(x, y) dA.

This established the desired result.

Theorem 4.4. Let E be a bounded set and G an open set in IR2. Suppose that G ⊆ Eand E \G is a null set. If f is a bounded function on E and if f is continuous on G, thenf is integrable on E.

Proof. Choose a closed rectangle R such that R◦ ⊃ E. Let f̃ be the function on R givenby f̃(x, y) := f(x, y) for (x, y) ∈ E and f̃(x, y) := 0 for (x, y) ∈ R \ E. Let K be theset of those points in R where f̃ is discontinuous. By our assumption f is continuous onthe open set G. Moreover, f̃(x, y) = 0 for all (x, y) ∈ R◦ \ E. Hence, f̃ is continuous onthe open set R◦ \ E. Furthermore, f̃ is continuous on Bd(R). Therefore, K ⊆ E \G. Byour assumption, E \ G is a null set. Consequently, K is a null set. By Theorem 4.1, f̃ isintegrable 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 torepeated 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 easilyseen that E is a Riemann domain. Let f be a bounded function on E. If f is continuouson E◦, then f is integrable on E and∫∫

E

f(x, y) dA =∫ ba

[∫ φ2(x)φ1(x)

f(x, y) dy]dx.

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 f̃ be the function on R defined by f̃(x, y) = f(x, y) for (x, y) ∈ E and f̃(x, y) = 0 forx ∈ R \ E. By Theorem 2.1 we have∫∫

R

f̃(x, y) dA =∫ ba

[∫ dc

f̃(x, y) dy]dx.

For each fixed x ∈ [a, b], f̃(x, y) is an integrable function of y on [c, d]. But f̃(x, y) = f(x, y)for φ1(x) ≤ y ≤ φ2(x) and f̃(x, y) = 0 for c ≤ y < φ1(x) or φ2(x) < y ≤ d. Hence,∫ d

c

f̃(x, y) dy =∫ φ2(x)φ1(x)

f(x, y) dy.

Therefore we obtain∫∫E

f(x, y) dA =∫∫

R

f̃(x, y) dA =∫ ba

[∫ φ2(x)φ1(x)

f(x, y) dy]dx.

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 abounded function on E. If f is continuous on E◦, then f is integrable on E and∫∫

E

f(x, y) dA =∫ dc

[∫ ψ2(y)ψ1(y)

f(x, y) dx]dy.

10

Example 1. Evaluate the double integral∫∫E

(2xy + y2) dA,

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∫∫E

(2xy + y2) dA =∫ 1

0

[∫ 2x0

(2xy + y2) dy]dx =

∫ 10

[xy2 +

y3

3

]y=2xy=0

dx

=∫ 1

0

203x3 dx =

[53x4

]10

=53.

Example 2. Evaluate the repeated integral∫ 10

∫ 1y2yex

2dx dy.

Solution. Note that the integral∫ex

2dx cannot be expressed in a closed form. We may

write this repeated integral as a double integral:∫ 10

∫ 1y2yex

2dx dy =

∫∫E

yex2dA,

where E = {(x, y) ∈ IR2 : 0 ≤ y ≤ 1, y2 ≤ x ≤ 1}. The domain E is also y-simple and canbe described as

E = {(x, y) ∈ IR2 : 0 ≤ x ≤ 1, 0 ≤ y ≤√x}.

We therefore have ∫ 10

∫ 1y2yex

2dx dy =

∫∫E

yex2dA

=∫ 1

0

∫ √x0

yex2dy dx =

∫ 10

[y22ex

2]y=√xy=0

dx

=∫ 1

0

x

2ex

2dx =

[14ex

2]10

=14(e− 1).

Example 3. Let r1 and r2 be two real numbers such that 0 < r1 < r2. Evaluate thedouble integral

∫∫Exy dA, where E = {(x, y) ∈ IR2 : r21 ≤ x2 + y2 ≤ r22, y ≥ 0}.

11

Solution. We observe that E = E1 ∪ E2 ∪ E3, where

E1 :={(x, y) ∈ IR2 : −r2 ≤ x ≤ −r1, 0 ≤ y ≤

√r22 − x2

},

E2 :={(x, y) ∈ IR2 : −r1 ≤ x ≤ r1,

√r21 − x2 ≤ y ≤

√r22 − x2

},

E3 :={(x, y) ∈ IR2 : r1 ≤ x ≤ r2, 0 ≤ y ≤

√r22 − x2

}.

By Theorem 4.3, ∫∫E

xy dA =∫∫

E1

xy dA+∫∫

E2

xy dA+∫∫

E3

xy dA.

Each of the domains E1, E2, and E3 is y-simple. We have∫∫E1

xy dA =∫ −r1−r2

[∫ √r22−x20

xy dy

]dx =

∫ −r1−r2

12x(r22 − x2) dx =

−(r22 − r21)2

8,

∫∫E2

xy dA =∫ r1−r1

[∫ √r22−x2√r21−x2

xy dy

]dx =

∫ r1−r1

12x(r22 − r21) dx = 0,∫∫

E3

xy dA =∫ r2r1

[∫ √r22−x20

xy dy

]dx =

∫ r2r1

12x(r22 − x2) dx =

(r22 − r21)2

8.

The value of∫∫Exy dS is the sum of these three numbers. Therefore

∫∫Exy dA = 0.

§6. Area

The area of a Riemann domain E in IR2 is defined to be

A(E) :=∫∫

E

1 dA.

This integral is well defined, since the constant function 1 on a Riemann domain is inte-grable. Clearly, if E is a null set, then A(E) = 0. Moreover, if E1 and E2 are Riemanndomains, 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)◦ = ∅. IfE1 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). Moregenerally, if E1, . . . , Em are mutually nonoverlapping Riemann domains and E = ∪mi=1Ei,then

A(E) =m∑i=1

A(Ei).

12

Theorem 6.1. Let E be a Riemann domain in IR2. For any given ε > 0 there exist twononoverlapping 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 anull set. Hence, there exists a finite collection {R1, R2, . . . , Rm} of closed rectangles suchthat Bd(E) ⊆ ∪mk=1Rk and

∑mk=1A(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)∈IkQij . If(x, y) ∈ Qij and Qij ∩ Rk 6= ∅, then ak − h ≤ x ≤ bk + h and ck − h ≤ y ≤ dk + h. Thisshows 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 := ∪mk=1Hk. Then

A(H) ≤m∑k=1

A(Hk) ≤ 4m∑k=1

A(Rk) < ε.

Now consider those indices (i, j) /∈ I := ∪mk=1Ik. If (i, j) /∈ I, then

Qij ∩ Bd(E) ⊆ Qij ∩ (∪mk=1Rk) = ∅.

Thus, Qij does not contain any boundary point of E. If Qij ∩ E◦ 6= ∅, then Qij does notcontain any exterior point, for otherwise the line segment joining an interior point of Eand 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 Gand 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 byTvx = x + v, x ∈ IR2. We call Tv the translation by v. The following theorem assertsthat 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 aRiemann domain and

A(Tv(E)

)= A(E).

Proof. We observe that Tv is a one-to-one continuous mapping from IR2 onto IR2 andits inverse mapping is T−v. Thus, Tv(E◦) is the interior of Tv(E) and Tv(Bd(E)) is theboundary of Tv(E). We write E+ v for Tv(E). Let ε > 0 be given. By Theorem 6.1, thereexist two nonoverlapping closed sets G and H in IR2 such that each of the sets G and His the union of finitely many nonoverlapping squares in IR2, G ⊆ E◦ ⊆ E ⊆ G ∪H, andthat A(H) < ε. If Q is a square, then Q+ v is a square and A(Q+ v) = A(Q). Since eachof the sets G and H is the union of finitely many nonoverlapping squares in IR2, we haveA(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

L

[x1x2

]=

[a11 a12a21 a22

] [x1x2

],

[x1x2

]∈ IR2.

We have

detL =∣∣∣∣ a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21.If detL 6= 0, then we say that L is nonsingular. Clearly, L is nonsingular if and only ifL is a bijective mapping on IR2.

If there is a real number λ 6= 0 such that

L

[x1x2

]=

[λ 00 1

] [x1x2

]or L

[x1x2

]=

[1 00 λ

] [x1x2

],

14

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 orL(R) = [λb, λa]× [c, d] for λ < 0. In both cases we obtain

A(L(R)

)= |λ|(b− a)(d− c) = |detL|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

L

[x1x2

]=

[1 µ0 1

] [x1x2

]or L

[x1x2

]=

[1 0µ 1

] [x1x2

].

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) = |detL|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) = |detL|A(R).

To summarize, we have proved that for every elementary linear mapping L and everyrectangle R in IR2,

A(L(R)

)= |detL|A(R).

Note that a nonsingular linear mapping can be represented as a composition of finitelymany elementary linear mappings.

Theorem 6.3. Let E be a Riemann domain in IR2. If L is a linear mapping on IR2, thenL(E) is a Riemann domain and

A(L(E)

)= |detL|A(E).

Proof. If detL = 0, then L(E) is included in a line segment. Hence, L(E) is a null setand A(L(E)) = 0 = |detL|A(E). In what follows, we assume that detL 6= 0, i.e., L is anonsingular 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 thateach 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)) = |detL|A(Q).Since each of the sets G and H is the union of finitely many nonoverlapping squaresin IR2, we have A(L(G)) = |detL|A(G) and A(L(H)) = |detL|A(H) < |detL|ε. ButBd(L(E)) = L(Bd(E)) ⊆ L(H). Consequently, Bd(L(E)) is a null set and hence L(E) isa Riemann domain. Moreover, we have

|detL|A(G) ≤ |detL|A(E) < |detL|A(G) + |detL|ε

and

|detL|A(G) = A(L(G)) ≤ A(L(E)) ≤ A(L(G) ∪ L(H)

)< |detL|A(G) + |detL|ε.

We deduce that for any ε > 0,

−|detL|ε < A(L(E))− |detL|A(E) < |detL|ε.

Therefore, A(L(E)) = |detL|A(E).Finally, suppose that L is a nonsingular linear mapping on IR2. Then L can be

represented asL = Lk · · ·L2L1,

where L1, L2, . . . , Lk are elementary linear mappings on IR2. Let Ej := Lj · · ·L1(E) forj = 1, 2, . . . , k. An inductive argument shows that E1, E2, . . . , Ek are Riemann domains.Moreover, by what has been proved

A(Ej) = |detLj |A(Ej−1), j = 1, . . . , k,

where E0 := E. Consequently,

A(L(E)) = A(Ek) = |detLk| · · · |detL2||detL1|A(E) = |detL|A(E),

where we have used the fact detL = detLk · · ·detL2 detL1 to derive the last equality.

A mapping S from IR2 to IR2 is called an affine mapping if there exist a linearmapping 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 Riemanndomain and

A(S(E)

)= |detL|A(E).

16

§7. Smooth Mappings

In this section we investigate the action of a continuously differentiable mapping onRiemann domains.

Let φ = (φ1, φ2) be a mapping from an open set U in IR2 to IR2. Suppose that thepartial derivatives D1φ1, D2φ1, D1φ2, and D2φ2 exist and are continuous on U . TheJacobian matrix of φ at x ∈ U is

Dφ(x) :=[D1φ1(x) D2φ1(x)D1φ2(x) D2φ2(x)

].

If x, y ∈ U and ‖Dφ(z)‖ ≤ M for all z in the line segment [x, y], then the mean valuetheorem tells us that ‖φ(x)− φ(y)‖ ≤M‖x− y‖.

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 verifiedthat

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 domainand

A(φ(E)

)≤

∫∫E

∣∣Jφ(x1, x2)∣∣ dA.Proof. Choose r > 0 such that F := Br(E) ⊂ U . Let M := sup{‖Dφ(x)‖ : x ∈ F}.Then M < ∞ because ‖Dφ‖ is continuous on the compact set F . For x, y ∈ F we have‖φ(x)− φ(y)‖ ≤M‖x− y‖. Let ε > 0 be given. There exists δ > 0 such that

x, y ∈ F and ‖x− y‖ < δ imply ‖Dφ(x)−Dφ(y)‖ < ε.

Let Q be a square of side length h such that 0 < h < δ/√

2 and Q ⊆ F . Choose anarbitrary point a ∈ Q. Let Sa be the affine mapping on IR2 given by

Sax := φ(a) +Dφ(a)(x− a), x ∈ IR2.

17

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‖x− a‖ ≤

√2h < δ; hence,

‖Dψ(x)‖ = ‖Dφ(x)−Dφ(a)‖ < ε.

It follows that

‖φ(x)− Sa(x)‖ = ‖ψ(x)− ψ(a)‖ ≤ ε‖x− a‖ < 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 theunion of four line segments of length ≤Mh. But Bd(Sa(Q)) = Sa(Bd(Q)). Therefore, thearea of the Riemann domain Bε√2h(Bd(Sa(Q))) is less than 4(Mh + 2ε

√2h)2ε

√2h. We

thereby obtain

A(φ(Q)) ≤ |Jφ(a)|h2 +M ′εh2 =[|Jφ(a)|+M ′ε

]A(Q), (∗)

where M ′ := 8√

2(M + 2√

2ε). Let N := sup{|Jφ(x)| : x ∈ F}. Then N

Since E is a Riemann domain, E \ E◦ = Bd(E) is a null set. Therefore, h > 0 can bechosen so small that A(H) < ε. Consequently,∑

(i,j)∈Γ1

A(φ(Cij)) ≤∑

(i,j)∈Γ1

[|Jφ(aij)|+M ′ε

]A(Cij) ≤ (N +M ′ε)A(H) < ε(N +M ′ε).

It follows that A(φ(Bd(E))) < ε(N +M ′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,

A(φ(E)) ≤∑

(i,j)∈Γ

A(φ(Cij)) =∑

(i,j)∈Γ1

A(φ(Cij)) +∑

(i,j)∈Γ2

A(φ(Cij)).

The first sum was estimated above. For the second sum we have the following estimate:∑(i,j)∈Γ2

A(φ(Cij)) ≤∑

(i,j)∈Γ2

|Jφ(aij)|A(Cij) +M ′ε∑

(i,j)∈Γ2

A(Cij).

Note that∑

(i,j)∈Γ2 A(Cij) ≤ A(E). Since |Jφ(aij)| ≤ |Jφ(x)| for all x ∈ Cij , we have∑(i,j)∈Γ2

|Jφ(aij)|A(Cij) ≤∑

(i,j)∈Γ2

∫∫Cij

|Jφ(x1, x2)| dA ≤∫∫

E

|Jφ(x1, x2)| dA.

Combining the above estimates, we obtain

A(φ(E)) ≤∫∫

E

|Jφ(x1, x2)| dA+M ′εA(E) + ε(N +M ′ε).

Since this estimate is valid whenever ε > 0, we conclude that

A(φ(E)) ≤∫∫

E

|Jφ(x1, x2)| dA.

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 openmapping, by the inverse mapping theorem. Thus, φ(E◦ \K) is an open set contained inφ(E). It follows that φ(E◦ \K) ⊆ (φ(E))◦. Moreover, since φ is a continuous mapping onthe 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 ′εA(Cij).Hence,

A(φ(K)) ≤∑

(i,j)∈Γ3

A(φ(Cij)) ≤M ′ε∑

(i,j)∈Γ3

A(Cij) ≤M ′εA(F ).

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 suchthat E ⊂ U . Suppose that φ is a continuously differentiable mapping from U to IR2. If fis a nonnegative continuous function on the domain φ(E), then∫∫

φ(E)

f(u1, u2) du1du2 ≤∫∫

E

f(φ(x1, x2)

)∣∣Jφ(x1, x2)∣∣ dx1dx2.Proof. Let ε > 0 be given. Since f ◦ φ is continuous on the compact set E, there existsδ > 0 such that

x, y ∈ E and ‖x− y‖ < δ imply |f(φ(x))− f(φ(y))| < ε.

Partition the domain E into mutually disjoint Riemann domains E1, E2, . . . , En such thatE = ∪nj=1Ej 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=1φ(Ej). Hence,∫∫

φ(E)

f(u1, u2) du1du2 ≤n∑j=1

∫∫φ(Ej)

f(u1, u2) du1du2 ≤n∑j=1

MjA(φ(Ej)).

By Theorem 7.1 we assert that

n∑j=1

MjA(φ(Ej)) ≤n∑j=1

∫∫Ej

Mj |Jφ(x1, x2)| dx1dx2.

Since Mj ≤ mj + ε ≤ f(φ(x1, x2)) + ε for all (x1, x2) ∈ Ej , we obtain∫∫φ(E)

f(u1, u2) du1du2 ≤n∑j=1

MjA(φ(Ej)) ≤∫∫

E

[f(φ(x1, x2)

)+ ε

]∣∣Jφ(x1, x2)∣∣ dx1dx2.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 ontoan 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 ∫∫

φ(E)

f(u1, u2) du1du2 =∫∫

E

f(φ(x1, x2)

)∣∣Jφ(x1, x2)∣∣ dx1dx2.Proof. First, consider the case that f is nonnegative. By Theorem 8.1,∫∫

φ(E)

f(u1, u2) du1du2 ≤∫∫

E

f(φ(x1, x2)

)∣∣Jφ(x1, x2)∣∣ dx1dx2.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∫∫

φ−1(φ(E))

f(φ(x1, x2)

)∣∣Jφ(x1, x2)∣∣ dx1dx2≤

∫∫φ(E)

f(φ ◦ φ−1(u1, u2)

)∣∣Jφ(φ−1(u1, u2))∣∣∣∣Jφ−1(u1, u2)∣∣ du1du2.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,∫∫

E

f(φ(x1, x2)

)∣∣Jφ(x1, x2)∣∣ dx1dx2 ≤ ∫∫φ(E)

f(u1, u2) du1du2.

Thus, the change of variable formula as stated in the theorem is valid for the case that fis nonnegative.

For the general case, we may write f = f+ − f−, where f+ := (|f | + f)/2 andf− := (|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 thatit is also valid for f .

For the special case that f = 1 on φ(E), Theorem 8.2 yields the following result:

A(φ(E)) =∫∫

E

∣∣Jφ(x1, x2)∣∣ dx1dx2.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 functionon V . If both φ and φ−1 are continuously differentiable, and if |Jφ| is bounded on U , then∫∫

V

f(u1, u2) du1du2 =∫∫

U

f(φ(x1, x2)

)∣∣Jφ(x1, x2)∣∣ dx1dx2.Proof. Let I1 and I2 denote the integral on the left and the right of the above equa-tion 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 compactRiemann domain K of U such that A(U \K) < ε and A(V \ φ(K)) < ε. By Theorem 8.2we obtain ∫∫

φ(K)

f(u1, u2) du1du2 =∫∫

K

f(φ(x1, x2)

)∣∣Jφ(x1, x2)∣∣ dx1dx2.On the other hand we have∣∣∣∣I1 − ∫∫

φ(K)

f(u1, u2) du1du2

∣∣∣∣ = ∣∣∣∣∫∫V \φ(K)

f(u1, u2) du1du2

∣∣∣∣ ≤MA(V \ φ(K)) < Mε.A similar argument shows that∣∣∣∣I2 − ∫∫

U

f(φ(x1, x2)

)∣∣Jφ(x1, x2)∣∣ dx1dx2∣∣∣∣ ≤MA(U \K) < Mε.Consequently, |I1 − I2| < 2Mε for any ε > 0. Therefore, I1 = I2.

We are in a position to investigate double integrals in polar coordinates. Consider themapping 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

Jφ(r, θ) =∣∣∣∣ ∂x∂r ∂x∂θ∂y∂r

∂y∂θ

∣∣∣∣ = ∣∣∣∣ cos θ −r sin θsin θ r cos θ∣∣∣∣ = r.

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 realnumbers such that 0 ≤ r1 < r2 and θ1 < θ2 ≤ θ1 +2π. It is easily seen that φ is one-to-one

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on the open domain U := {(r, θ) : r1 < r < r2, θ1 < θ < θ2}. Hence, if f is a boundedcontinuous function on V := φ(U), then by Theorem 8.3 we obtain∫∫

V

f(x, y) dx dy =∫ θ2θ1

∫ r2r1

f(r cos θ, r sin θ) r dr dθ.

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 + y2 ≤ r22}. Hence, for a bounded continuousfunction f on the disk V1 we have∫∫

V1

f(x, y) dx dy =∫∫

V

f(x, y) dx dy =∫ 2π

0

∫ r20

f(r cos θ, r sin θ) r dr dθ.

Example 1. Evaluate the double integral∫∫V

e−x2−y2 dx dy,

where V is the open disk {(x, y) : x2 + y2 < b2} with b > 0.

Solution. By using polar coordinates we obtain∫∫V

f(x, y) dx dy =∫ 2π

0

∫ b0

e−r2r dr dθ = 2π

[−e−r

2/2

]b0

= π(1− e−b2).

Example 2. Evaluate the double integral∫∫E

xy(x2 + y2) dx dy,

where E is the domain in the first quadrant bounded by the curves x2−y2 = 1, x2−y2 = 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 − y2, 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 + y2 =

√u2 + v2. Hence, if x > 0 and y > 0, then

x =√

(√u2 + v2 + u)/2 and y =

√(√u2 + v2 − u)/2.

Let φ be the mapping (u, v) 7→ (x, y) as given above. Then φ : ψ(E) → E is the inverse ofthe mapping ψ : E → ψ(E). Note that ψ(E) = {(u, v) : 1 ≤ u ≤ 4, 2 ≤ v ≤ 4}. We have

Jψ(x, y) =∣∣∣∣ ∂u∂x ∂u∂y∂v∂x

∂v∂y

∣∣∣∣ = ∣∣∣∣ 2x −2y2y 2x∣∣∣∣ = 4(x2 + y2).

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It follows thatJφ(u, v) =

1Jψ(x, y)

=1

4(x2 + y2)=

14√u2 + v2

.

Therefore,∫∫E

xy(x2 + y2) dx dy =∫∫

ψ(E)

v

2

√u2 + v2

14√u2 + v2

du dv =18

∫ 41

[∫ 42

v dv

]du =

94.

Example 3. For a > 0, find the area of the domain

Q := {(x, y) ∈ IR2 : x2/3 + y2/3 ≤ a2/3}.

Solution. Consider the mapping φ : (r, t) 7→ (x, y) given by

x = r cos3 t, y = r sin3 t, (r, t) ∈ IR2.

Let E := {(r, t) ∈ IR2 : 0 ≤ r ≤ a, 0 ≤ t ≤ 2π}. It is easily seen that φ(E) = Q. TheJacobian determinant of φ is

Jφ(r, t) =∣∣∣∣ ∂x∂r ∂x∂t∂y∂r

∂y∂t

∣∣∣∣ = ∣∣∣∣ cos3 t −3r cos2 t sin tsin3 t 3r sin2 t cos t∣∣∣∣ = 3r sin2 t cos2 t.

Therefore, the area of the domain Q is

A(Q) = A(φ(E)) =∫∫

E

|Jφ(r, t)| dr dt =∫ 2π

0

[∫ a0

3r dr]

sin2 t cos2 t dt.

It follows that

A(Q) =3a2

2

∫ 2π0

sin2(2t)4

dt =3a2

8

∫ 2π0

1− cos 4t2

dt =3a2

16[t− sin 4t/4

]2π0

=38πa2.

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