THE DOUBLE INTEGRAL OVER A REGION

THE DOUBLE INTEGRAL OVER A REGION 1
THE DOUBLE INTEGRAL OVER A REGION
We start with a closed bounded region Ω in the xy-plane. We assume that Ω is a
basic region; that is, we assume that the boundary of Ω consists of a finite number
of continuous arcs y = φ(x), x = ψ(y). Now let’s suppose that f is some function
continuous on Ω. We want to define the double integral
∫∫
f(x, y) dxdy.
Ω
To do this, we surround Ω by a rectangle R. We now extend f to all of R by
setting f equal to 0 outside of Ω. This extended function f is bounded on R, and it is
continuous on all of R except possibly at the boundary of Ω. In spite of these possible
discontinuities, f is still integrable on R; that is, there still exists a unique number I
such that
L f (P ) ≤ I ≤ U f (P ) for all partitions P of R.
This number I is by definition the double integral
∫∫
R
f(x, y) dxdy.
We define the double integral over Ω by setting
∫∫
Ω
∫∫
f(x, y) dxdy =
R
f(x, y) dxdy.
If f is nonnegative over Ω, the extended f is nonnegative on all of R. The double
integral gives the volume of the solid trapped between the surface z = f(x, y) and the
rectangle R. But since the surface has height 0 outside of Ω, the volume outside of Ω
is 0. It follows then that
∫∫
f(x, y) dxdy
Ω
gives the volume of the solid T bounded above by z = f(x, y) and below by Ω:
volume of T = ∫∫ Ω f(x, y) dxdy.
The double integral
∫∫
∫∫
1 dxdy =
dxdy
Ω
Ω
March 17, 2001

THE DOUBLE INTEGRAL OVER A REGION 2
gives the volume of a solid of constant height 1 over Ω. In square units this is the area
of Ω:
∫∫
area of Ω = dxdy.
Ω
Below we list four elementary properties of the double integral. They are all analogous
to what was in the one-variable case. The Ω referred to is a basic region. The
functions f and g are assumed to be continuous on Ω.
I. The double integral is linear:
∫∫
Ω
∫∫
[αf(x, y) + βg(x, y)] dxdy = α
II. The double integral preserves order:
Ω
∫∫
f(x, y) dxdy + β g(x, y) dxdy.
Ω
∫∫
if f ≥ 0 on Ω, then
∫∫
if f ≤ g on Ω, then
III. The double integral is additive:
Ω
Ω
f(x, y) dxdy ≥ 0;
∫∫
f(x, y) dxdy ≤ g(x, y) dxdy.
if Ω is broken up into a finite number of basic regions Ω 1 , · · · , Ω n , then
∫∫
∫∫
∫∫
f(x, y) dxdy = f(x, y) dxdy + · · · + f(x, y) dxdy.
Ω
Ω 1
IV. The double integral satisfies a mean-value condition; namely, there is a point
(x 0 , y 0 ) for which
∫∫
Ω
Ω
Ω n
f(x, y) dxdy = f(x 0 , y 0 ) · (the area of Ω).
We call f(x 0 , y 0 ) the average value of f on Ω.
The notion of average given in (IV) enables us to write
∫∫
Ω
f(x, y) dxdy = (the arerage value of f on Ω) · (the area of Ω).
This is a powerful intuitive way of viewing the double integral. We will capitalize on
it as we go on.
March 17, 2001

THE EVALUATION OF DOUBLE INTEGRALS BY REPEATED INTEGRALS 3
THEOREM MEAN-VALUE THEOREM FOR DOUBLE INTEGRALS.
Let f and g be functions continuous on a basic region Ω. If g is nonnegative on Ω,
then there exists a point (x 0 , y 0 ) in Ω for which
∫∫
∫∫
f(x, y)g(x, y) dxdy = f(x 0 , y 0 ) g(x, y) dxdy.
Ω
We call f(x 0 , y 0 ) the g-weighted average of f on Ω.
Proof. Since f is continuous on Ω and Ω is closed and bounded, we know that f
takes on a minimum value m and a maximum value M. Since g is nonnegative on Ω,
and
Therefore
∫∫
Ω
∫∫
m
mg(x, y)≤ f(x, y)g(x, y) ≤ Mg(x, y)
∫∫
∫∫
mg(x, y) dxdy ≤ f(x, y)g(x, y) dxdy ≤ Mg(x, y) dxdy
Ω
Ω
∫∫
∫∫
g(x, y) dxdy ≤ f(x, y)g(x, y) dxdy ≤ M g(x, y) dxdy.
Ω
We know that ∫∫ Ω g(x, y) dxdy ≥ 0. If ∫∫ Ω g(x, y) dxdy = 0, then from the last
equation we have ∫∫ Ω f(x, y)g(x, y) dxdy = 0 and the theorem holds for all choices of
(x 0 , y 0 ) in Ω. If ∫∫ Ω g(x, y) dxdy > 0, then
∫∫
Ω f(x, y)g(x, y) dxdy
m ≤ ∫∫Ω g(x, y) dxdy ≤ M
and by the intermediate-value theorem there exists (x 0 , y 0 ) in Ω for which
∫∫
Ω f(x, y)g(x, y) dxdy
f(x 0 , y 0 ) = ∫∫Ω g(x, y) dxdy .
Ω
Ω
Ω
Obviously then
∫∫
f(x 0 , y 0 )
∫∫
g(x, y) dxdy =
f(x, y)g(x, y) dxdy.
Ω
Ω
THE EVALUATION OF DOUBLE INTEGRALS
BY REPEATED INTEGRALS
The Reduction Formulas
If an integral
∫ b
a
f(x) dx
March 17, 2001

THE EVALUATION OF DOUBLE INTEGRALS BY REPEATED INTEGRALS 4
proves difficult to evaluate, it is not because of the interval [a, b] but because of the
integrand f. Difficulty in evaluating a double integral
∫∫
f(x, y) dxdy
Ω
can come from two sources: from the integrand f and from the base region Ω. Even
such a simple looking integral as ∫∫ Ω 1 dxdy is difficult to evaluate if Ω is complicated.
In this section we introduce a technique for evaluating double integrals of continuous
functions over Ω’s that have a simple structure. The fundamental idea of this section
is that double integrals over sets of this structure can be reduced to a pair of ordinary
integrals.
Type I The projection of Ω onto the x-axis is a closed interval [a, b] and Ω consists
of all points (x, y) with
Then
Here we first calculate
∫∫
Ω
a ≤ x ≤ b and φ 1 (x) ≤ y ≤ φ 2 (x).
f(x, y) dxdy =
∫ φ2 (x)
φ 1 (x)
∫ ( b ∫ φ2 (x)
a
φ 1 (x)
f(x, y) dy
f(x, y) dy
by integrating f(x, y) with respect to y from y = φ 1 (x) to y = φ 2 (x). The resulting
expression is a function of x alone, which we then integrate with respect to x from
x = a to x = b.
Type II. The projection of Ω onto the y-axis is a closed interval [a, b] and Ω consists
of all points (x, y) with
Then
∫∫
This time we first calculate
Ω
a ≤ y ≤ b and φ 1 (y) ≤ x ≤ φ 2 (y).
f(x, y) dxdy =
∫ φ2 (y)
φ 1 (y)
∫ ( b ∫ φ2 (y)
a
φ 1 (y)
f(x, y) dx
f(x, y) dx
by integrating f(x, y) with respect to x from x = φ 1 (y) to x = φ 2 (y). The resulting
expression is a function of y alone, which we then integrate with respect to y from
y = a to y = b. These formulas are easy to understand geometrically.
)
)
dx
dy.
The Reduction Formulas March 17, 2001

Computations 5
The Reduction Formulas Viewed Geometrically
Let’s take f as nonnegative and Ω of type I. The double integral over Ω gives the
volume of the solid T that is trapped between Ω and the surface z = f(x, y) :
∫∫
Ω
f(x, y) dxdy = volume of T.
We can also compute the volume of T by the method of parallel cross sections. Let
A(x) be the area of that cross section of T that has first coordinate x. Then
∫ b
Since
A(x) =
we have
a
∫ ( b ∫ φ2 (x)
a φ 1 (x)
A(x)dx = volume of T.
∫ φ2 (x)
φ 1 (x)
f(x, y) dy
)
f(x, y) dy,
dx = volume of T.
Combining the two equations, we have the first reduction formula
∫∫
∫ ( b ∫ )
φ2 (x)
f(x, y) dxdy =
f(x, y) dy dx.
a φ 1 (x)
Ω
The other reduction formula can be obtained in a similar manner.
Computations
Problem 1. Evaluate
with Ω the region bounded by
∫∫
Ω
(x 2 − y) dxdy
y = x 2 , y = −x 2 , x = −1, x = 1.
Solution. By projecting Ω onto the x-axis we obtain the interval [−1, 1]. The region Ω
consists of all points (x, y) with
−1 ≤ x ≤ 1 and − x 2 ≤ y ≤ x 2 .
March 17, 2001

Computations 6
This is a region of type I.
∫∫
∫ 1
( ∫ x 2
)
(x 2 − y) dxdy = (x 2 − y) dy dx
−1 −x 2
Ω
=
∫ 1
−1
[x 2 y − 1 2 y2 ] x 2
Problem 2. Evaluate
with Ω bounded by
=
−x 2 dx =
∫ 1
−1
∫ 1
−1
[
(x 4 − 1 2 x4 ) − (−x 4 − 1 ]
2 x4 ) dx
[ ] 2 1
2x 4 dx =
5 x5 = 4
−1 5 .
∫∫
(xy − y 3 ) dxdy
Ω
y = 0, x = y, y = 1, x = −1.
Solution. By projecting Ω onto the y-axis we obtain the interval [0, 1]. The region
Ω consists of all points (x, y) with
This is a region of type II.
=
0 ≤ y ≤ 1 and −1 ≤ x ≤ y.
∫∫
∫ 1 (∫ y
)
(xy − y 3 ) dxdy = (xy − y 3 ) dx dy
0 −1
Ω
∫ 1 [ ] 1 y ∫ 1
0 2 x2 y − xy 3 dy = (− 1
−1 0 2 y − 1 2 y3 − y 4 )dy
[ 1
=
4 y2 − 1 8 y4 − 1 ] 1
5 y5 0
= − 23
40 .
We can also project Ω onto the x-axis and express Ω as a region of type I, but
then the lower boundary is defined piecewise and the calculations are somewhat more
complicated: setting
φ(x) =
we have Ω as the set of all points (x, y) with
⎧
⎪ ⎨
⎪ ⎩
0, if − 1 ≤ x ≤ 0
x if 0 ≤ x ≤ 1
−1 ≤ x ≤ 1 and φ(x) ≤ y ≤ 1;
,
thus
∫∫
Ω
∫ ( 1 ∫ )
1
(xy − y 3 ) dxdy = (xy − y 3 ) dy dx
−1 φ(x)
March 17, 2001

Computations 7
=
∫ 0
=
−1
∫ 0
( ∫ 1
−1
Repeated integrals
)
(xy − y 3 ) dy
φ(x)
(∫ 1
)
(xy − y 3 ) dy
0
∫ ( b ∫ φ2 (x)
a
φ 1 (x)
dx +
dx +
∫ ( 1 ∫ 1
0
∫ 1 (∫ 1
= (− 1 2 ) + (− 3 40 ) = −23 40 .
f(x, y) dy
)
dx and
0
(xy − y 3 ) dy
φ(x)
)
x
∫ ( b ∫ φ2 (y)
a
(xy − y 3 ) dy
φ 1 (y)
f(x, y) dx
can be written in more compact form by omitting the large parentheses. From now on
we will simply write
)
)
dx
dx
dy
∫ b ∫ φ2 (x)
a φ 1 (x)
f(x, y) dydx and
∫ b ∫ φ2 (y)
a φ 1 (y)
f(x, y) dxdy.
Problem 3. Evaluate
with Ω the region bounded by
∫∫
Ω
(x 1/2 − y 2 ) dxdy
y = x 2 and x = y 4 .
Solution. The projection of Ω onto the x-axis is the closed interval [0, 1], and Ω can
be characterized as the set of all (x, y) with
0 ≤ x ≤ 1 and x 2 ≤ y ≤ x 1/4 .
Thus
∫∫
∫ 1 ∫ x 1/4
(x 1/2 − y 2 ) dxdy = (x 1/2 − y 2 ) dydx
0 x 2
Ω
∫ 1
=
[x 1/2 y − 1 ] x 1/4 ∫ 1
0 3 y3 dx = ( 2
x 2 0 3 x3/4 − x 5/2 + 1 3 x6 ) dx
[ 8
=
21 x7/4 − 2 7 x7/2 + 1 ] 1
21 x7 = 1
0 7 .
We can also integrate in the other order. The projection of Ω onto the y-axis is the
closed interval [0, l], and Ω can be characterized as the set of all (x, y) with
0 ≤ y ≤ 1 and y 4 ≤ x ≤ y 1/2 .
March 17, 2001

Computations 8
This gives the same result:
∫∫
∫ 1 ∫ y 1/2
(x 1/2 − y 2 ) dxdy = (x 1/2 − y 2 ) dxdy
0 y 4
Ω
=
∫ 1
0
( 2 3 y3/4 − y 5/2 + 1 3 y6 )dy = 1 7 .
Problem 4. Calculate by double integration the area of the region Ω that lies
between
√ √ √ x + y = a and x + y = a.
Solution. The area of Ω is given by the double integral
∫∫
Ω
dxdy.
Here again we can integrate in either order. We can project Ω onto the x- axis and
write the boundaries as functions of x:
Ω is then the set of all (x, y) with
This gives
y = ( √ a − √ x) 2 and y = a − x;
0 ≤ x ≤ a and ( √ a − √ x) 2 ≤ y ≤ a − x.
∫∫
Ω
∫ a
=
0
dxdy =
∫ a ∫ a−x
0
( √ a− √ x) 2 dydx
(−2x + 2 √ a √ x) dx = a2
3 .
We can also project Ω onto the y-axis and write the boundaries as functions of y:
Ω is then the set of all (x, y) with
x = ( √ a − √ y) 2 and x = a − y;
0 ≤ y ≤ a and ( √ a − √ y) 2 ≤ x ≤ a − y.
This gives
∫∫
Ω
dxdy =
∫ a ∫ a−y
0
( √ a− √ y) 2 dxdy,
which is also a2
3 . March 17, 2001

Symmetry in Double Integration 9
Symmetry in Double Integration
First we go back to the one-variable case. Let’s suppose that g is continuous on an
interval that is symmetric about the origin, say [−a, a].
If g is odd, then ∫ a
−a g(x)dx = 0.
If g is even, then ∫ a
−a g(x)dx = 2 ∫ a
0 g(x)dx.
We have similar results for double integrals.
Suppose that Ω is symmetric about the y-axis.
If f is odd in x [if f(−x, y) = −f(x, y)], then ∫∫ Ω f(x, y) dxdy = 0.
If f is even in x [if f(−x, y) = f(x, y)], then
∫∫
Ω f(x, y) dxdy = 2 ∫∫ right half of Ω f(x, y) dxdy.
Suppose that Ω is symmetric about the x-axis.
If f is odd in y [if f(x, −y) = −f(x, y)], then ∫∫ Ω f(x, y) dxdy = 0.
If f is even in y [if f(x, −y) = f(x, y)], then
∫∫
Ω f(x, y) dxdy = 2 ∫∫ upper half of Ω f(x, y) dxdy.
As an example take the region bounded by
x + y = 0, x − y = 0, x = a, x = −a.
Suppose we wanted to calculate
∫∫
Ω
(2x − sin x 2 y) dxdy.
Symmetry about the y-axis gives
∫∫
2x dxdy = 0.(integrand is odd in x)
Ω
Symmetry about the x-axis gives
∫∫
sin x 2 y dxdy = 0.(integrand is odd in y)
Ω
March 17, 2001

Symmetry in Double Integration 10
Therefore
∫∫
(2x − sin x 2 y) dxdy = 0.
Ω
As another example we go to Problem 1 and reevaluate
∫∫
Ω
(x 2 − y) dxdy,
this time capitalizing on the symmetry of Ω. Note that
Therefore
∫∫
Ω
∫∫
Ω
x 2 dxdy = 2
ydxdy = 0 (symmetric about x − axis)
∫∫
right half of Ω
= 4
x 2 dxdy = 0 (symmetric about y − axis)
∫∫
upper part
of the right half of Ω
x 2 dxdy.
∫∫
∫ 1 ∫ x 2
(x 2 − y) dxdy = 4 x 2 dydx = 4
0 0
5 .
Ω
Problem 5. Calculate the volume within the cylinder
x 2 + y 2 = b 2
between the planes
y + z = a 2
and
z = 0
given that
a 2 ≥ b > 0.
Solution. The solid in question is bounded below by the disc
Ω : 0 ≤ x 2 + y 2 ≤ b 2
and above by the plane
z = a 2 − y.
March 17, 2001

Concluding Remarks 11
The volume is given by the double integral
Since Ω is symmetric about the x-axis,
Thus
Ω
∫∫
(a 2 − y) dxdy.
Ω
∫∫
Ω
Ω
y dxdy = 0.
∫∫
∫∫
(a 2 − y) dxdy = a 2 dxdy = a 2 (area of Ω) = πa 2 b 2 .
Concluding Remarks
When two orders of integration are possible, one order may be easy to carry out, while
the other may present serious difficulties. Take as an example the double integral
with Ω bounded by
Projection onto the x-axis leads to
Projection onto the y-axis leads to
∫∫
Ω
cos 1 2 πx2 dxdy
y = x, y = 0, x = 1.
∫ 1 ∫ x
0
0
∫ 1 ∫ 1
0
y
cos 1 2 πx2 dydx.
cos 1 2 πx2 dxdy.
The first expression is easy to calculate, but the second is not. The only practical way
to handle the second expression is to expand cos 1 πx as a series in x and then integrate
2
term by term.
Finally, if Ω, the region of integration, is neither of type I nor of type II, it is usually
possible to break it up into a finite number of regions Ω 1 , · · · , Ω n , each of type I or type
II. Since the double integral is additive,
∫∫
Ω 1
∫∫
Ω n
∫∫
Ω
f(x, y) dxdy + · · · + f(x, y) dxdy = f(x, y) dxdy.
Each of the integrals on the left can be evaluated by the methods of this section.
March 17, 2001

Exercise 12
Exercise
1. Evaluate the integrals
(a) ∫ 3 ∫ 2
0 0 (4 − y2 )dydx [16]
(b) ∫ 0 ∫ 1
−1 −1 (x + y + 1)dxdy [1]
(c) ∫ π
0
(d) ∫ ln 8
1
∫ x
0 x sin ydydx [2 + 1 2 π2 ]
∫ ln y
0 e x+y dxdy [24 ln 2 − 16 + e]
(e) ∫ 1 ∫ y 2
0 0 3y 3 e xy dxdy [e − 2]
(f) ∫ 0 ∫ −v
−2 v 2dpdv [8]
(g) ∫ π/3 ∫ sec t
−π/3 0 3 cos tdudt [2π]
(h) ∫ π
0
∫ π
x
(i) ∫ 2 √ ln 3
0
sin y
y
dydx [2]
∫ √ ln 3
y/2 e x2 dxdy [2]
(j) ∫ 3 ∫
√ 1
0 x/3 ey2 dydx [e − 1]
2. Write an equivalent double integral with the order of integration reversed.
(a) ∫ 1 ∫ 4−2x
0 2 dydx [ ∫ 4 ∫ (4−y)/2
2 0 dxdy]
(b) ∫ 1 ∫ √ y
0 y dxdy [ ∫ 1 ∫ x
0 x dydx]
2
(c) ∫ 1 ∫ e x
0 0 dydx [ ∫ e ∫ 1
1 ln y dxdy]
(d) ∫ 3/2
0
(e) ∫ 1
0
∫ 9−4x 2
0 dydx [ ∫ 9
0
∫ √ ( 9−y)/2
0 dxdy]
∫ √ 1−y 2
−
√1−y dxdy [∫ 1 ∫ √ 1−x 2
2 −1 0 dydx]
3. Find the volume of the solid whose base is the region in the xy-plane that is
bounded by the parabola y = 4 − x 2 and the line y = 3x, while the top of the
solid is bounded by the plane z = x + 4. [ 625
12 ]
4. Find the volume of the solid in the first octant bounded by the coordinate planes,
the plane x = 3, and the parabolic cylinder z = 4 − y 2 . [16]
5. Find the volume of the region that lies under the paraboloid z = x 2 + y 2 and
above the triangle enclosed by the lines y = x, x = 0, and x + y = 2 on the
xy-plane. [4/3]
March 17, 2001