We start with a closed bounded region
in the
plane.
We assume that
is a basic region; that is, we assume that the boundary of
consists of a finite number of continuous arcs
.
Now let's suppose that
is some function continuous on
.
We want to define the double integral
To do this, we surround
by a rectangle
We now extend
to all of
by setting
equal to
outside of
.
This extended function
is bounded on
,
and it is continuous on all of
except possibly at the boundary of
.
In spite of these possible discontinuities,
is still integrable on
;
that is, there still exists a unique number
such that
This number
is by definition the double integral
We define the double integral over
by setting
If
is nonnegative over
,
the extended
is nonnegative on all of
.
The double integral gives the volume of the solid trapped between the surface
and the rectangle
.
But since the surface has height
outside of
,
the volume outside of
is
.
It follows then that
gives the volume of the solid
bounded above by
and below by
:
volume of =
The double integral
gives the volume of a solid of constant height
over
.
In square units this is the area of
:
Below we list four elementary properties of the double integral. They are all analogous to what was in the onevariable case. The referred to is a basic region. The functions and are assumed to be continuous on .
I. The double integral is linear:
II. The double integral preserves order:
III. The double integral is additive:
if
is broken up into a finite number of basic regions
,
then
IV. The double integral satisfies a meanvalue condition;
namely, there is a point
for which
We call the average value of on
The notion of average given in (IV) enables us to write
This is a powerful intuitive way of viewing the double integral. We will
capitalize on it as we go on.
THEOREM MEANVALUE THEOREM FOR DOUBLE INTEGRALS.
Let
and
be functions continuous on a basic region
.
If
is nonnegative on
,
then there exists a point
in
for which
We call the weighted average of on
Proof. Since is continuous on and is closed and bounded, we know that takes on a minimum value and a maximum value . Since is nonnegative on
Therefore
and
We know that
If
,
then from the last equation we have
and the theorem holds for all choices of
in
.
If
,
then
and by the intermediatevalue theorem there exists
in
for which
Obviously then
If an integral
proves difficult to evaluate, it is not because of the interval
but because of the integrand
.
Difficulty in evaluating a double integral
can come from two sources: from the integrand
and from the base region
.
Even such a simple looking integral as
is difficult to evaluate if
is complicated.
In this section we introduce a technique for evaluating double integrals of continuous functions over '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
onto the
axis
is a closed interval
and
consists of all points
with
Then
Here we first calculate
by integrating
with respect to
from
to
.
The resulting expression is a function of
alone, which we then integrate with respect to
from
to
.
Type II. The projection of
onto the
axis
is a closed interval
and
consists of all points
with
Then
This time we first calculate
by integrating
with respect to
from
to
.
The resulting expression is a function of
alone, which we then integrate with respect to
from
to
.
These formulas are easy to understand geometrically.
Let's take
as nonnegative and
of type I. The double integral over
gives the volume of the solid
that is trapped between
and the surface
We can also compute the volume of
by the method of parallel cross sections. Let
be the area of that cross section of
that has first coordinate
.
Then
Since
we have
Combining the two equations, we have the first reduction formula
The other reduction formula can be obtained in a similar manner.
Problem 1. Evaluate
with
the region bounded by
Solution. By projecting
onto the
axis
we obtain the interval
The region
consists of all points
with
This is a region of type I.
Problem 2. Evaluate
with
bounded by
Solution. By projecting onto the axis we obtain the interval . The region consists of all points with
and
This is a region of type II.
We can also project
onto the
axis
and express
as a region of type I, but then the lower boundary is defined piecewise and
the calculations are somewhat more complicated: setting
we have
as the set of all points
with
thus
Repeated integrals
can be written in more compact form by omitting the large parentheses. From
now on we will simply write
Problem 3. Evaluate
with
the region bounded by
Solution. The projection of
onto the
axis
is the closed interval
,
and
can be characterized as the set of all
with
Thus
We can also integrate in the other order. The projection of
onto the
axis
is the closed interval
,
and
can be characterized as the set of all
with
This gives the same result:
Problem 4. Calculate by double integration the area of the
region
that lies between
Solution. The area of
is given by the double integral
Here again we can integrate in either order. We can project
onto the

axis and write the boundaries as functions of
:
is then the set of all
with
This gives
We can also project
onto the
axis
and write the boundaries as functions of
:
is then the set of all
with
This gives
which is also
First we go back to the onevariable case. Let's suppose that is continuous on an interval that is symmetric about the origin, say
If is odd, then .
If is even, then
We have similar results for double integrals.
Suppose that is symmetric about the axis .
If is odd in if , then
If is even in if , then
Suppose that is symmetric about the axis.
If is odd in if , then
If is even in if , then
As an example take the region bounded by
Suppose we wanted to calculate
Symmetry about the
axis
gives
Symmetry about the
axis
gives
Therefore
As another example we go to Problem 1 and reevaluate
this time capitalizing on the symmetry of
.
Note that
Therefore
Problem 5. Calculate the volume within the cylinder
between the planes
and
given that
Solution. The solid in question is bounded below by the disc
and above by the plane
The volume is given by the double integral
Since
is symmetric about the
axis,
Thus
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
bounded by
Projection onto the
axis
leads to
Projection onto the
axis
leads to
The first expression is easy to calculate, but the second is not. The only
practical way to handle the second expression is to expand
as a series in
and then integrate term by term.
Finally, if
,
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
,
each of type I or type II. Since the double integral is additive,
Each of the integrals on the left can be evaluated by the methods of this section.
Evaluate the integrals
Write an equivalent double integral with the order of integration reversed.
Find the volume of the solid whose base is the region in the xyplane that is bounded by the parabola and the line , while the top of the solid is bounded by the plane
Find the volume of the solid in the first octant bounded by the coordinate planes, the plane and the parabolic cylinder
Find the volume of the region that lies under the paraboloid and above the triangle enclosed by the lines and on the xyplane.