THE DOUBLE INTEGRAL OVER A REGION

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 one-variable 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 mean-value 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 MEAN-VALUE 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 intermediate-value theorem there exists in for which

Obviously then

The Reduction Formulas

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.

The Reduction Formulas Viewed 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.

Computations

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

Symmetry in Double Integration

First we go back to the one-variable 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

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

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 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.

Exercise

1. Evaluate the integrals

2. Write an equivalent double integral with the order of integration reversed.

3. Find the volume of the solid whose base is the region in the xy-plane that is bounded by the parabola and the line , while the top of the solid is bounded by the plane

4. Find the volume of the solid in the first octant bounded by the coordinate planes, the plane and the parabolic cylinder

5. Find the volume of the region that lies under the paraboloid and above the triangle enclosed by the lines and on the xy-plane.

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