In the one-variable case we can write the integral as the limit of Riemann
sums:

The same approach works with double integrals. To explain it we need to
explain what we mean by the **diameter of a set.** Suppose that
is a bounded closed set (on the line, in the plane, or in three-space). For
any two points
and
of
we can measure their separation,
The greatest separation between points of
is called the **diameter of **
:

diam = .

For a circle, a circular disc, a sphere, or a ball, this sense of diameter agrees with the usual one.

Now let's start with a basic region and decompose it into a finite number of basic subregions

If
is continuous on
,
then
is continuous on each
.
Now from each
,
we pick an arbitrary point
and form the Riemann sum

The double integral over
can be obtained as the limit of such sums; namely, given any
,
there exists
such that, if the diameters of the are all
less than
then

no matter how the
are chosen within the
We express this by writing

Here we explain how to calculate double integrals

using polar coordinates
.
Throughout we take
and restrict
to

We will work with the type of region consisting of all points
that have polar coordinates
in the set

According to the formula in one-variable case the area of
is given by

We can write this as a double integral over
:

Proof. Simply note that

and therefore

Now let's suppose that
is some function continuous at each point
of
.
Then the composition

is continuous at each point
of
.
We will show that

Proof. Our first step is to place a grid on
by using a finite number of rays
and a finite number of continuous curves
.
This grid decomposes
into a finite number of little regions

with polar coordinates in sets
.
Note that

Writing
for
,
we have

This last expression is a Riemann sum for the double integral

and, as such, differs from that integral by less than any preassigned positive
provided only that the diameters of all the
are sufficiently small. This we can guarantee by making our grid sufficiently
fine.

**Problem 1.** Use polar coordinates to evaluate

where
is the portion of the unit disc that lies in the first quadrant.

Solution. Here

Therefore

**Problem 2.** Use polar coordinates to calculate the volume of a
sphere of radius
.

Solution. In rectangular coordinates

where
is the disc of radius
centered at the origin. Here

Therefore

**Problem 3.** Calculate the volume of the solid bounded above by
the lower nappe of the cone
and bounded below by the disc

Solution

The first integral is
(area
of
)
.
We evaluate the second integral by changing to polar coordinates.

The equation
simplifies to
.
In polar coordinates this becomes
,
which simplifies to
.
The disc
is the set of all points with polar coordinates in the set

Therefore

We then have

**Problem 4.** Evaluate

where
is the triangle bounded by

Solution. The vertical side of the triangle is part of the line
In polar coordinates this is
,
which can be written
.
Therefore

where

The double integral over
reduces to

For

Therefore the integral can be written

The function
has no elementary antiderivative. Nevertheless, by taking a circuitous route
and then using polar coordinates, we can show that

Proof. The circular disc
is the set of all
with polar coordinates
in the set
Therefore

Let
be the square
Since
and
is positive,

It follows that

As
and
Therefore

But

Therefore

This integral comes up frequently in probability theory and plays an important role in what is called ``statistical mechanics.''

Suppose that a thin distribution of matter, what we call a *plate*, is
laid out in the
-plane
in the form of a basic region
.
If the mass density of the plate (the mass per unit area) is constant, then
the total mass
of the plate is simply the density
times the area of the plate:

the area of

If the density varies continuously from point to point, say , then the mass of the plate is the average density of the plate times the area of the plate:

average density the area of

This is an integral:

The center of mass of a rod
is a density-weighted average taken over the interval occupied by the rod:

The center of mass of a plate
is determined by two density-weighted averages, each taken over the region
occupied by the plate:

**Problem 1.** A plate is in the form of a half-disc of radius
.
Find the mass of the plate and the center of mass given that the mass density
of the plate varies directly as the distance from the center of the straight
edge of the plate.

*Solution.* Place the plate over the x-axis. The mass density can then
be written
Here

Thus

Since
we have

If the plate is homogeneous, then the mass density
is constantly
where
is the area of the base region
.
In this case the center of mass of the plate falls on the
**centroid** of the base region . The centroid
depends only on the geometry of
:

Dividing by
and multiplying through by
we have

Thus
is the average
-coordinate
on
and
is the average
-coordinate
on
.
The mass of the plate does not enter into this at all.

**Problem 2.** Find the centroid of the region

*Solution.*

A particle of mass
at a distance
from a given line rotates about that line with angular speed
.
The speed
of the particle is then
and the kinetic energy is given by the formula

Imagine now a rigid body composed of a finite number of point masses
located at distances
from some fixed line. If the rigid body rotates about that line with angular
speed
,
then all the point masses rotate about that same line with that same angular
speed
.
The kinetic energy of the body can be obtained by adding up the kinetic
energies of all the individual particles:

The expression in parentheses is called the **moment of inertia**
(or *rotational inertia*) of the body and is denoted by the letter
:

For a rigid body in straight-line motion

For a rigid body in rotational motion

Suppose that a plate in the shape of a basic region
rotates about an axis. The moment of inertia of the plate about that axis is
given by the formula

where
is the mass density function and
is the distance from the axis to the point

*Derivation.* Decompose the plate into
pieces in the form of basic regions
From each
,
choose a point
and view all the mass of the plate as concentrated there. The moment of
inertia of this piece is then approximately

area of area of

The sum of these approximations,

is a Riemann sum for the double integral

As the maximum diameter of the
tends to zero, the Riemann sum tends to this integral.

**Problem 3.** A rectangular plate of mass
,
length
,
and width
with lower left-corner situated at
rotates about the
-axis.
Find the moment of inertia of the plate about that axis (a) given that the
plate has uniform mass density. (b) given that the mass density of the plate
varies directly as the square of the distance from the rightmost side.

Solution. Let the plate be placed in the rectangle with corners

(a) Here
and
Thus

(b) In this case
but we still have
Therefore

We can eliminate the constant of proportionality
by noting that

Therefore
and

If the mass of an object is all concentrated at a distance from a given axis, then the moment of inertia about that axis is given by the product .

Suppose now that we have a plate of mass
(actually any object of mass
will do here), and suppose that
is some line. The object has some moment of inertia
about
.
Its **radius of gyration** about
is the distance
for which

Namely, the radius of gyration about
is the distance from
at which all the mass of the object would have to be concentrated to effect
the same moment of inertia. The formula for radius of gyration is usually
written

**Problem 4.** A homogeneous circular plate of mass
and radius
rotates about an axle that passes through the center of the plate and is
perpendicular to the plate. Calculate the moment of inertia and the radius of
gyration.

*Solution.* Take the axle as the
-axis
and let the plate rest on the circular region
The density of the plate is
and
Hence

The radius of gyration
is
The circular plate of radius
has the same moment of inertia about the central axle as a circular wire of
the same mass with radius
.
The circular wire is a more efficient carrier of moment of inertia than the
circular plate.

Suppose we have an object of mass
and a line
that passes through the center of mass of the object. The object has some
moment of inertia about that line; call it
.
If
is any line parallel to
,
then the object has a certain moment of inertia about
;
call that
The parallel axis theorem states that

where
is the distance between the axes.

We will prove the theorem under somewhat restrictive assumptions. Assume that
the object is a plate of mass
in the shape of a basic region
,
and assume that
is perpendicular to the plate. Call
the
-axis.
Call the plane of the plate the
-plane.
Denoting the points of
by
we have

An obvious consequence of the parallel axis theorem is that for all lines parallel to . To minimize moment of inertia we must pass our axis through the center of mass.

Once we are familiar with double integrals

it is not hard to understand triple integrals

Basically the only difference is this: instead of working with functions of
two variables continuous on a plane region
we will be working with functions of three variables continuous on some
portion
of three-space.

For double integration we began with a rectangle

For triple integration we begin with a box

To partition this box, we first partition the edges. Taking a partition

a partition

and a partition

we form the Cartesian product

and call this a partition of
.
breaks up
into
nonoverlapping boxes

Taking

as the maximum value of on

and

as the minimum value of on

we form the **upper sum
**and the

**DEFINITION** THE TRIPLE INTEGRAL OVER A BOX

The unique number
that satisfies the inequality

is called the **triple integral** of
over
and is denoted by

We start with a three-dimensional, bounded, open, connected set and adjoin to
it the boundary. We now have a three-dimensional, bounded, closed, connected
set
.
We assume that
is a *basic solid*; that is, we assume that the boundary of
consists of a finite number of continuous surfaces

Now let's suppose that
is some function continuous on
.
To define the triple integral of
over
we first encase
in a rectangular box
.
We then extend
to all of
by defining
to be zero 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 over
;
that is, there still exists a unique number
such that

The number
is by definition the triple integral

We define the triple integral over
by setting

The simplest triple integral of interest is the triple integral of the function that is constantly one on T. This gives the volume of :

volume of =

Proof. Set for all in . Encase in a box . Define to be zero outside of An arbitrary partition of breaks up into little boxes Note that

the sum of the volumes of all the that are contained in

the sum of the volumes of all the that intersect .

It follows that

the volume of .

The arbitrariness of gives the formula.

Below we give without proof the salient elementary properties of the triple integral. They are all analogous to what you saw in the one- and two-variable cases. The referred to is a basic solid. The functions and are assumed to be continuous on .

I. The triple integral is linear:

II. It preserves order:

if on , then

if on , then

III. It is additive:

if is broken up into a finite number of basic solids then

IV. It satisfies a mean-value condition: namely, there is a point
in
for which

We call
the **average value** of
on
.

The notion of average given above enables us to write

We can also take weighted averages: if
and
are continuous and
is nonnegative on
,
then there is a point
in
for which

We call
the
**-weighted
average** of
on
.

The formulas for mass, center of mass, and moments of inertia derived in the previous section for two-dimensional plates are easily extended to three-dimensional objects.

Suppose that
is an object in the form of a basic solid. If
has constant mass density
(here density is mass per unit volume), then the mass of
is the density
times the volume of
:

If the mass density varies continuously over T, say
,
then the mass of
is the average density of
times the volume of
:

The coordinates of the *center of mass*
are density-weighted averages:

If the object
is homogeneous (constant mass density
),
then the center of mass of
depends only on the geometry of
and falls on the *centroid*
of the space occupied by
.
The density is irrelevant. The coordinates of the centroid are simple averages
over
:

The moment of inertia of
about a line is given by the formula

Here
is the mass density of
at
and
is the distance of
from the line in question. The moments of inertia about the
axes are again denoted by
.