Let be a solid in The projection of onto the -plane is denoted by The solid is then the set of all with

in and

The triple integral over
can be evaluated by setting

Suppose that
takes the form

itself is then the set of all
with

The triple integral over
can then be expressed by three ordinary integrals:

It is customary to omit the brackets and parentheses and write

Here we first integrate with respect to
[from
to
],
then with respect to
[from
to
],
and finally with respect to
[from
to
].

There is nothing sacred about this order of integration. Other orders of
integration are possible and in some cases more convenient. Suppose, for
example, that the projection of
onto the
-plane
is a region of the form

If
is the set of all
with

then

In this case we integrate first with respect to
,
then with respect to
,
and finally with respect to
.
Still four other orders of integration are possible.

**Problem 1.** Use triple integration to find the volume of the
tetrahedron
bounded by the planes

Where is the centroid?

*Solution.* The volume of
is given by the triple integral

To evaluate this triple integral we can project
onto any one of the three coordinate planes. We will project onto the
-plane.
The base region is then the triangle

Since the inclined face is part of the plane
,
we have
as the set of all
with

It follows that

By symmetry
.
We can calculate
as follows:

Since
we have
The centroid is the point

**Problem 2.** Find the mass of a solid right circular cylinder
of radius
and height
given that the mass density varies directly with the distance from one of the
bases.

*Solution.* Call the solid
.
We can characterize
by the following inequalities:

The first two inequalities define the base region
.
Supposing that the density varies directly with the distance from the lower
base, we have
Then

**Remark.** In Problem 2 we would have profited by not skipping
the double integral stage; namely, we could have written

**Problem 3.** Integrate
over that part of the first octant
that is cut off by the ellipsoid

**Solution.** Call the solid
.
The upper boundary of
has equation

This surface intersects the
-plane
in the curve

We can take

as the base region and characterize
as the set of all
with

We can therefore calculate the triple integral by evaluating

A straightforward computation gives an answer of
.

*Another Solution.* We return to Problem 3 but this time carry out the
integration in a different order. The same solid is now projected onto the
-plane.
In terms of
and
the curved surface has equation

This surface intersects the
-plane
in the curve

We can take

as the base region and characterize
as the set of all
with

This leads to the repeated integral

which also gives

**Problem 4.** Use triple integration to find the volume of the
solid
bounded above by the parabolic cylinder
and bounded below by the elliptic paraboloid
.

*Solution.* Solving the two equations simultaneously, we have

and thus

This tells us that the two surfaces intersect in a space curve that lies along the elliptic cylinder The projection of this intersection onto the -plane is the ellipse .

The projection of
onto the
-plane
is the region

The solid
is then the set of all
with

Its volume is given by

The first two coordinates,
and
,
of the cylindrical coordinates
of a point in
-space
are the usual plane polar coordinates, except that
is taken as nonnegative and
is restricted to the interval
.
The third coordinate is the third rectangular coordinate
.
In rectangular coordinates the coordinate surfaces

are three mutually perpendicular planes. In cylindrical coordinates the
coordinate surfaces take the form

The first surface is a circular cylinder of radius . The central axis of the cylinder is the -axis. The surface is a vertical half-plane hinged at the -axis. The plane stands at an angle of radians from the positive -axis. The last coordinate surface is the plane of rectangular coordinates.

The
-solids
easiest to describe in cylindrical coordinates are the *cylindrical
wedges.* A wedge consists of all points
that have cylindrical coordinates
in the box

Suppose that
is some basic solid in
-space,
not necessarily a wedge. If
is the set of all
with cylindrical coordinates in some basic solid
in
-space,
then

**Derivation**. We will carry out the argument on the assumption
that
is projectable onto some basic region
of the
-plane.
(It is for such solids that the formula is most useful.)
has some lower boundary
and some upper boundary
is then the set of all
with

The region
has polar coordinates in some set
(which we assume is a basic region). Then
is the set of all
with

Therefore

If
for all
in
,
then the formula reduces to

The triple integral on the left is the volume of
.
In summary, if
is a basic solid in
-space
and the cylindrical coordinates of
constitute a basic solid
in
-space,
then the volume of
is given by the formula

Cylindrical coordinates are particular]y useful when an axis of symmetry is present. The axis of symmetry is then taken as the -axis.

**Problem 1.** Find the mass of a solid cylinder
of radius
and height
given that the density varies directly with the distance from the axis of the
cylinder.

*Solution.* Place the cylinder
on the
-plane
so that the axis of
coincides with the
-axis.
The density function then takes the form
and
consists of all points
with cylindrical coordinates
in the set

Therefore

**Problem 2.** Use cylindrical coordinates to find the volume of
the solid
bounded above by the plane
and below by the paraboloid

*Solution.* In cylindrical coordinates the plane has equation
and the paraboloid has equation
.
Solving these two equations simultaneously, we have
.
This tells us that the two surfaces intersect in a space curve that lies along
the circular cylinder
.
The projection of this intersection onto the
-plane
is the circle with polar equation
.
The base region
is thus the set of all
with polar coordinates in the set

itself is the set of all
with cylindrical coordinates in the set

Therefore

**Problem 3.** Locate the centroid of the solid in Problem 2.

*Solution.* Since
is symmetric about the
-plane,
we see that
.
To get
we begin as usual:

Since
, we have
Now for
:

Division by
gives
The centroid is thus the point

We have seen how single integrals and double integrals can be obtained as limits of Riemann sums. The same holds true for triple integrals.

Start with a basic solid
in
-space
and decompose it into a finite number of basic solids
.
If
is continuous on
,
then
is continuous on each
.
From each
pick an arbitrary point
and form the Riemann sum

The triple integral over
is 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

The spherical coordinates
of a point
in
-space
have the following meaning. The first coordinate,
,
is the distance from the origin; thus
.
The second coordinate, the angle
,
is the longitude;
ranges from
to
.
The third coordinate, the angle marked
,
ranges only from
to
is the **colatitude**, or more simply the **polar
angle**. (The complement of
would be the *latitude* on a globe.)

The coordinate surfaces

have the following meaning. The surface
is a sphere; the radius is
and the center is the origin. The second surface,
,
is the same as in cylindrical coordinates: the vertical half-plane hinged at
the
-axis
and standing at an angle of
radians from the positive
-axis.
The surface
requires detailed explanation. If
or
,
the surface is a nappe of a cone; it is generated by revolving about the
-axis
any ray that emerges from the origin at an angle of
radians from the positive
-axis.
The surface
is the
-plane.
(The nappe of the cone has opened up completely.) The equation
gives the positive
-axis,
and the equation
gives the negative
-axis.
(When
or
,
the nappe of the cone has closed up completely.)

Rectangular coordinates
are related to spherical coordinates
by the following equations:

Conversely, with obvious exclusions, we have

A *spherical wedge*
consists of all points
that have spherical coordinates in the box

The volume of this wedge is given by the formula

Proof. Note first that is a solid of revolution. One way to obtain is to rotate the face of , call it , about the -axis for radians. On that face and play the role of polar coordinates. The face is the set of all with polar coordinates in the set The centroid of is at a distance from the -axis where

(area of )

As the face
is rotated from
to
the centroid travels through a circular arc of length

From Pappus formula we know that

the volume of (area of ) =

Suppose that
is a basic solid in
-space
with spherical coordinates in some basic solid
of
-space.
Then

Derivation. Assume first that
is a spherical wedge
.
is then a box
.
Now decompose
into
boxes
.
This induces a subdivision of
into
spherical wedges
.
Writing
for
to save space, we have

Here

and this expression is a Riemann sum for

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

This verifies the formula for the case where is a spherical wedge.

If
for all
in
,
then the change of variables formula reduces to

The integral on the left is the volume of
.
It follows that the volume of
is given by the formula

Spherical coordinates are commonly used in applications where there is a
center of symmetry. The center of symmetry is then taken as the origin.

**Problem 1.** Calculate the mass
of a solid ball of radius
given that the density varies directly with the square of the distance from
the center of the ball.

*Solution.* Center the ball at the origin. The ball, call it
,
is now the set of all
with spherical coordinates
in the box

Therefore

**Problem 2.** Find the volume of the solid
enclosed by the surface

*Solution.* In spherical coordinates the bounding surface takes the
form

This equation places no restriction on
;
thus
can range from
to
.
Since
remains nonnegative,
can range only from
to
Thus the solid
is the set of all
with spherical coordinates
in the set

The rest is straightforward: