REDUCTION TO REPEATED INTEGRALS

Let $T$ be a solid in $\QTR{bf}{R^3.}$ The projection of $T$ onto the $xy$-plane is denoted by $\Omega _{xy}.$ The solid $T$ is then the set of all $(x,y,z)$ with

$(x,y)$ in $\Omega _{xy}$ and MATH

The triple integral over $T$ can be evaluated by setting
MATH
Suppose that $\Omega _{xy}$ takes the form
MATH
$T$ itself is then the set of all $(x,y,z)$ with
MATH
The triple integral over $T$ can then be expressed by three ordinary integrals:
MATH
It is customary to omit the brackets and parentheses and write
MATH
Here we first integrate with respect to $z$ [from $z=\psi _1(x,y)$ to $z=\psi _2(x,y)$], then with respect to $y$ [from $y=\phi _1(x)$ to $y=\phi _2(x)$], and finally with respect to $x$ [from MATH to $x=a_2$].

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 $T$ onto the $xz$-plane is a region of the form
MATH
If $T$ is the set of all $(x,y,z)$ with
MATH
then
MATH
In this case we integrate first with respect to $y$, then with respect to $x$, and finally with respect to $z$. Still four other orders of integration are possible.

Problem 1. Use triple integration to find the volume of the tetrahedron $T$ bounded by the planes
MATH
Where is the centroid?

Solution. The volume of $T$ is given by the triple integral
MATH

To evaluate this triple integral we can project $T$ onto any one of the three coordinate planes. We will project onto the $xy$-plane. The base region is then the triangle
MATH
Since the inclined face is part of the plane $z=1-x-y$, we have $T$ as the set of all $(x,y,z)$ with
MATH
It follows that
MATH

MATH
By symmetry MATH. We can calculate $\bar x$ as follows:
MATH
Since $V=\frac 16$ we have $\bar x=\frac 14.$ The centroid is the point MATH

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

Solution. Call the solid $T$. We can characterize $T$ by the following inequalities:
MATH
The first two inequalities define the base region $\Omega _{xy}$. Supposing that the density varies directly with the distance from the lower base, we have MATH Then
MATH

MATH

MATH

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

MATH

Problem 3. Integrate $f(x,y,z)=yz$ over that part of the first octant MATH that is cut off by the ellipsoid
MATH

Solution. Call the solid $T$. The upper boundary of $T$ has equation
MATH
This surface intersects the $xy$-plane in the curve
MATH
We can take
MATH
as the base region and characterize $T$ as the set of all $(x,y,z)$ with
MATH
We can therefore calculate the triple integral by evaluating
MATH
A straightforward computation gives an answer of $\frac 1{15}ab^2c^2$ .

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 $yz$-plane. In terms of $y$ and $z$ the curved surface has equation
MATH
This surface intersects the $yz$-plane in the curve
MATH
We can take
MATH
as the base region and characterize $T$ as the set of all $(x,y,z)$ with
MATH
This leads to the repeated integral
MATH
which also gives MATH

Problem 4. Use triple integration to find the volume of the solid $T$ bounded above by the parabolic cylinder $z=4-y^2$ and bounded below by the elliptic paraboloid $z=x^2+3y^2$.

Solution. Solving the two equations simultaneously, we have

$4-y^2=x^2+3y^2$ and thus $x^2+4y^2=4.$

This tells us that the two surfaces intersect in a space curve that lies along the elliptic cylinder $x^2+4y^2=4.$ The projection of this intersection onto the $xy$-plane is the ellipse $x^2+4y^2=4$.

The projection of $T$ onto the $xy$-plane is the region
MATH
The solid $T$ is then the set of all $(x,y,z)$ with
MATH
Its volume is given by
MATH

MATH

CYLINDRICAL COORDINATES

Introduction to Cylindrical Coordinates

The first two coordinates, $r$ and $\theta $, of the cylindrical coordinates $(r,\theta ,z)$ of a point in $xyz$-space are the usual plane polar coordinates, except that $r$ is taken as nonnegative and $\theta $ is restricted to the interval $[0,2\pi ]$. The third coordinate is the third rectangular coordinate $z$. In rectangular coordinates the coordinate surfaces
MATH
are three mutually perpendicular planes. In cylindrical coordinates the coordinate surfaces take the form
MATH

The first surface is a circular cylinder of radius $r_0$. The central axis of the cylinder is the $z$-axis. The surface $\theta =\theta _0$ is a vertical half-plane hinged at the $z$-axis. The plane stands at an angle of $\theta _0$ radians from the positive $x$-axis. The last coordinate surface is the plane $z=z_0$ of rectangular coordinates.

The $xyz$-solids easiest to describe in cylindrical coordinates are the cylindrical wedges. A wedge consists of all points $(x,y,z)$ that have cylindrical coordinates $(r,\theta ,z)$ in the box
MATH

Evaluating Triple Integrals Using Cylindrical Coordinates

Suppose that $T$ is some basic solid in $xyz$-space, not necessarily a wedge. If $T$ is the set of all $(x,y,z)$ with cylindrical coordinates in some basic solid $S$ in $r\theta z$-space, then
MATH

Derivation. We will carry out the argument on the assumption that $T$ is projectable onto some basic region $\Omega _{xy}$ of the $xy$-plane. (It is for such solids that the formula is most useful.) $T$ has some lower boundary $z=\psi _1(x,y)$ and some upper boundary $z=\psi _2(x,y).$ $T$ is then the set of all $(x,y,z)$ with
MATH
The region $\Omega _{xy}$ has polar coordinates in some set $\Omega _{r\theta }$ (which we assume is a basic region). Then $S$ is the set of all $(r,\theta ,z)$ with
MATH
Therefore
MATH

MATH

MATH

Volume Formula

If $f(x,y,z)=1$ for all $(x,y,z)$ in $T$, then the formula reduces to
MATH
The triple integral on the left is the volume of $T$. In summary, if $T$ is a basic solid in $xyz$-space and the cylindrical coordinates of $T$ constitute a basic solid $S$ in $r\theta z$-space, then the volume of $T$ is given by the formula
MATH

Calculations

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

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

Solution. Place the cylinder $T$ on the $xy$-plane so that the axis of $T$ coincides with the $z$-axis. The density function then takes the form MATH and $T$ consists of all points $(x,y,z)$ with cylindrical coordinates $(r,\theta ,z)$ in the set
MATH
Therefore
MATH

MATH

Problem 2. Use cylindrical coordinates to find the volume of the solid $T$ bounded above by the plane $z=y$ and below by the paraboloid $z=x^2+y^2.$

Solution. In cylindrical coordinates the plane has equation $z=r\sin \,\theta $ and the paraboloid has equation $z=r^{2}$. Solving these two equations simultaneously, we have $r=\sin \,\theta $. This tells us that the two surfaces intersect in a space curve that lies along the circular cylinder $r=\sin \,\theta $. The projection of this intersection onto the $xy $-plane is the circle with polar equation $r=\sin \,\theta $. The base region $\Omega _{xy}$ is thus the set of all $(x,y)$ with polar coordinates in the set
MATH
$T$ itself is the set of all $(x,y,z)$ with cylindrical coordinates in the set
MATH
Therefore
MATH

MATH

MATH

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

Solution. Since $T$ is symmetric about the $yz$-plane, we see that $\bar{x}=0$. To get $\bar{y}$ we begin as usual:
MATH

MATH

MATH
Since $V=\frac{\pi }{32}$ , we have MATH Now for $\bar{z}$:
MATH
Division by $V=\frac{\pi }{32}$ gives MATH The centroid is thus the point MATH

THE TRIPLE INTEGRAL AS THE LIMIT OF RIEMANN SUMS; SPHERICAL COORDINATES

The Triple Integral as the Limit of Riemann Sums

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 $T$ in $xyz$-space and decompose it into a finite number of basic solids $T_{1},...,T_{N}$. If $f$ is continuous on $T$, then $f$ is continuous on each $T_{i}$. From each $T_{i}$ pick an arbitrary point MATH and form the Riemann sum
MATH
The triple integral over $T$ is the limit of such sums; namely, given any $\epsilon >0,$ there exists $\delta >0$ such that, if the diameters of the $T_{i}$ are all less than $\delta $, then
MATH
no matter how the MATH are chosen within the $T_{i}$. We express this by writing
MATH

Introduction to Spherical Coordinates

The spherical coordinates MATH of a point $P$ in $xyz$-space have the following meaning. The first coordinate, $\rho $, is the distance from the origin; thus $\rho \ge 0$. The second coordinate, the angle $\theta $, is the longitude; $\theta $ ranges from $0$ to $2\pi $. The third coordinate, the angle marked $\phi $, ranges only from $0$ to $\pi $ is the colatitude, or more simply the polar angle. (The complement of $\phi $ would be the latitude on a globe.)

The coordinate surfaces
MATH
have the following meaning. The surface $\rho =\rho _0$ is a sphere; the radius is $\rho _0$ and the center is the origin. The second surface, $\theta =\theta _0$, is the same as in cylindrical coordinates: the vertical half-plane hinged at the $z$-axis and standing at an angle of $\theta _0$ radians from the positive $x$-axis. The surface $\phi =\phi _0$ requires detailed explanation. If MATH or MATH, the surface is a nappe of a cone; it is generated by revolving about the $z$-axis any ray that emerges from the origin at an angle of $\phi _0$ radians from the positive $z$-axis. The surface $\phi =\frac \pi 2$ is the $xy$-plane. (The nappe of the cone has opened up completely.) The equation $\phi =0$ gives the positive $z$-axis, and the equation $\phi =\pi $ gives the negative $z$-axis. (When $\phi =0$ or $\phi =\pi $, the nappe of the cone has closed up completely.)

Rectangular coordinates $(x,y,z)$ are related to spherical coordinates MATH by the following equations:
MATH
Conversely, with obvious exclusions, we have
MATH

The Volume of a Spherical Wedge

A spherical wedge $W$ consists of all points $(x,y,z)$ that have spherical coordinates in the box
MATH
The volume of this wedge is given by the formula
MATH

Proof. Note first that $W$ is a solid of revolution. One way to obtain $W$ is to rotate the $\theta =b_1$ face of $W$, call it $\Omega $, about the $z$-axis for MATH radians. On that face $\rho $ and MATH play the role of polar coordinates. The face $\Omega $ is the set of all $(z,X)$ with polar coordinates $(\rho ,\alpha )$ in the set MATH The centroid of $\Omega $ is at a distance $\bar X$ from the $z$-axis where

$\bar X$(area of MATH) MATH

MATH

As the face $\Omega $ is rotated from $\theta =b_1$ to $\theta =b_2,$ the centroid travels through a circular arc of length
MATH

From Pappus formula we know that

the volume of $W=s$(area of $\Omega $) =MATH


MATH

MATH

Evaluating Triple Integrals Using Spherical Coordinates

Suppose that $T$ is a basic solid in $xyz$-space with spherical coordinates in some basic solid $S$ of $\rho \theta \phi $-space. Then
MATH

Derivation. Assume first that $T$ is a spherical wedge $W$. $S$ is then a box $\Pi $. Now decompose $\Pi $ into $N$ boxes $\Pi _1,...,\Pi _N$. This induces a subdivision of $W$ into $N$ spherical wedges $W_1,...,W_N$. Writing MATH for MATH to save space, we have
MATH

MATH

MATH

MATH
Here
MATH

MATH

MATH
and this expression is a Riemann sum for
MATH
and, as such, will differ from that integral by less than any preassigned positive number $\epsilon $ provided only that the diameters of all the $W_i$ are sufficiently small. This we can guarantee by making the diameters of all the $\Pi _i$ sufficiently small.

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

Volume Formula

If $f(x,y,z)=1$ for all $(x,y,z)$ in $T$, then the change of variables formula reduces to
MATH
The integral on the left is the volume of $T$. It follows that the volume of $T$ is given by the formula
MATH
Spherical coordinates are commonly used in applications where there is a center of symmetry. The center of symmetry is then taken as the origin.

Calculations

Problem 1. Calculate the mass $M$ of a solid ball of radius $1$ 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 $T$, is now the set of all $(x,y,z)$ with spherical coordinates MATH in the box
MATH
Therefore
MATH

MATH

MATH

MATH

MATH

Problem 2. Find the volume of the solid $T$ enclosed by the surface
MATH

Solution. In spherical coordinates the bounding surface takes the form
MATH
This equation places no restriction on $\theta $; thus $\theta $ can range from $0$ to $2\pi $. Since $\rho $ remains nonnegative, $\QTR{group}{\phi }$ can range only from $0$ to $\frac{\pi }{2}.$ Thus the solid $T$ is the set of all $(x,y,z)$ with spherical coordinates MATH in the set
MATH
The rest is straightforward:
MATH

MATH

MATH

This document created by Scientific WorkPlace 4.0.