# The Double Integral as the Limit of Riemann Sums

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

# Evaluating Double Integrals Using Polar Coordinates

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.''

# The Mass of a Plate

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 Plate

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

# Centroids

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.

# Kinetic Energy and Moment of Inertia

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

# The Moment of Inertia of a Plate

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.

# The Parallel Axis Theorem

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.

# TRIPLE INTEGRALS

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.

# The Triple Integral over a Box

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 lower sum

As in the case of functions of one and two variables, it turns out that, with continuous on , there is one and only one number that satisfies the inequality

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

# The Triple Integral over a More General Solid

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

# Volume as a Triple Integral

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.

# Some Properties of the Triple Integral

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

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 .

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