THE DOUBLE INTEGRAL AS THE LIMIT OF RIEMANN SUMS; POLAR COORDINATES

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:
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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 $S$ is a bounded closed set (on the line, in the plane, or in three-space). For any two points $P$ and $Q$ of $S$ we can measure their separation, $d(P,Q).$ The greatest separation between points of $S$ is called the diameter of $\QTR{bf}{S}$:

diam $S$ = MATH $d(P,Q)$.

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 $\Omega $ and decompose it into a finite number of basic subregions MATH

If $f$ is continuous on $\Omega $, then $f$ is continuous on each $\Omega _i$. Now from each $\Omega _i$, we pick an arbitrary point $(x_i^{*},y_i^{*})$ and form the Riemann sum
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The double integral over $\Omega $ can be obtained as the limit of such sums; namely, given any $\epsilon >0$, there exists $\delta >0$ such that, if the diameters of the are all $\Omega _{i}$ less than $\delta $ then
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no matter how the MATH are chosen within the $\Omega .$ We express this by writing
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Evaluating Double Integrals Using Polar Coordinates

Here we explain how to calculate double integrals
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using polar coordinates $(r,\theta )$. Throughout we take $r\ge 0$ and restrict $\theta $ to $[0,2\pi ].$

We will work with the type of region consisting of all points $(x,y)$ that have polar coordinates $(r,\theta )$ in the set
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According to the formula in one-variable case the area of $\Omega $ is given by
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We can write this as a double integral over $\Gamma $:
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Proof. Simply note that
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and therefore
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Now let's suppose that $f$ is some function continuous at each point $(x,y)$ of $\Omega $. Then the composition
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is continuous at each point $(r,\theta )$ of $\Gamma $. We will show that
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Proof. Our first step is to place a grid on $\Omega $ by using a finite number of rays $\theta =\theta _j$ and a finite number of continuous curves $r=\rho _k(\theta )$. This grid decomposes $\Omega $ into a finite number of little regions
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with polar coordinates in sets MATH. Note that
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Writing $F(r,\theta )$ for MATH, we have
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MATH

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This last expression is a Riemann sum for the double integral
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and, as such, differs from that integral by less than any preassigned positive $\epsilon $ provided only that the diameters of all the $\Omega _i$ are sufficiently small. This we can guarantee by making our grid sufficiently fine.

Problem 1. Use polar coordinates to evaluate
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where MATH is the portion of the unit disc that lies in the first quadrant.

Solution. Here
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Therefore
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Problem 2. Use polar coordinates to calculate the volume of a sphere of radius $R$.

Solution. In rectangular coordinates
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where $\Omega $ is the disc of radius $R$ centered at the origin. Here
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Therefore
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Problem 3. Calculate the volume of the solid bounded above by the lower nappe of the cone $z=2-\sqrt{x^2+y^2}$ and bounded below by the disc MATH

Solution
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The first integral is $2$(area of $\Omega $ ) $=2\pi $. We evaluate the second integral by changing to polar coordinates.

The equation $(x-1)^2+y^2=1$ simplifies to $x^2+y^2=2x$. In polar coordinates this becomes MATH, which simplifies to $r=2\cos \,\theta $. The disc MATH is the set of all points with polar coordinates in the set
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Therefore
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We then have MATH

Problem 4. Evaluate
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where $\Omega $ is the triangle bounded by
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Solution. The vertical side of the triangle is part of the line $x=1.$ In polar coordinates this is $r\cos \,\theta =1$, which can be written $r=\sec \,\theta $. Therefore
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where
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The double integral over $\Gamma $ reduces to
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For MATH
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Therefore the integral can be written
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The function $f(x)=e^{-x^2}$ has no elementary antiderivative. Nevertheless, by taking a circuitous route and then using polar coordinates, we can show that
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Proof. The circular disc MATH is the set of all $(x,y)$ with polar coordinates $(r,\theta )$ in the set MATH Therefore
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Let $S_{a}$ be the square MATH Since MATH and $e^{-(x^{2}+y^{2})}$ is positive,
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It follows that
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As MATH and MATH Therefore
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But
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Therefore
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This integral comes up frequently in probability theory and plays an important role in what is called ``statistical mechanics.''

SOME APPLICATIONS OF DOUBLE INTEGRATION

The Mass of a Plate

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

$M=\lambda \times $the area of $\Omega .$

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

$M=$ average density $\times $ the area of $\Omega .$

This is an integral:
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The Center of Mass of a Plate

The center of mass of a rod $x_M$ is a density-weighted average taken over the interval occupied by the rod:
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The center of mass of a plate $(x_{M},y_{M})$ is determined by two density-weighted averages, each taken over the region occupied by the plate:
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Problem 1. A plate is in the form of a half-disc of radius $a$. 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 MATH Here
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MATH

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Thus $x_{M}=0.$
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Since MATH we have MATH

Centroids

If the plate is homogeneous, then the mass density $\lambda $ is constantly $M/A$ where $A$ is the area of the base region $\Omega $. In this case the center of mass of the plate falls on the centroid of the base region . The centroid $(\bar x,\bar y)$ depends only on the geometry of $\Omega $:
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Dividing by $M$ and multiplying through by $A$ we have
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Thus $\bar x$ is the average $x$-coordinate on $\Omega $ and $\bar y$ is the average $y$-coordinate on $\Omega $. The mass of the plate does not enter into this at all.

Problem 2. Find the centroid of the region
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Solution.
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Kinetic Energy and Moment of Inertia

A particle of mass $m$ at a distance $r$ from a given line rotates about that line with angular speed $\omega $. The speed $v$ of the particle is then $r\omega $ and the kinetic energy is given by the formula
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Imagine now a rigid body composed of a finite number of point masses $m_{i}$ located at distances $r_{i}$ from some fixed line. If the rigid body rotates about that line with angular speed $\omega $, then all the point masses rotate about that same line with that same angular speed $\omega $. The kinetic energy of the body can be obtained by adding up the kinetic energies of all the individual particles:
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The expression in parentheses is called the moment of inertia (or rotational inertia) of the body and is denoted by the letter $I$:
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For a rigid body in straight-line motion
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For a rigid body in rotational motion
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The Moment of Inertia of a Plate

Suppose that a plate in the shape of a basic region $\Omega $ rotates about an axis. The moment of inertia of the plate about that axis is given by the formula
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where MATH is the mass density function and $r(x,y)$ is the distance from the axis to the point $(x,y).$

Derivation. Decompose the plate into $N$ pieces in the form of basic regions MATH From each $\Omega _i$, choose a point $(x_i^{*},y_i^{*})$ and view all the mass of the plate as concentrated there. The moment of inertia of this piece is then approximately

MATHarea of MATHarea of $\Omega _i)$

The sum of these approximations,
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is a Riemann sum for the double integral
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As the maximum diameter of the $\Omega _i$ tends to zero, the Riemann sum tends to this integral.

Problem 3. A rectangular plate of mass $M$, length $L$, and width $W $ with lower left-corner situated at $(0,0)$ rotates about the $y$-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 $R$ with corners MATH

(a) Here $\lambda (x,y)=M/LW$ and $r(x,y)=x.$ Thus
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(b) In this case MATH but we still have $r(x,y)=x.$ Therefore
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We can eliminate the constant of proportionality $k$ by noting that
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Therefore $k=3M/L^{3}W$ and
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Radius of Gyration

If the mass $M$ of an object is all concentrated at a distance $r$ from a given axis, then the moment of inertia about that axis is given by the product $Mr^2$.

Suppose now that we have a plate of mass $M$ (actually any object of mass $M$ will do here), and suppose that $l$ is some line. The object has some moment of inertia $I$ about $l$. Its radius of gyration about $l$ is the distance $K$ for which
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Namely, the radius of gyration about $I$ is the distance from $l$ 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
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Problem 4. A homogeneous circular plate of mass $M$ and radius $R$ 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 $z$-axis and let the plate rest on the circular region MATH The density of the plate is $M/A=M/\pi R^2$ and MATH Hence
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The radius of gyration $K$ is MATH The circular plate of radius $R$ has the same moment of inertia about the central axle as a circular wire of the same mass with radius $R/\sqrt{2}$. 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 $M$ and a line $l_M$ that passes through the center of mass of the object. The object has some moment of inertia about that line; call it $I_M$. If $l$ is any line parallel to $l_M$, then the object has a certain moment of inertia about $l$; call that $I.$ The parallel axis theorem states that
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where $d$ is the distance between the axes.

We will prove the theorem under somewhat restrictive assumptions. Assume that the object is a plate of mass $M$ in the shape of a basic region $\Omega $, and assume that $l_{M}$ is perpendicular to the plate. Call $l$ the $z$-axis. Call the plane of the plate the $xy$-plane. Denoting the points of $\Omega $ by $(x,y)$ we have
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MATH

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An obvious consequence of the parallel axis theorem is that $I_{M}\le I$ for all lines $l$ parallel to $l_{M}$. To minimize moment of inertia we must pass our axis through the center of mass.

TRIPLE INTEGRALS

Once we are familiar with double integrals
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it is not hard to understand triple integrals
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Basically the only difference is this: instead of working with functions of two variables continuous on a plane region $\Omega ,$ we will be working with functions of three variables continuous on some portion $T$ of three-space.

The Triple Integral over a Box

For double integration we began with a rectangle
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For triple integration we begin with a box
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To partition this box, we first partition the edges. Taking a partition
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a partition
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and a partition
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we form the Cartesian product
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and call this a partition of $\Pi $. $P$ breaks up $\Pi $ into $m\times n\times q$ nonoverlapping boxes
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Taking

$M_{ijk}$ as the maximum value of $f$ on $\Pi _{ijk}$

and

$m_{ijk}$ as the minimum value of $f$ on $\Pi _{ijk}$

we form the upper sum
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and the lower sum
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As in the case of functions of one and two variables, it turns out that, with $f$ continuous on $\Pi $, there is one and only one number $I$ that satisfies the inequality
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DEFINITION THE TRIPLE INTEGRAL OVER A BOX $\Pi $

The unique number $I$ that satisfies the inequality
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is called the triple integral of $f$ over $\Pi $ and is denoted by
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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 $T$. We assume that $T$ is a basic solid; that is, we assume that the boundary of $T$ consists of a finite number of continuous surfaces MATH

Now let's suppose that $f$ is some function continuous on $T$. To define the triple integral of $f$ over $T$ we first encase $T$ in a rectangular box $\Pi $. We then extend $f$ to all of $\Pi $ by defining $f$ to be zero outside of $T$. This extended function $f$ is bounded on $\Pi $, and it is continuous on all of $\Pi $ except possibly at the boundary of $T$. In spite of these possible discontinuities, $f$ is still integrable over $\Pi $; that is, there still exists a unique number $I$ such that
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The number $I$ is by definition the triple integral
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We define the triple integral over $T$ by setting
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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 $T$:

volume of $T$ =$\iiint_T\,dxdydz.$

Proof. Set $f(x,y,z)=1$ for all $(x,y,z)$ in $T$. Encase $T$ in a box $\Pi $. Define $f$ to be zero outside of $T.$ An arbitrary partition $P$ of $\Pi $ breaks up $T$ into little boxes $\Pi _{ijk}.$ Note that

$L_f(P)=$ the sum of the volumes of all the $\Pi _{ijk}$ that are contained in $T$

$U_f(P)=$ the sum of the volumes of all the $\Pi _{ijk}$ that intersect $T$.

It follows that

$L_f(P)\le $ the volume of $T$ $\le U_f(P)$.

The arbitrariness of $P$ 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 $T$ referred to is a basic solid. The functions $f$ and $g$ are assumed to be continuous on $T$.

I. The triple integral is linear:
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II. It preserves order:

if $f\geq 0$ on $T$, then MATH

if $f\le g$ on $T$, then MATH

III. It is additive:

if $T$ is broken up into a finite number of basic solids $T_1,\cdots ,T_n,$ then

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IV. It satisfies a mean-value condition: namely, there is a point $(x_0,y_0,z_0)$ in $T$ for which
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We call $f(x_0,y_0,z_0)$ the average value of $f$ on $T$.

The notion of average given above enables us to write
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We can also take weighted averages: if $f$ and $g$ are continuous and $g$ is nonnegative on $T$, then there is a point $(x_0,y_0,z_0)$ in $T$ for which
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We call $f(x_0,y_0,z_0)$ the $g$-weighted average of $f$ on $T$.

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 $T$ is an object in the form of a basic solid. If $T$ has constant mass density $\lambda $ (here density is mass per unit volume), then the mass of $T$ is the density MATH times the volume of $T$:
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If the mass density varies continuously over T, say MATH, then the mass of $T$ is the average density of $T$ times the volume of $T$:
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The coordinates of the center of mass MATH are density-weighted averages:
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If the object $T$ is homogeneous (constant mass density $M/V$), then the center of mass of $T$ depends only on the geometry of $T$ and falls on the centroid MATH of the space occupied by $T$. The density is irrelevant. The coordinates of the centroid are simple averages over $T$:
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The moment of inertia of $T$ about a line is given by the formula
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Here $\lambda (x,y,z)$ is the mass density of $T$ at $(x,y,z)$ and $r(x,y,z)$ is the distance of $(x,y,z)$ from the line in question. The moments of inertia about the $x,y,z$ axes are again denoted by $I_x,I_y,I_z$.

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