# Multiple integrals ​

See also the graded exercises and additional exercises associated to this part of the course. If you want more, Chapter 5 (plus sections 6.3 and 6.4) of OpenStax Calculus Volume 3 is a good option.

The extension to higher dimension of differentiation was established in the previous chapters. We then defined line integrals which are, in a sense, one dimensional integrals which exist in a high dimensional setting. We now take the next step and define higher dimensional integrals in the sense of how to integrate a scalar field defined on a subset of ${\mathbb{R}}^{n}$. The first step will be to rigorously define which scalar fields are integrable and to define the integral. Then we need to find reasonable ways to evaluate such integrals. Among other applications we will use this multiple integrals to calculate volumes and moment of inertia. In Green's Theorem we find a connection between multiple integrals and line integrals. We also develop the important topic of change of variables which takes advantage of the Jacobian determinant and is often invaluable for actually working with a given problem.

## Definition of the integral ​

First we need to find a definition of integrability and the integral. Then we will proceed to study the properties of this higher dimensional integral. Recall that, in the one-dimensional case integration was defined using the following steps:

1. Define the integral for step functions,

2. Define integral for "integrable functions",

3. Show that continuous functions are integrable.

For higher dimensions we follow the same logic. We will then show that we can evaluate higher dimensional integrals by repeated one-dimensional integration.

Definition (partition)

Let $R=\left[{a}_{1},{b}_{1}\right]×\left[{a}_{2},{b}_{2}\right]$ be a rectangle. Suppose that ${P}_{1}=\left\{{x}_{0},\dots ,{x}_{m}\right\}$ and ${P}_{2}=\left\{{y}_{0},\dots ,{y}_{n}\right\}$ such that ${a}_{1}={x}_{0}<{x}_{1}<\cdots <{x}_{m}={b}_{1}$ and ${a}_{2}={y}_{0}<{y}_{1}<\cdots <{y}_{n}={b}_{2}$. $P={P}_{1}×{P}_{2}$ is said to be a partition of $R$.

Observe that a partition divides $R$ into $nm$ sub-rectangles. If $P\subseteq Q$ then we say that $Q$ is a finer partition than $P$. Partitions are constructed in higher dimension, for ${\mathbb{R}}^{n}$, in an analogous way. Before defining integration for general functions it is convenient to make the definition for a special class of functions called step functions.

Definition (step function)

A function $f:R\to \mathbb{R}$ is said to be a step function if there is a partition $P$ of $R$ such that $f$ is constant on each sub-rectangle of the partition.

If $f$ and $g$ are step functions and $c,d\in \mathbb{R}$, then $cf+dg$ is also a step function. Also note that the area of the sub-rectangle ${Q}_{jk}:=\left[{x}_{j},{x}_{j+1}\right]×\left[{y}_{k},{y}_{k+1}\right]$ is equal to $\left({x}_{j+1}-{x}_{j}\right)\left({y}_{k+1}-{y}_{k}\right)$.

We can now define the integral of a step function in a reasonable way. The definition here is for 2D but the analogous definition holds for any dimension.

Suppose that $f$ is a step function with value ${c}_{jk}$ on the sub-rectangle $\left({x}_{j},{x}_{j+1}\right)×\left({y}_{k},{y}_{k+1}\right)$. Then we define the integral as

This should remind you of Riemann sums from Analysis I. Observe that the value of the integral is independent of the partition, as long as the function is constant on each sub-rectangle. In this sense the integral is well-defined (not dependent on the choice of partition used to calculate it).

Theorem

Let $f,g$ be step functions. Then

Proof

All properties follow from the definition by basic calculations.

We are now in the position to define the set of integrable functions. In order to define integrability we take advantage of "upper" and "lower" integrals which "sandwich" the function we really want to integrate.

Definition (integrability on a rectangle)

Let $R$ be a rectangle and let $f:R\to \mathbb{R}$ be a bounded function. We call $f$ an integrable function if there is one and only one number $I\in \mathbb{R}$ such that

for every pair of step functions $g,h$ such that, for all $\left(x,y\right)\in R$,

$g\left(x,y\right)\le f\left(x,y\right)\le h\left(x,y\right).$

This number $I$ is called the integral of $f$ on $R$ and is denoted .

All the basic properties of the integral of step functions, as stated in the above Theorem, also hold for the integral of any integrable functions. This can be shown by considering the limiting procedure of the upper and lower integral of step functions which are part of the definition of integrability.

The most important words in the definition are "only one number": that's what we need to check to verify that a function is integrable. That still isn't immediately easy to check and so it is convenient to now investigate the integrability of continuous functions.

Theorem

Suppose that $f$ is a continuous function defined on the rectangle $R$. Then $f$ is integrable.

Proof

Continuity implies boundedness and so upper and lower integrals exist. Let $ϵ>0$. Exists $\delta >0$ such that $|f\left(\mathbf{x}\right)-f\left(\mathbf{y}\right)|\le ϵ$ whenever $‖\mathbf{x}-\mathbf{y}‖\le \delta$. We can choose a partition such that $‖\mathbf{x}-\mathbf{y}‖\le \delta$ whenever $\mathbf{x},\mathbf{y}$ are in the same sub-rectangle ${Q}_{jk}$. We then define the step functions $g,h$ s.t. $g\left(\mathbf{x}\right)=\underset{Qjk}{inf}f$, $h\left(\mathbf{x}\right)=\underset{Qjk}{sup}f$ when $\mathbf{x}\in {Q}_{jk}$. To finish the proof we observe that $|\underset{Qjk}{inf}f-\underset{Qjk}{sup}f|\le ϵ$ and $ϵ>0$ can be made arbitrarily small, so we can make the upper and lower integrals as close as we want.

## Evaluation of multiple integrals ​

Now we have a definition, so we know what a multidimensional integral is, and we also know that some interesting ones exist, but it is essential to also have a way to practically evaluate any given integral. It turns out we can do that by integrating in one variable at a time:

Theorem (Fubini)

Let $f$ be an integrable function on the rectangle $R=\left[{a}_{1},{b}_{1}\right]×\left[{a}_{2},{b}_{2}\right]$. Then

Proof

To see this, think about any pair of step functions $g,h$ such that $g\le f\le h$. Since these are step functions,

since these are all just different names for the same sum, and the same is true for $h$. Using this,

in other words the iterated integral in the middle is bounded from above and below by the same upper and lower integrals as the integral of $f$, which leaves only one possible value

and the other equality holds for the same reason.

This integral naturally allows us to calculate the volume of a solid. Let $f\left(x,y\right)\le z\le g\left(x,y\right)$ be defined on the rectangle $R\subset {\mathbb{R}}^{2}$ and consider the 3D set defined as

$V=\left\{\left(x,y,z\right):\left(x,y\right)\in R,f\left(x,y\right)\le z\le g\left(x,y\right)\right\}.$

The volume of the set $V$ is equal to .

Up until now we have considered step function and continuous functions. As with one-dimensional integrals we can permit some discontinuities and we introduce the following concept to be able to control the functions with discontinuities sufficiently to guarantee that the integrals are well-defined.

Definition (Content zero sets)

A bounded subset $A\subset {\mathbb{R}}^{2}$ is said to have content zero if, for every $ϵ>0$, there exists a finite set of rectangles whose union includes $A$ and the sum of the areas of the rectangles is not greater than $ϵ$.

Examples of content zero sets include: finite sets of points; bounded line segments; continuous paths.

Theorem

Suppose that $\phi$ is a continuous function on $\left[a,b\right]$. Then the graph $\left\{\left(x,y\right):x\in \left[a,b\right],y=\phi \left(x\right)\right\}$ has zero content.

Proof

By continuity, for every $ϵ>0$, there exists $\delta >0$ such that $|\phi \left(x\right)-\phi \left(y\right)|\le ϵ$ whenever $|x-y|\le \delta$. We then take partition of $\left[a,b\right]$ into subintervals of length less than $\delta$. Using this partition we generate a cover of the graph which has area not greater than $2ϵ|b-a|$.

Theorem

Let $f$ be a bounded function on $R$ and suppose that the set of discontinuities $A\subset R$ has content zero. Then the double integral exists.

Proof

Take a cover of $A$ by rectangles with total area not greater than $\delta >0$. Let $P$ be a partition of $R$ which is finer than the cover of $A$. We may assume that $|\underset{Qjk}{inf}f-\underset{Qjk}{sup}f|\le ϵ$ on each sub-rectangle of the partition which doesn't contain a discontinuity of $f$. The contribution to the integral of bounding step functions from the cover of $A$ is bounded by $\delta sup|f|$.

## Regions bounded by functions ​

A major limitation is that we have only integrated over rectangles whereas we would like to integrate over much more general different shaped regions. This we develop now.

Suppose $S\subset R$ and $f$ is a bounded function on $S$. We extend $f$ to $R$ by defining

We use this notation in the following definition.

Definition

We say that $f:S\to \mathbb{R}$ is integrable if ${f}_{R}$ is integrable and define

Suppose that there are continuous functions ${\phi }_{1}$, ${\phi }_{2}$ on $\mathbb{R}$ and consider the set

$S=\left\{\left(x,y\right):a\le x\le b,{\phi }_{1}\left(x\right)\le y\le {\phi }_{2}\left(x\right)\right\}\subset {\mathbb{R}}^{2}.$

Not all sets can be written in this way but many can and such a way of describing a subset of ${\mathbb{R}}^{2}$ is convenient for evaluating integrals. We will call sets that can be written this way Type 1 sets in this section.

Let $S=\left\{\left(x,y\right):x\in \left[a,b\right],{\phi }_{1}\left(x\right)\le y\le {\phi }_{2}\left(x\right)\right\}$ where ${\phi }_{1},{\phi }_{2}$ are continuous and let $f$ be a bounded continuous function of $S$. Then $f$ is integrable on $S$ and

the set of discontinuity of ${f}_{R}$ is the boundary of $S$ in $R=\left[a,b\right]×\left[\stackrel{~}{a},\stackrel{~}{b}\right]$ which consists of the graphs of ${\phi }_{1}$, ${\phi }_{2}$. These graphs have zero content as we proved before. For each $x$, $f\left(x,y\right)$ is integrable since it has only two discontinuity points. Additionally . Observe that it doesn't make a different to the integral if we use $<$ or $\le$ in the definition of $S$ since the difference would be a content zero set.

We could also consider the following set

$R=\left\{\left(x,y\right):a\le y\le b,{\phi }_{1}\left(y\right)\le x\le {\phi }_{2}\left(y\right)\right\},$

which we will call a Type 2 set. This is just the same situation as above with the roles of $x$ and $y$ switched, that is

In the first case we could describe the representation as projecting along the $y$-coordinate whereas in the second case we are projecting along the $x$-coordinate. Many interesting sets are both Type 1 and Type 2, although very often one form is more obvious or more useful than the other.

For higher dimensions we need to also have an understanding of how to represent subsets of ${\mathbb{R}}^{n}$. Take for example a 3D solid; then we would hope to be able to "project" along one of the coordinate axis and so describe it using the 2D "shadow" and a pair of continuous functions. For example, consider the upside-down cone in this figure which has base of radius $5$ lying in the plane $\left\{z=5\right\}$ and has its tip at the origin.

In order to describe this set it is convenient to imagine how it projects down onto the $xy$-axis. We then describe it as

$V=\left\{\left(x,y,z\right):\left(x,y\right)\in S,{\gamma }_{1}\left(x,y\right)\le z\le {\gamma }_{2}\left(x,y\right)\right\}$

where $S\subset {\mathbb{R}}^{2}$ is the "shadow" and the functions represent the control we need in the vertical direction. In this case we must choose $S=\left\{\left(x,y\right):{x}^{2}+{y}^{2}\le {5}^{2}\right\}$since the base of the cone, at the top of the picture, it the largest part in terms of the shadow. We also must choose