\documentclass[11pt]{amsart}
\usepackage{graphicx}
%\usepackage{amssymb}
\usepackage{epstopdf}
\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `basename #1 .tif`.png}
\def\Dx{\Delta x}
\def\Dy{\Delta y}
\def\Dz{\Delta z}
\def\trint{\mathop{\int\!\int\!\int}}
\def\ds{\displaystyle}
\textwidth = 6.5 in
\textheight = 9 in
\oddsidemargin = 0.0 in
\evensidemargin = 0.0 in
\topmargin = 0.0 in
\headheight = 0.0 in
\headsep = 0.0 in
\parskip = 0.2in
\parindent = 0.0in
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}{Definition}
\theoremstyle{definition}
\newtheorem{example}{Example}
%\title{Brief Article}
%\author{The Author}
\begin{document}
%\maketitle
\begin{centering}
\textbf{\large Notes on Triple integrals: Wednesday, November 26}
\end{centering}
These are some notes for my lecture on triple integrals. The idea of a triple integral is similar to the idea of a double integral. We are given some solid region $E$ in 3-space, and a function $f(x,y,z)$, and we want to know `how much of $f$ is there in the region $E$'. A good example to think about is when $f$ represents the density, say in mass/(unit volume) of a solid object $E$, and we want to know the the total mass. (We did something analogous for double integrals, where the function represents mass/(unit area).)
The idea, that should be familiar from our discussion of double integrals, is to break the region $E$ into small cubes, whose sides are given by $\Dx$, $\Dy$, and $\Dz$. The volume of such a small cube is given by $\Delta V = \Dx\Dy\Dz$.
\vspace{-.2in}
\begin{center}
\includegraphics[width=1.5in]{deltav}
\end{center}
%\vspace{.2in}
Assuming that the function $f$ is continuous, ie doesn't vary too much over a small region, this small cube contributes roughly $f(x,y,z)\Delta V = f(x,y,z)\Dx\Dy\Dz$ worth of mass. The total mass is approximately a sum of such terms; in the limit as the size of the cube goes to $0$, the sum converges to what we call the triple integral of $f$ over $E$, denoted
$$
\mathop{\int\!\int\!\int}_{E} f(x,y,z)dV =\mathop{\int\!\int\!\int}_E f(x,y,z)dxdydz
$$
If the function $f$ is the constant function $1$, then the integral represents the volume of the region $E$.
We want to learn how to calculate these triple integrals; the procedure is analogous to what we did with double integrals. The main complication is that $3$-dimensional shapes are even more complicated that $2$-dimensional ones. Let's start with the easiest kind, where $E$ is a box $[a,b] \times [c,d] \times [e,f]$. Then we can express the integral as an iterated integral (and carry it out in any of the 6 possible orders).
\begin{example}
Compute the integral of the function $f(x,y,z) = \frac{yz}{x}$ over the rectangular solid $1\le x\le 2$, $0\le y\le 1$, $0\le z\le 2$, also known as $[1,2]\times [0,1][\times[0,2]$, or in other words
$\ds
\int_0^2\int_0^1\int_1^2 \frac{yz}{x} dx\; dy\; dz$. As the notation suggests, we integrate from inside out, and so do the x-integral, then the y-integral, and finally the z-integral.
\begin{align*}
\int_0^2\int_0^1\int_1^2 \frac{yz}{x} dx\; dy\; dz&= \int_0^2\int_0^1 \left[yz \ln x\right]_1^2 dy\; dz =
\ln2 \int_0^2\int_0^1 yz dy\; dz\\
& = \ln2 \int_0^2 y \left[\frac{z}{2}\right]_0^1\; dy = \frac{\ln 2}{2}\int_0^2 y \; dy =
\frac{\ln 2}{2}\left[\frac{y}{2}\right]_0^2= \ln 2.
\end{align*}
\end{example}
What about more complicated regions?
\begin{example}Consider the solid $S$ that lies underneath the paraboloid $z=1+x^2 +y^2$, above the $xy$--plane, and inside the cylinder $x^2+y^2=1$. Let's find the integral over $S$ of the function $f(x,y,z) = z$.
\vspace{-.2in}
\begin{center}
\includegraphics[width=3.5in]{paraboloid}
\end{center}
We need to describe the solid, in terms that we can use for the integral. Remember that we wrote the circle in the following way: $-1\le x \le 1$, and for each $x$ in that interval, $y$ ranges over the interval $[-\sqrt{1-x^2},\sqrt{1-x^2}]$. Now think about how to describe the solid in $3$-dimensional terms: for each point $(x,y)$ inside the circle, the height $z$ can range from $0$ to $1+x^2 +y^2$. So we write the triple integral as
$$
\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_0^{1+x^2 +y^2} z\ dz\; dy\; dx
$$
This looks pretty complicated, so we'd better think about how to carry it out before we start integrating. The integral is set up to do the z-integral first; it's clear that we will get $z^2/2$, and evaluate with limits $0$ and $1+x^2 +y^2$. Thus we will be integrating some function of $x$ and $y$ (in fact it's $(1+x^2 +y^2)^2/2$) over a region in the $xy$-plane. This is a double integral, of the sort that we've done before. In symbols, the integral is
$$
\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \left(\left[\frac{z^2}{2}\right]_0^{1+x^2 +y^2}\right) \ dy\; dx =
\frac12\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (1+x^2 +y^2)^2\ dy\; dx
$$
Let's pause again to think before we go any further. The next integral is the $y$-integral; to carry this out we should expand out $(1+x^2 +y^2)^2$, getting a polynomial in $y$ (remember $x$ acts like a constant during this integration) that we can integrate. Then we'd stick in $\pm \sqrt{1-x^2}$ for limits, leaving us with something very messy to integrate. There's got to be a better way!
Well, of course there is. The double integral above is over a circular region in the plane, so we should immediately think about doing that integral using polar coordinates, rather than rectangular coordinates. (Remember the old saying about putting a square peg in a round hole?) In polar coordinates, the region over which we are integrating is described by $0 \leq r \leq1$, $0\leq \theta\leq 2\pi$. The integrand $(1+x^2 +y^2)^2$, when we substitute $x=r\cos\theta$, $y=r\sin\theta$, becomes $(1+r^2)^2$. Finally, remember that in polar coordinates, the element of area $dA = dx\; dy$ becomes $rdr\;d\theta$. Putting this all together, the integral is
$$
\frac12\int_0^{2\pi}\int_0^1 (1+r^2)^2 rdr\;d\theta = \frac12\int_0^{2\pi} \left[\frac16(1+r^2)^3\right]_0^1d\theta =
\frac8{12}\int_0^{2\pi} d\theta = \frac43\pi
$$
\end{example}
If there's time, we'll do one more example.
\begin{example}
Consider the solid bounded by the cylinder $y=x^2$, the $xy$-plane, the vertical plane $y=4$, and the plane $z=y$, as drawn below. We want to calculate the integral of the function $f(x,y,z) =y$ over this solid. Again the main issue is setting it up. The solid sits over the region in the $xy$ plane bounded by the parabola and $y=4$.
\vspace{-.2in}
\begin{center}
\includegraphics[width=3.5in]{slice.eps}
\end{center}
This region could be described in two different ways, giving two different integrals. (In each of them, $z$ will range from $0$ to $y$.) The first is to say that
$x$ ranges from $-2$ to $2$, and for each $x$ value, $y$ ranges from $x^2$ to $4$. This leads to writing the integral as
$$
\int_{-2}^2\int_{x^2}^4\int_0^y y dz \; dy\; dx = \int_{-2}^2\int_{x^2}^4 y^2\; dy\; dx = \frac13 \int_{-2}^2 (64-x^6)\; dx= \frac13\left[64x-\frac{x^7}{7}\right]_{-2}^2 = \frac{512}{7}.
$$
The second way to write the integral is to say that $y$ ranges from $0$ to $4$, and for each $y$ value, $x$ ranges between $-\sqrt{y}$ and $\sqrt{y}$. We would then write the integral as
$$
\int_0^4\int_{-\sqrt{y}}^{\sqrt{y}}\int_0^y y dz \; dx\; dy =\int_0^4\int_{-\sqrt{y}}^{\sqrt{y}}y^2\; dx\; dy= \int_0^4 y^2\left[x\right]_{-\sqrt{y}}^{\sqrt{y}}= 2\int_0^4y^{5/2}\; dy = \frac47\left[y^{7/2}\right]_0^4 = \frac{512}{7}.
$$
\end{example}
\end{document}