\documentclass{article}
%The following imports commands created for formatting for SIAST lecture notes.
\usepackage{lectures}
%Define your course info here
\institution{SIAST Palliser Campus}
\department{Mathematics}
\course{MAT 235}
\unittitle{Applications of the Definite Integral}
\unitnumber{3}
\date{September 2011}
\title{STUDY GUIDE}
\author{Alan Dill and Robert G. Petry}
%Note that the publisher is just the distributor of the printed copies and
%has no other claim on the document.  (See the GFDL license.)  It can be
%either a person or entity.
\publisher{the Department of Mathematics, SIAST Palliser Campus}
\begin{document}
%Create title page
\makeunittitle
%Include the copyright, etc. page.  Placed in a separate latex file for
%convenience.
\input{copyright_unit3}
%Reset the pages to start the actual content and change to arabic numbering
%since \makeunittitle set it to roman numerals.
\setcounter{page}{1}
\pagenumbering{arabic}
\section{Area Between Curves}
The area bounded by the vertical lines $x=a$ and $x=b$ and the curves
$y_1=f_1(x)$ on the bottom and $y_2=f_2(x)$ on top may be written
\begin{boxedequation}
A=\int_a^b (y_2-y_1)\;dx
\end{boxedequation}

To find the area between curves by integration, follow the steps given
on page~892 of the text.

It is also a good idea to do a rough estimate of the area by sketching a
rectangle or triangle and estimating what fraction the given area is of
the area of the rectangle or triangle.

\textbf{Examples}
\begin{enumerate}
\item Find the area bounded by the curve $y = 3x^2$, the $x$-axis and
  the line $x = 3$.
\item Find the first-quadrant area bounded by the curve $y = x^2 + 2$,
  $x = 0$ and $y = 4$.
\item Find the area bounded by the curves $y = 9 - x^2$ and $y = 2x + 1$.
\item Find the area bounded by the curves $y = 4x$ and $y = x^3$.
\item Find the area bounded by the curve $y = x^3 - 3$ and the lines
  $x = 2$, $y = -1$, and $y = 3$.
\end{enumerate}

\textbf{Additional Examples} (Text, p. 900)
\begin{itemize}
\item Problem 32
\item Problem 36
\end{itemize}

\textassignment{Sec. 31-3, pp. 892-899}{Ex. 3, p. 899 \# 1-23 (odd), 31, 33, 35}

\section{Volume of Solids of Revolution}

When an area bounded by two or more curves is rotated about a horizontal
or vertical axis, it sweeps out a solid of revolution. The volume of
such a solid can be found by one or more of the following methods.

% Tex Note:  I used a minipage here rather than a parbox because in the
% minipage environment the \textwidth is redefined to be the width of the
% minipage.  That means my boxedeqnarray functions correctly.
\subsection{Disk Method}
\begin{minipage}{.53\textwidth}
\begin{boxedeqnarray}
dV & = & \pi r^2\;dh \\
V & = & \int_a^b\pi r^2\;dh \\
& & \nonumber
\end{boxedeqnarray}

\vspace{1em}
Note that, for a specific problem, $r$ and $h$ must be expressed in
terms of $x$ and $y$.
\end{minipage}\hspace{.02\textwidth}
\raisebox{-5em}{\includegraphics[width=.45\textwidth,keepaspectratio=,clip=]{Unit3figs-img1.pdf}}

\textbf{Examples}
\begin{itemize}
\item Find the volume generated by rotating the first-quadrant area bounded
by $y = 4 - x^2$, the $x$-axis and the $y$-axis, about the $x$-axis.
\item Show that the volume of a sphere of radius $R$ is $\displaystyle V=\frac{4}{3}\pi R^3$\ .
\end{itemize}

\subsection{Shell Method}
\begin{minipage}{.53\textwidth}
\begin{boxedeqnarray}
dV & = & 2\pi r h\;dr \\
V & = & \int_a^b 2\pi r h\;dr \\
& & \nonumber
\end{boxedeqnarray}

\vspace{1em}
Again, $r$ and $h$ must be expressed in terms of
$x$ and $y$ for a particular problem.
\end{minipage}\hspace{.02\textwidth}
\raisebox{-5em}{\includegraphics[width=.45\textwidth,keepaspectratio=,clip=]{Unit3figs-img2.pdf}}

\textbf{Example}
\begin{itemize}
\item Find the volume generated by rotating the first area in the previous
  examples about the $y$-axis (use the shell method).
\end{itemize}
For a solid of revolution with a hole, the shell method can often be
used. Another option is the ring method, which is similar to the disk
method, with the central part of the disk missing.

\subsection{Ring Method}
\begin{minipage}{.53\textwidth}
\begin{boxedeqnarray}
dV & = &(\pi r_0^2-\pi r_i^2)\;dh \\
   & = & \pi (r_0^2-r_i^2)\;dh \nonumber\\
V  & = & \pi\int_a^b(r_0^2-r_i^2)\;dh \\
& & \nonumber
\end{boxedeqnarray}
\end{minipage}\hspace{.02\textwidth}
\raisebox{-5em}{\includegraphics[width=.45\textwidth,keepaspectratio=,clip=]{Unit3figs-img3.pdf}}

\textbf{Example}
\begin{itemize}
\item Find the volume generated by rotating the area bounded by $y=\sqrt{x}$
and $y=\frac{x}{2}$ about the line $y = 4$
\end{itemize}
\textbf{Additional Examples} (Text, p. 908)
\begin{itemize}
\item Problem 16
\item Problem 18
\item Problem 20
\end{itemize}

\textassignment{Sec. 31-4, pp. 902-907}{Ex. 4, p. 908 \# 1, 3, 7-21 (odd)}
\clearpage
\section{Arc Length}
%Tex Note the following silliness regarding paragraph skipping is required
%because in a parbox the paragraph skipping is set to zero.  My workaround is
%to create a temporary variable outside of the parbox (defined in siast.sty),
%set its value to the current paragraph skipping and the set parskip back to
%this within the parbox. In this way the correct paragraph skipping occurs
%before "By Pythagorus" and will not be a problem if the paragraph skip is
%defined differently within the document.
\parbox{.53\textwidth}{\setlength{\parskip}{\docparskip}In the graph shown,
consider a small segment of the curve of length $\Delta s$, short enough to
be considered a straight line.

By Pythagoras,
\[\left(\Delta s\right)^2=\left(\Delta x\right)^2+\left(\Delta
  y\right)^2\] }\hspace{.02\textwidth}
\raisebox{-5em}{\includegraphics[width=.45\textwidth,keepaspectratio=,clip=]{Unit3figs-img4.pdf}}

Dividing by $(\Delta x)^2$ gives
\[\frac{\left(\Delta s\right)^2}{\left(\Delta x\right)^2}
=\frac{\left(\Delta x\right)^2}{\left(\Delta x\right)^2}
+\frac{\left(\Delta y\right)^2}{\left(\Delta x\right)^2}\]

Taking the square root gives
\[\frac{\Delta s}{\Delta x}=\sqrt{1+\frac{\left(\Delta y\right)^2}
{\left(\Delta x\right)^2}}\]

Taking the limit as $\Delta x$ approaches zero, we get
\[\lim_{\Delta x \rightarrow 0} \frac{\Delta s}{\Delta x}
=\lim_{\Delta x \rightarrow 0}\sqrt{1+\frac{\left(\Delta y\right)^2}{\left(\Delta x\right)^2}}\]

This is equivalent to the following equation written in terms of
derivatives:
\[\frac{ds}{dx}=\sqrt{1+\left(\frac{dy}{dx}\right)^2}\]

Multiplying by $dx$ gives us the following expression for the
differential $ds$:
\begin{boxedequation}
ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx
\label{eq:ds}
\end{boxedequation}

To find the total arc length between two $x$-values, $a$
and $b$, we integrate this expression between the limits of
$a$ and $b$.
\begin{boxedequation}
s=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx
\end{boxedequation}

If the equation of the curve is written as $x = f(y)$,
the arc length between $y = c$ and $y = d$ can be found
from the following formula:
\begin{boxedequation}
s=\int_c^d\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy
\end{boxedequation}

\clearpage
\textbf{Examples} (Text, p. 914)
\begin{itemize}
\item Problem 2
\item Problem 6
\item Problem 12
\end{itemize}

\textassignment{Sec. 32-1, pp. 912-914}{Ex. 1, p. 914 \# 1-15 (odd)}
\clearpage

\section{Surface Area of Solids of Revolution}
\parbox{.53\textwidth}{\setlength{\parskip}{\docparskip}If a curve is
  rotated about an axis, it generates of surface of revolution.

For rotation about the $x$-axis, the circular element of
surface area $dS$ is given by
\[dS=2\pi y\;ds\]
where $ds$ is an element of arc length on the curve. But we have
already calculated $ds$ in Equation~(\ref{eq:ds}) in the previous section.
}\hspace{.02\textwidth}\raisebox{-5em}{\includegraphics[width=.45\textwidth,keepaspectratio=,clip=]{Unit3figs-img5.pdf}}

Substituting that expression for $ds$ into the expression for $dS$, we get
\begin{boxedequation}
dS=2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx
\end{boxedequation}

We get the total surface area between $x = a$ and $x = b$
by integrating $dS$ between these limits.
\begin{boxedequation}
S=2\pi\int_a^b y\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx
\end{boxedequation}

For rotation about the $y$-axis, the total surface area is
given by
\begin{boxedequation}
S=2\pi\int_a^b x\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx
\end{boxedequation}

\textbf{Examples} (Text, p. 917)
\begin{itemize}
\item Problem 2
\item Problem 12
\item Problem 14
\item Problem 16
\end{itemize}

\textassignment{Sec. 32-2, pp. 915-917}{Ex. 2, p. 917 \# 1-15 (odd)}
\clearpage

\section{Centroids}

\subsection{Centroid of an Area by Integration}
\begin{minipage}{.53\textwidth}
  \setlength{\parskip}{\docparskip}Suppose we
  want to find the \emph{centroid} $(\bar{x},\bar{y})$ of the area
  bounded by the curves \mbox{$y_1 = f_1(x)$}, $y_2 = f_2(x)$, and the lines
  $x = a$ and $x = b$, as shown.

  The moment of the element of area $dA$ about the $y$-axis is given
  by\footnotemark
  \begin{eqnarray}
    dM_y & = & x\;dA \nonumber \\
    & = & x(y_2-y_1)dx \nonumber
  \end{eqnarray}
\end{minipage}\hspace{.02\textwidth}
\raisebox{-5em}{\includegraphics[width=.45\textwidth,keepaspectratio=,clip=]{Unit3figs-img6.pdf}}
%Tex Note: I used a minipage rather than a parbox so I could use a footnote.
%To prevent the footnote text actually appearing within the minipage but rather
%at the bottom of the page, I had to use a \footnotemark and a \footnotetext
%rather than a \footnote command which one would use elsewhere.
\footnotetext{If one wanted to find the \emph{centre of mass} rather
  than the centroid, the moment would also be multipled by a density
  $m$ and rather than using the area $A$ we would use the total
  mass which would be $M=\int dM=\int m\; dA$.  The centroid is effectively
  just the centre of mass in the special case when the density
  $m$ is constant.}

The moment of the whole bounded area about the $y$-axis is found by
integrating this expression between $a$ and $b$.
\begin{boxedequation}
M_y=\int_a^b x(y_2-y_1)\;dx
\end{boxedequation}

But the moment about the y-axis is just equal to the product of the
area and the distance to the centroid, i.e.
\[M_y=A\bar{x}\]

Thus we get
\[A\bar{x}=\int_a^b x(y_2-y_1)\;dx\]

Solving for $\bar{x}$,we get the following formula for the
$x$-coordinate of the centroid:
\begin{boxedequation}
\bar{x}=\frac{1}{A}\int_a^b x(y_2-y_1)\;dx
\label{eq:centroidx}
\end{boxedequation}

The moment of the element of area $dA$ about the $x$-axis is
\[dM_x=\frac{y_1+y_2}{2}(y_2-y_1)\; dx\]
so the moment about the $x$-axis, $M_x$, is the integral of $dM_x$
\begin{boxedequation}
M_x=\int_a^b\frac{y_1+y_2}{2}(y_2-y_1)\;dx
\end{boxedequation}

The $y$-coordinate of the centroid, $\bar{y}$, is similarly
defined through $M_x=A\bar{y}$, so solving for $\bar{y}$ and inserting
$M_x$ we have the result
\begin{boxedequation}
\bar{y}=\frac{1}{2A}\int_a^b (y_1+y_2)(y_2-y_1)\;dx
\label{eq:centroidy}
\end{boxedequation}

\emph{Note: If horizontal elements are used, interchange $x$ and $y$ in formulas~(\ref{eq:centroidx}) and~(\ref{eq:centroidy}).}
\clearpage

\textbf{Examples} (Text, p. 925)
\begin{itemize}
\item Problem 6
\item Problem 12
\item Find the centroid of a right triangle with base $b$ and height $h$.
\end{itemize}
\clearpage

\subsection{Centroid of a Solid of Revolution}
%Tex Note: Once again I used a minipage rather than a parbox so my
%boxedequation formula would work.
\begin{minipage}{.53\textwidth}
  \setlength{\parskip}{\docparskip} For a solid of revolution about
  the $x$-axis, the element of volume $dV$ is
\[dV = \pi y^2\,dx\]

The moment of this element of volume about the $y$-axis is
\[dM_y=x(\pi y^2\,dx)\]

The moment of the entire volume about the $y$-axis is found by
integrating from \mbox{$x = a$} to $x = b$.
\begin{boxedequation}
M_y=\pi\int_a^b xy^2\;dx
\end{boxedequation}
\end{minipage}
\hspace{.02\textwidth}
\raisebox{-5em}{\includegraphics[width=.45\textwidth,keepaspectratio=,clip=]{Unit3figs-img7.pdf}}

Since $M_y=V\bar{x}$, we get the following for the
$x$-coordinate of the centroid:
\begin{boxedequation}
\bar{x}=\frac{\pi}{V}\int_a^b xy^2\;dx
\end{boxedequation}

Because the $x$-axis is the axis of symmetry, the $y$-coordinate of
the centroid is zero.

For a solid of revolution about the $y$-axis, the $x$-coordinate of the
centroid is zero, and the $y$-coordinate is given by
\begin{boxedequation}
\bar{y}=\frac{\pi}{V}\int_c^d yx^2\;dy
\end{boxedequation}

\textbf{Example} (Text, p. 926)
\begin{itemize}
\item Problem 16
\end{itemize}

\textassignment{Sec. 32-3, pp. 918-925}{Ex. 3, p. 925 \# 5-21 (odd)}
\clearpage
%
%\section{Fluid Pressure}
%Pressure is defined as force per unit area perpendicular to a surface.
%In a fluid, the pressure is the same in all directions, so the
%pressure exerted by a fluid on a submerged surface will be the same
%regardless of whether the surface is vertical, horizontal or oblique.
%The pressure at a depth $y$ is given by
%\[P=\delta y\] where $\delta$ is the weight density of the
%fluid.\footnote{The \emph{weight density} $\delta$ is just the density
%  (mass per unit volume) $m$ times the gravitational acceleration
%  $g=9.8$ $\mathrm{m}/\mathrm{s}^2$, i.e. $\delta=m\cdot g$.}  This
%result comes from a consideration of the weight of the column of the
%fluid over a unit of area at depth $y$.
%
%The force exerted by a fluid on a surface of area $A$ at depth $y$ is
%equal to this pressure times the area.
%\[F=PA=\delta y A\]
%
%\begin{minipage}{.53\textwidth}\setlength{\parskip}{\docparskip}
%If the surface is not all at the same depth, we need to divide the
%surface into thin horizontal strips of area $dA$, each of which
%is at the same depth $y$. The force exerted on this strip is
%\begin{boxedequation}
%dF=\delta y\,dA
%\end{boxedequation}
%
%The total force on the surface is found by integrating this expression
%\begin{boxedequation}
%F=\delta\int y\;dA
%\end{boxedequation}
%\end{minipage}\hspace{.02\textwidth}
%\raisebox{-8em}{\includegraphics[width=.45\textwidth,keepaspectratio=,clip=]{Unit3figs-img8.pdf}}
%
%The expression $\int y\;dA$ is the first moment of the area about the
%$x$-axis, $M_x$, which is equal to the product of the area and the
%vertical distance $\bar{y}$ to the centroid, so we can also write
%\begin{boxedequation}
%F=\delta \bar{y}A
%\end{boxedequation}
%
%Either this equation or the previous equation can be used to find the
%force exerted by a fluid on a submerged surface.
%
%\textbf{Example}
%\begin{itemize}
%\item A vertical rectangular plate in the wall of a water tank is
%  $3.00$ m long and $2.00$ m high, and has its upper edge $1.50$ m
%  below the surface of the water.  The weight density of water is
%  $9800$ $\mathrm{N}/\mathrm{m}^3$. Find the force on the plate (in
%  Newtons).
%\end{itemize}
%
%\clearpage
%\textbf{Additional Examples} (Text, p. 929)
%\begin{itemize}
%\item Problem 4
%\item Problem 6
%\item Problem 8
%\end{itemize}
%
%\textassignment{Sec. 32-4, pp. 927-929}{Ex. 4, p. 929 \# 1-7 (odd)}
%\clearpage
%
%\section{Work}
%When a constant force $F$ acts on an object, the work,
%$W$, done on the object is the product of the force and the
%distance $x$ moved by the object in the direction of the force.
%\[W=Fx\]
%
%If a variable force is acting, the work done in moving the object a
%small distance $dx$ (small enough so the force can be considered
%constant over that distance) is given by
%\[dW=F\;dx\]
%
%The total work done in moving from $x = a$ to $x = b$ is given by the
%definite integral
%\begin{boxedequation}
%W=\int_a^b F\;dx
%\end{boxedequation}
%
%One example of a variable force is the force exerted by a spring when it
%is stretched or compressed. The force required to stretch or compress
%a spring by an amount $x$ is proportional to the deflection
%$x$, i.e.
%\[F=kx\] where $k$ is a constant for any given spring, called the
%\emph{spring constant}.
%
%\textbf{Example}
%\begin{itemize}
%\item A certain spring has an unstretched length of 20.0 cm. When a
%  stretching force of 1.50 N is applied to it, its length increases to
%  25.0 cm. How much work must be done to stretch the spring to a
%  length of 35.0 cm?
%\end{itemize}
%\textbf{Additional Examples} (Text, p. 932)
%\begin{itemize}
%\item Problem 4
%\item Problem 6
%\item Problem 8
%\end{itemize}
%
%\textassignment{Sec. 32-5, pp. 930-932}{Ex. 5, p. 932 \# 1-13 (odd)}
%\clearpage
%
\section{Moment of Inertia}
\subsection{Moment of Inertia of an Area}
\begin{minipage}{.53\textwidth}\setlength{\parskip}{\docparskip}
  The \emph{moment of inertia} (or \emph{second moment}), $dI$, of an
  element of area $dA$ about a line at a distance $r$ from the element
  of area is defined by
\[dI=r^2\,dA\]

For the element of area shown on the right, the moment of inertia
about the $y$-axis is given by
\[dI_y=x^2\,dA=x^2y\,dx\]
\end{minipage}\hspace{.02\textwidth}
\raisebox{-4em}{\includegraphics[width=.45\textwidth,keepaspectratio=,clip=]{Unit3figs-img9.pdf}}

The moment of inertia of the whole area between $x=a$ and $x=b$ about
the $y$-axis is obtained by integration\footnote{Technically the
  moment of inertia requires a factor of $m$, the area mass density.
  We will consider only areas of constant density so any such factor
  will factor out of the integrals.  When we consider the moment of
  inertia of a volume we will include the volume mass density
  explicitly.  This arbitrariness is for consistency with the textbook.}
\begin{boxedequation}
I_y=\int_a^b x^2y\;dx
\end{boxedequation}

\begin{minipage}{.53\textwidth}\setlength{\parskip}{\docparskip}
For the area shown on the right, if we take horizontal elements, the
moment of inertia of the element $dA$ about the
$x$-axis is given by
\[dI_x=y^2\,dA=y^2x\,dy\]
\end{minipage}\hspace{.02\textwidth}
\raisebox{-4em}{\includegraphics[width=.45\textwidth,keepaspectratio=,clip=]{Unit3figs-img10.pdf}}

The moment of inertia of the whole area between $y = c$ and $y = d$ is
then
\begin{boxedequation}
I_x=\int_c^d y^2x\;dy
\end{boxedequation}

\begin{minipage}{.78\textwidth}\setlength{\parskip}{\docparskip}
\textbf{Example}
\begin{itemize}
\item Show that the moment
of inertia of the rectangle shown about an $x$-axis through
its base is $\frac{b^3a}{3}$.
\end{itemize}
\end{minipage}\hspace{.02\textwidth}
\raisebox{-4em}{\includegraphics[width=.2\textwidth,keepaspectratio=,clip=]{Unit3figs-img11.pdf}}
\clearpage
Using the result of the previous example, we can find a
formula for the moment of inertia of the area at the top of the page
about the $x$-axis (by summing/integrating vertical elements of height
$y$ and width $dx$).
\[dI_x = \frac{y^3\,dx}{3}\]
\begin{boxedequation}
I_x = \frac{1}{3}\int_a^b y^3\;dx
\end{boxedequation}

If the area whose moment of inertia has been found were stretched into
a thin strip and placed at a distance $r$ from the axis such that the
moment of inertia of the strip is the same as that of the original
area, the distance $r$ is called the \emph{radius of
  gyration}.\footnote{The radius of gyration may be thought of as the
  rotational equivalent of the centroid (centre of mass).  It gives
  the point at which all the mass of the system could be placed to
  have an equivalent mechanical system.} The radius of gyration is
found from the equation
\[I=Ar^2\]

Solving for $r$, we get
\begin{boxedequation}
r=\sqrt{\frac{I}{A}}
\end{boxedequation}

\textbf{Examples} (Text, p. 939)
\begin{itemize}
\item Problem 2
\item Problem 4
\item Problem 6
\end{itemize}
\clearpage

\subsection{Polar Moment of Inertia}
The \emph{polar moment of inertia} (also called the \emph{mass moment
  of inertia}) is a quantity that is required when studying the
rotation of solid rigid bodies, such as flywheels.

\subsubsection{Polar Moment of Inertia by the Shell Method}
\begin{minipage}{.53\textwidth}\setlength{\parskip}{\docparskip}
Using the shell method, the volume of the
cylindrical shell shown is given by
\[dV = 2\pi rh\,dr\]

If $m$ is the density (mass per unit volume), the mass of the
cylindrical shell is then
\[dM=2\pi mrh\,dr\]
\end{minipage}\hspace{.02\textwidth}
\raisebox{-6em}{\includegraphics[width=.45\textwidth,keepaspectratio=,clip=]{Unit3figs-img12.pdf}}

The polar moment of inertia of the cylindrical shell about its axis of
rotation is defined as the product of its mass and the square of its
distance $r$ from the axis.
\[dI=r^2\,dM=2\pi mr^3h\,dr\]

The polar moment of inertia of the entire solid is then the integral
of this quantity over the whole range of $r$-values.
\begin{boxedequation}
I=2\pi m\int r^3h\;dr
\label{eq:inertiashell}
\end{boxedequation}

It should be noted that, in a particular problem, $r$ and $h$ must be
expressed in terms of $x$ and $y$.
\clearpage

\subsubsection{Polar Moment of Inertia of a Solid Cylinder}
\begin{minipage}{.53\textwidth}\setlength{\parskip}{\docparskip}
As an example of applying the shell method, suppose we have a solid
cylinder of radius $R$ and length $H$.

If we orient the cylinder along the $x$-axis as shown, then the shell
radius of Equation~(\ref{eq:inertiashell}) is $r=y$ (so $dr=dy$), and
we integrate from shells of radius $0$ to $R$.  The shell length $h$
is constant in this example, $h=H$.  Making these substitutions in
Equation~(\ref{eq:inertiashell}) gives
\end{minipage}\hspace{.02\textwidth}
\raisebox{-4em}{\includegraphics[width=.45\textwidth,keepaspectratio=,clip=]{Unit3figs-img13.pdf}}

%
%The moment of inertia of the cylindrical element of volume is
%\[dI=2\pi my^3h\,dy\]
%
%Integrating from $y = 0$ to $y = r$, we get
\[I=2\pi m\int_0^R y^3H\;dy=2\pi mH\int_0^R y^3\;dy=2\pi mH\left[\frac{y^4}{4}\right]_0^R
=\frac{\pi mHR^4}{2}\]

We can also write this in terms of the volume of the cylinder, given by
\[V=\pi R^2H\]

Using this gives the following formula for polar moment of inertia of a
cylinder:
\[
I=\frac{\pi mHR^4}{2}=\frac{1}{2}mVR^2
\]

\subsubsection{Polar Moment of Inertia Using the Disk Method}
\begin{minipage}{.53\textwidth}\setlength{\parskip}{\docparskip}
  We can use the result of the previous equation to
  derive an alternative formula for finding polar moment of inertia
  called the disk method.

  In the diagram shown, we have a disk-shaped element of volume which
  is effectively a thin cylinder of radius $R=r$ and height $H=dh$.
  Inserting these into the equation for the polar moment of inertia of a
  cylinder gives the disk's moment of inertia $dI$
\[dI=\frac{m\pi r^4\,dh}{2}\]

Integrating gives the total moment of inertia of the volume, $I=\int dI$, to be
\end{minipage}\hspace{.02\textwidth}
\raisebox{-4em}{\includegraphics[width=.45\textwidth,keepaspectratio=,clip=]{Unit3figs-img14.pdf}}
\begin{boxedequation}
I=\frac{m\pi}{2}\int_a^b r^4\;dh
\end{boxedequation}

Note that a radius of gyration for solids may be defined as for areas,
namely $r=\sqrt{\frac{I}{M}}$, where now the total mass $M=mV$, where
V is the volume of constant density $m$.
\clearpage \textbf{Example}
(Text, p. 940)
\begin{itemize}
\item Problem 10
\end{itemize}

\textassignment{Sec. 32-6, pp. 933-939}{Ex. 6, p. 939 \# 1-11 (odd)}
\clearpage

\section{Average Value of a Function}
\begin{minipage}{.53\textwidth}\setlength{\parskip}{\docparskip}
Suppose we want to find the average value
of a function $y=f(x)$ between two $x$-values,
$a$ and $b$. We can start by dividing the interval
between $a$ and $b$ into $n$ equal intervals of
width  $\Delta x=\frac{b-a}{n}$, then summing the
$y$-values and dividing by $n$. The average $y$-value
would then by approximated by
\[y_{\mathit{avg}}\approx\frac{1}{n}\sum_{i=1}^n y_i\]
\end{minipage}\hspace{.02\textwidth}
\raisebox{-6em}{\includegraphics[width=.45\textwidth,keepaspectratio=,clip=]{Unit3figs-img15.pdf}}

Solving the equation $\Delta x=\frac{b-a}{n}$ for 
$\frac{1}{n}$, we get
\[\frac{1}{n}=\frac{\Delta x}{b-a}\]

We can substitute this for $\frac{1}{n}$ in the previous equation to
get
\[y_{\mathit{avg}}\approx\frac{\Delta x}{b-a}\sum_{i=1}^n y_i\]

Since $\Delta x$ is a constant, we can move it inside the summation
sign and get
\[y_{\mathit{avg}}\approx\frac{1}{b-a}\sum_{i=1}^n y_i\Delta x\]

Of course, the larger the number of $y$-values, the more
accurate the average value of $y$ will be. To get an exact value for
the average, we take the limit as $n$ approaches infinity.
\begin{eqnarray}
y_{\mathit{avg}} & = & \lim_{n\rightarrow\infty}\frac{1}{b-a}\sum_{i=1}^n y_i\;\Delta x
\nonumber\\
& = & \frac{1}{b-a}\left(\lim_{n\rightarrow\infty}\sum_{i=1}^n y_i\;\Delta x\right)\nonumber\\
& = & \frac{1}{b-a}\left(\lim_{n\rightarrow\infty}\sum_{i=1}^n f(x_i)\;\Delta x\right)\nonumber
\end{eqnarray}

The quantity in parentheses is just the definite integral of $f(x)$
from $x = a$ to $x = b$, so we can write the following formula for the
average value of $y=f(x)$ from $a$ to $b$:
\begin{boxedequation}
y_{\mathit{avg}}=\frac{1}{b-a}\int_a^b f(x)\;dx
\end{boxedequation}

\begin{minipage}{.53\textwidth}\setlength{\parskip}{\docparskip}
We can give a geometric interpretation of $y_{\mathit{avg}}$.  Since
the definite integral is equal to the area $A_{ab}$ under the curve
between $a$ and $b$, we can write
\[y_{\mathit{avg}}=\frac{1}{b-a}A_{ab}\]

Solving for $A_{ab}$, we get
\[{A_{ab}=y_{\mathit{avg}}(b-a)}\]
\end{minipage}\hspace{.02\textwidth}
\raisebox{-6em}{\includegraphics[width=.45\textwidth,keepaspectratio=,clip=]{Unit3figs-img16.pdf}}

Since $b-a$ is the width of the rectangle shown, whose height is
$y_{\mathit{avg}}$, we see that $y_{\mathit{avg}}$ is just the height of a rectangle
that is equal in area to $A_{ab}$ under the curve from $x = a$ to $b$.

\textbf{Examples} (Give answers to 3 significant digits)
\begin{enumerate}
\item Find the average value of the function $y=\sqrt{2x+5}$ between
  $x = 0$ and $x = 2$.
\item If an object is falling freely in the earth's gravitational
  field, its velocity as a function of time is $v =-9.8t$. Find its
  average velocity over the first $10.0$ seconds.
\end{enumerate}

\textassignment{Sec. 34-3, pp. 975-976}{Ex. 3, p. 976 \# 1, 3~~}
\clearpage

\section{Review Problems: Applications of the \mbox{Definite} Integral}
\textbf{Questions}

\emph{Round all numerical answers to 3 significant digits where applicable.}
\begin{enumerate}

\item \parbox[t]{.53\textwidth}{A tile manufacturing company plans to
    produce enamel tiles with the design shown. Find the area of the
    inner lenticular-shaped tile and the area of the surrounding tile
    that will be
    required.}\hspace{.02\textwidth}\raisebox{-8em}{\includegraphics[width=.35\textwidth,keepaspectratio=,clip=]{Unit3figs-img17.pdf}}

\item A hemispherical water tank has a radius of $4.00$ m. The water is
  $1.50$ m deep at the centre. How much water is in the tank?

\item Find the length of the curve $y=\sqrt{8x^3}$ from $x = 0$ to $x
  = 2$. Do a rough estimate, and then find the length by
  integration. Give the answer in exact form and in approximate
  decimal form.

\item Find the area of the surface generated by rotating the curve
  $y=2\sqrt{x}$ from $x = 0$ to $8$ about the $x$-axis. Give the
  answer in exact form and in approximate decimal form.

\item Locate the centroid of the plane region bounded by the curves
  $y = 2 - x$ and $y = x^2$.

\item Find the moment of inertia, about the $y$-axis, of the plane
  region bounded by the curve $y = x^2$, the $x$-axis, and the line
  $x = 2$.

\item Find the average value of the function $f(t)=4.9t^2-4.2t+9.0$
  over the interval $t = 0$ to $t = 1.6$\ .

%\item A hemispherical tank of radius 2.00 m is full of water (weight
%  density $980~\mathrm{N}/\mathrm{m}^3$) Find the work done (in
%  joules) in pumping the water to the top of the tank. (Note: 1 joule
%  (J) = 1 $\mathrm{N}\cdot\mathrm{m}$).
%
%\item A right triangular plate with a base of 2.00 m and a height of
%  1.00 m is submerged vertically in water, with the top vertex 3.00 m
%  below the surface. Find the force on one side of the plate.
\end{enumerate}

\clearpage

\textbf{Answers}
\begin{enumerate}
\item inner: $0.333~\mathrm{ft}^2$, outer: $0.667~\mathrm{ft}^2$
\item $24.7~\mathrm{m}^3$
\item $\displaystyle \frac{1}{27}(37^{\frac{3}{2}}-1)\approx 8.30~\mathrm{units}$
\item $\displaystyle \frac{208\pi}{3}\approx 218~\mathrm{u}^2$
\item $\left(-\frac{1}{2},\frac{8}{5}\right)$
\item $6.4$
\item $9.82$
%\item $1.23\times 10^5$ J
%\item $3.59\times 10^4$ N
\end{enumerate}

\clearpage
\section{Additional Review Problems: Applications of the Definite Integral}
\textbf{Questions}
\begin{itemize}
\item Chapter 31 Review Problems (p. 909) \# 1, 2, 5, 7, 13, 15 (Note question 7 requires use of integration Rule 52 of Appendix C or numerical integration.)
\item Chapter 32 Review Problems (p. 940) \# 1, 3, 6, 8, 10, 12,
  16, 17, 18
\end{itemize}
\textbf{Answers to Even-Numbered Review Problems}
\begin{itemize}
\item Chapter 31
  \begin{description}
  \item[2.]$\frac{8}{3}$
  \end{description}
\item Chapter 32
  \begin{description}
  \item[6.]$\frac{3Mr^2}{10}$
  \item[8.]$\frac{8\pi mr^5}{15}$
  \item[10.] $(0.800,1.00)$
  \item[12.] 13.75 (Requires Rule 66 of Appendix C or numerical integration.)
  \item[16.] 102.6 m (Requires Rule 66 of Appendix C or numerical integration.)
  \item[18.] 2p/3
  \end{description}
\end{itemize}
\clearpage
\section{Formulas}
\setcounter{equation}{0}
\begin{center}
\textbf{\large Area between Curves}
\end{center}
Area bounded by the vertical lines $x=a$ and $x=b$ and the curves
$y_1=f_1(x)$ on the bottom and $y_2=f_2(x)$ on top:
\begin{boxedequation}
A=\int_a^b (y_2-y_1)\;dx
\end{boxedequation}

\begin{center}
\textbf{\large Volumes of Solids of Revolution}
\end{center}

\textbf{Disk Method:}
\begin{boxedeqnarray}
dV & = & \pi r^2\;dh \\
V & = & \int_a^b\pi r^2\;dh \\
& & \nonumber
\end{boxedeqnarray}
\textbf{Shell Method:}
\begin{boxedeqnarray}
dV & = & 2\pi r h\;dr \\
V & = & \int_a^b 2\pi r h\;dr \\
& & \nonumber
\end{boxedeqnarray}
\textbf{Ring Method:}
\begin{boxedeqnarray}
dV & = &(\pi r_0^2-\pi r_i^2)\;dh \\
V  & = & \pi\int_a^b(r_0^2-r_i^2)\;dh \\
& & \nonumber
\end{boxedeqnarray}

\begin{center}
\textbf{\large Arc Length}
\end{center}

\begin{boxedequation}
ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx
\end{boxedequation}
Arc length between $x=a$ and $x=b$ of $y=f(x)$:
\begin{boxedequation}
s=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx
\end{boxedequation}
Arc length between $y=c$ and $y=d$ of $x=f(y)$:
\begin{boxedequation}
s=\int_c^d\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy
\end{boxedequation}

\clearpage
\begin{center}
\textbf{\large Surface Area of Solids of Revolution}
\end{center}

\begin{boxedequation}
dS=2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx
\end{boxedequation}

Surface area between $x=a$ and $x=b$ of $y=f(x)$ rotated about the $x-axis$:
\begin{boxedequation}
S=2\pi\int_a^b y\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx
\end{boxedequation}

Surface area between $x=a$ and $x=b$ of $y=f(x)$ rotated about the $y-axis$:
\begin{boxedequation}
S=2\pi\int_a^b x\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx
\end{boxedequation}

\begin{center}
\textbf{\large Centroid of an Area}
\end{center}

In the following the area is bounded by the vertical lines $x=a$ and
$x=b$ and the curves $y_1=f_1(x)$ on the bottom and $y_2=f_2(x)$ on
top.  The area of the region is $A$. $(\bar{x},\bar{y})$ is the
centroid and $M_x$ and $M_y$ are the moments about the $x-$ and
$y-$axes respectively.
\begin{boxedequation}
M_y=\int_a^b x(y_2-y_1)\;dx
\end{boxedequation}
\begin{boxedequation}
\bar{x}=\frac{1}{A}\int_a^b x(y_2-y_1)\;dx
\end{boxedequation}
\begin{boxedequation}
M_x=\int_a^b\frac{y_1+y_2}{2}(y_2-y_1)\;dx
\end{boxedequation}
\begin{boxedequation}
\bar{y}=\frac{1}{2A}\int_a^b (y_1+y_2)(y_2-y_1)\;dx
\end{boxedequation}

\begin{center}
\textbf{\large Centroid of a Solid of Revolution}
\end{center}

The following two equations apply for a solid of revolution about the
$x-$axis.  Here $M_y$ is the moment about the $x-$axis and $\bar{x}$
is the the $x-$coordinate of the centroid.
\begin{boxedequation}
M_y=\pi\int_a^b xy^2\;dx
\end{boxedequation}
\begin{boxedequation}
\bar{x}=\frac{\pi}{V}\int_a^b xy^2\;dx
\end{boxedequation}
%(Continued next page.)
%\clearpage
The equation for the $y-$coordinate of the centroid for a solid of
revolution about the $y-$axis is:
\begin{boxedequation}
\bar{y}=\frac{\pi}{V}\int_c^d yx^2\;dy
\end{boxedequation}
\clearpage
\begin{center}
\textbf{\large Moment of Inertia of an Area}
\end{center}

The following gives the moment of inertia of the area bounded by $y=f(x)$ and the $x-$axis, and the vertical lines $x=a$ and $x=b$ about the $y-$axis:
\begin{boxedequation}
I_y=\int_a^b x^2y\;dx
\end{boxedequation}
The following gives the moment of inertia of the area bounded by $x=f(y)$ and the $y-$axis, and the horizontal lines $y=c$ and $y=d$ about the $x-$axis:
\begin{boxedequation}
I_x=\int_c^d y^2x\;dy
\end{boxedequation}
The following gives the moment of inertia of the area bounded by $y=f(x)$ and the $x-$axis, and the vertical lines $x=a$ and $x=b$ about the $x-$axis:
\begin{boxedequation}
I_x = \frac{1}{3}\int_a^b y^3\;dx
\end{boxedequation}
The \textbf{Radius of gyration} of an area about an axis is given by the following where $I$ is the moment of inertia and $A$ is the area.
\begin{boxedequation}
r=\sqrt{\frac{I}{A}}
\end{boxedequation}

\begin{center}
\textbf{\large Moment of Inertia of a Solid}
\end{center}

Using the \textbf{Shell Method} where $r$ is the radius and $h$ is the height
of the cylindrical shell about the axis of rotation, the moment of inertia of the solid of density $m$ is:
\begin{boxedequation}
I=2\pi m\int r^3h\;dr
\end{boxedequation}
Using the \textbf{Disk Method} where the disk is of radius $r$ the
moment of inertia is:
\begin{boxedequation}
I=\frac{m\pi}{2}\int_a^b r^4\;dh
\end{boxedequation}

\begin{center}
\textbf{\large Average Value of a Function}
\end{center}
\begin{boxedequation}
y_{\mathit{avg}}=\frac{1}{b-a}\int_a^b f(x)\;dx
\end{boxedequation}
% Print the license
\gfdlsinglenewpage
\end{document}
