\section{Areas Between Curves}

We have seen this topic in a previous course.  We review it here for the purpose of reminding ourselves how to interpret differentials in integration formulae.

We have already calculated an area between curves.  Namely, when $y=f(x)\ge0$ on $[a,b]$ then
\[A=\int_a^b f(x)\,dx\]
is the area below the curve $y=f(x)$ and above the $x$-axis.  But since the $y$-axis is just the curve $y=0$, the above integral is also the area between the curve $y=f(x)$ and the curve $y=0$.  If the lower curve is $y=g(x)$ instead of $y=0$ we get the following result.

\theorem{If $y=f(x)$ and $y=g(x)$ are integrable on the $x$-interval $[a,b]$ with $f(x)\ge g(x)$ on the interval then the \textbf{area between the curves}\index{area between the curves} $y=f(x)$ and $y=g(x)$ and the lines $x=a$ and $x=b$ is 
\[A=\int_a^b \left[f(x)-g(x)\right]\,dx~.\]
}

The situation is illustrated in the following diagram.
\begin{center}
\includegraphics{graphics/integration_applications1}%
\end{center}

A simple way to remember the theorem is using differential notation.  In the picture is shown a vertical differential area element at $x$.  The endpoints have this same $x$ coordinate but different $y$ coordinates.  The height of the area element is $y_2-y_1$ where we subtract the lower value from the higher value to ensure a positive height. The width of the element is $dx$.  The differential area $dA$ is then
\[dA=\underbrace{\left(y_2-y_1\right)}_{\mbox{\tiny height}}\cdot\underbrace{\rule[-.25em]{0em}{1em}dx}_{\mbox{\tiny width}}\,.\]
We require the differential area to be written in terms of the differential variable, here $x$.  Since $(x,y_2)$ lies on the upper curve it satisfies that equation so $y_2=f(x)$ and similarly $(x,y_1)$ lies on the lower curve and satisfies that equation so $y_1=g(x)$\,. The differential area element becomes
\[dA=\left[f(x)-g(x)\right]dx\,.\]
We are finding the sum $A=\int dA$ of these infinitesimal rectangular areas each of area $dA$ starting at $x=a$ and ending at $x=b$ which returns us to our formula.  Note that the limits must be in terms of the same variable as the differential, here $x$.

Sometimes the region for which one wants an area is better described by functions $x=f(y)$ and $x=g(y)$ denoting the right and left boundaries of the region respectively.  One then has the following result.

\theorem{If $x=f(y)$ and $x=g(y)$ are integrable on the $y$-interval $[c,d]$ with $f(y)\ge g(y)$ on the interval then the \textbf{area between the curves} $x=f(y)$ and $x=g(y)$ and the lines $y=c$ and $y=d$ is 
\[A=\int_c^d \left[f(y)-g(y)\right]\,dy~.\]
}

The situation is depicted below.
\begin{center}
\includegraphics{graphics/integration_applications2}%
\end{center}

Here we have a horizontal area element at $y$, so the $y$ coordinate at its endpoints is this value, but the $x$ coordinates must be labelled distinctly.  The width of the area element is $x_2-x_1$ where we subtract the leftmost value from the rightmost one to ensure a positive width.  The height of the area element is $dy$ and therefore the differential area element is
\[dA=\underbrace{\left(x_2-x_1\right)}_{\mbox{\tiny width}}\cdot\underbrace{\rule[-.25em]{0em}{1em}dy}_{\mbox{\tiny height}}\,.\]
We require the area element to be in terms of the differential variable, here $y$.  Since $(x_2,y)$ lies on the curve to the right it satisfies that equation and $x_2=f(y)$. Similarly $(x_1,y)$ lies on the curve on the left so $x_2=g(y)$.  The differential area element becomes 
\[dA=\left[f(y)-g(y)\right]dy\,.\]
In this case the integral represents the addition $A=\int dA$ of horizontal area elements of area $dA$ starting at $y=c$ and ending at $y=d$ which returns us to the formula in our theorem.  Note that once again the limits must be in terms of the variable found in the differential, here $y$.

Sometimes no interval $[a,b]$ or $[c,d]$ is given.  If one is asked to find the region bounded by $y=h_1(x)$ and $y=h_2(x)$ one uses vertical area elements (first theorem) to find the area. One must solve $h_1(x)=h_2(x)$ to find $x$-values $x_i$ which enclose the bounded region(s).  There may be multiple enclosed regions and these values determine the endpoints of the integration.  One must ascertain which function is the higher one on each region.

If one is asked to find the region bounded by $x=h_1(y)$ and $x=h_2(y)$ then one uses horizontal area elements (second theorem) to find the area. One must solve $h_1(y)=h_2(y)$ to find $y$-values $y_i$ which enclose the bounded region(s). Once again these determine the endpoints of integration.  One must determine which function is greater (on the right) on each region.

As we proceed through this chapter we will try to construct our integrals using these intuitive steps to build the necessary differentials.  This will limit the need to memorize many formulae.

\begin{example}
Find the area of the region bounded by the given curves.
\begin{enumerate}
\item $\displaystyle y=x^2-5x$ and $y=10x-x^2$
\item $\displaystyle x=y^2-4$ and $x=-y^2$
\end{enumerate}
\begin{solution*}
\begin{enumerate}
\item $\displaystyle y=x^2-5x$ and $y=10x-x^2$

Here both curves are written as functions of $x$ so we will use vertical area elements. The curves intersect when
\begin{align*}
x^2-5x =10x-x^2 \implies
2x^2-15x = 0 \implies
x(2x-15)=0 \implies x=0\ \text{or}\ x=\frac{15}{2}\,.
\end{align*}
Thus $x=0$ and $x=\frac{15}{2}$ are the $x$-coordinates of the points of intersection.  Evaluating the $y$-coordinates using either function gives the points of intersection to be $(0,0)$ and $(15/2,75/4)$. Consideration of the value $x=3$ (between the endpoints) gives $y=3^2-5(3)=-6$ for the first curve and $y=10(3)-3^2=21$ for the second curve showing that the second curve lies above the first over the interval of intersection.  A graph of the situation follows.
\begin{center}
\includegraphics[width=.7\textwidth]{graphics/integration_applications5}%
\end{center}
Noting that the height $h=y_2-y_1$, the differential area element is
\begin{align*}
  dA=(y_2-y_1)dx=[f(x)-g(x)]dx=\left[(10x-x^2)-(x^2-5x)\right]dx\,,
\end{align*}
where we used the points on the diagram to write $y_2$ and $y_1$ as functions of the variable $x$ found in the differential, $dx$.  The rectangles are summed starting at $x=0$ and ending at $x=\frac{15}{2}$.  Once again these limits must be in terms of the variable $x$ found in the differential.
The area is therefore, using this argument or by formula $A=\int_a^b \left[f(x)-g(x)\right]\,dx$,
\begin{align*}
A&=\int_{0}^{\frac{15}{2}}\left[(10x-x^2)-(x^2-5x)\right]\, dx
=\int_{0}^{\frac{15}{2}}(15x-2x^2)\,dx\\
&=\left[\frac{15}{2}x^{2}-\frac{2}{3}x^{3} \right|^{\frac{15}{2}}_{0} 
=\frac{15}{2}\left(\frac{15}{2}\right)^{2}-\frac{2}{3}\left(\frac{15}{2}\right)^3
=\frac{15^3}{8}\left(1-\frac{2}{3}\right) \\
&=\frac{15^3}{24}= \frac{3375}{24}=\frac{1125}{8} \,\,\ \text{square units.}
\end{align*}
\item $\displaystyle x=y^2-4$ and $x=-y^2$

Here both curves are written as functions of $y$ so we will use horizontal area elements. The curves intersect when
\begin{align*}
y^2-4=-y^2\implies 2y^2 = 4 \implies y^2 = 2 \implies y=\pm \sqrt{2}\,.
\end{align*}
Evaluating $x$ at these values shows the points of intersection are $(-2,-\sqrt{2})$ and $(-2,\sqrt{2})$\,. Evaluating $x$ at the value $y=0$ that lies within the interval $[-\sqrt{2},\sqrt{2}]$ gives $x=-4$ for the first curve and $x=0$ for the second curve showing that $f(y)=-y^2$ is the curve further to the right and $g(y)=y^2-4$ is the curve further to the left over the interval of interest. 
A graph of the situation follows.
\begin{center}
\includegraphics[width=.7\textwidth]{graphics/integration_applications6}%
\end{center}
Noting that the width $w=x_2-x_1$, the differential area element is
\begin{align*}
  dA=(x_2-x_1)dy=[f(y)-g(y)]dy=(-y^2-(y^2-4))\,dy\,,
\end{align*}
where we used the points on the diagram to write $x_2$ and $x_1$ as functions of the variable $y$ found in the differential, $dy$.  The rectangles are summed starting at $y=-\sqrt{2}$ and ending at $y=\sqrt{2}$.  Once again these limits must be in terms of the variable $y$ found in the differential.
The area is therefore, using this argument or by formula $A=\int_c^d \left[f(y)-g(y)\right]\,dy$,
\begin{align*}
A&=\int_{-\sqrt{2}}^{\sqrt{2}} (-y^2-(y^2-4))\,dy
=\int_{-\sqrt{2}}^{\sqrt{2}} (-2y^2+4)\,dy
=\left.-\frac{2}{3}y^{3} +4y \right|^{\sqrt{2}}_{-\sqrt{2}} \\
&=-\frac{2}{3}(\sqrt{2})^{3}+4(\sqrt{2})-\left[-\frac{2}{3}(-\sqrt{2})^{3}+4(-\sqrt{2})\right] \\
&=-\frac{4}{3}(2\sqrt{2})+8(\sqrt{2}) =\left(8-\frac{8}{3}\right)\sqrt{2}
=\frac{16\sqrt{2}}{3} \,\,\,\text{square units.}
\end{align*}
\end{enumerate}
\end{solution*}

\begin{furtherquestions}
Find the area of the region bounded by the following curves.
\begin{multicols}{2}
\begin{enumerate}
\item $y=6-x^2$ and $y=3-2x$
\item $x=y^2$ and $x-y-2=0$
\end{enumerate}
\end{multicols}
\end{furtherquestions}
\end{example}

\textbook{\clearpage}
\begin{exercise}
\input{INTEGRATION_APPLICATIONS/areas-between-curves.tex}
\end{exercise}

\section{Calculation of Volume}
\subsection{Volume as a Calculus Problem}
The \textbf{volume}\index{volume} of a three-dimensional object is the amount of space it occupies (expressed in cubic units).  If we had an odd-shaped jar and had a bag of caramels that were a cm on each side we could estimate the jar's volume by counting the number of caramels that would fit in the jar.  This would be the volume in cubic centimetres (i.e. millilitres).  In a more accurate estimate we would fill the jar with water and then measure the number of millilitres with a measuring cup.

We are interested in how we can use calculus to find such volumes.  An object for which we have a formula for volume is a \textbf{right cylinder} \index{volume! of right cylinders}. A cylinder is a three-dimensional geometric figure that has two congruent and parallel bases.  When the bases are aligned one directly above the other one has a right cylinder.  The following are three examples of right cylinders:
\begin{center}
\parbox[t]{\textwidth}{\vspace{0em}{\resizebox{\textwidth}{!}{\includegraphics{graphics/cylinders}}}}
\end{center}
The first shows the most general case, while the second and third are special cases, a rectangular parallelepiped and a right circular cylinder, respectively.  In general the area of a right cylinder is:
\[V=Ah\]
In the special case where the area is expressible in terms of other lengths we can further specialize the formula.  So a right circular cylinder has volume:
\[V=\pi r^2h\]

Returning to the general case of finding a volume of an arbitrary 3D object we may follow the same route we did for solving general areas.  In the area case we sliced the figure into increasingly thinner rectangles ($\Delta x\rightarrow 0$) and then added those up.  For a 3D object we can imagine slicing it into a large number of increasingly thinner (shorter) right cylinders (of height $\Delta h\rightarrow 0$) and adding them up.  To do this we take an abitrary axis through our solid $S$ and partition the axis into equal widths $\Delta h$.  At each point $h$ along the axis we will have a \textbf{cross-sectional area} \index{area!cross-sectional} $A$ in the plane perpendicular to our axis $h$.
\begin{center} 
\parbox[t]{.6\textwidth}{\vspace{0em}{\resizebox{.6\textwidth}{!}{\includegraphics{graphics/crosssection}}}}
\end{center}
 The area $A$ will depend upon the position $h$ along the axis and so it is a function  $A(h)$ of $h$.  The volume of our object will then be the limit of the Riemann Sum of the $n$ cylinders with volumes $V_i$:
\[V=\lim_{n\rightarrow\infty}\sum_{i=1}^n\underbrace{A(h_i^*)\Delta h}_{V_i}\]
The position $h_i^*$ is taken to be a point in the corresponding interval along $h$ of the $i^{\mbox{\tiny th}}$ cylinder.
This limit is the definite integral:
\[V=\int_a^bA(h)dh\ ,\]
where $a$ and $b$ denote the position of the first and last cylinders needed.
Thinking in differential notation we are summing $(\int dV)$ the volumes of infinitessimally thin cylinders given by:
\[dV=A\,dh\]

\subsection{Solids of Revolution}
In general the function $A(h)$ for a particular solid $S$ will be complicated.  However if the solid has a symmetry and the axis $h$ is directly along that axis of symmetry the function $A(h)$ may be calculated.  A certain class of solids, called \textbf{Solids of Revolution}\index{solids of revolution} have such a symmetry because they are generated by sweeping out a two dimensional area around just such an axis:
\begin{center}
\parbox[t]{.8\textwidth}{\vspace{0em}{\resizebox{.8\textwidth}{!}{\includegraphics{graphics/solids_of_revolution}}}}
\end{center}
As can be seen in the diagram some common shapes (spheres and cones) can be thought of as solids of revolution of simple areas (semicircles and triangles).  Other interesting shapes may be constructed by choosing other functions for the upper boundary.

\subsection{The Disk Method}\index{disk method} \index{solids of revolution! calculating volume using the disk method}
For a solid of revolution for which the planar area of the revolution extends from the axis of symmetry to a boundary curve describable by a function, the volume is now calculable because the area function $A(h)$ can be found:
\begin{center}
\parbox[t]{.65\textwidth}{\vspace{0em}{\resizebox{.65\textwidth}{!}{\includegraphics{graphics/disk_method}}}}
\end{center}
As shown in the diagram, the general right cylinders are now right \textbf{circular} cylinders with area $A=\pi r^2$.  Our volume element is the \textbf{disk} of differential volume:
\[\boxed{dV=\pi r^2 dh}\]
where the radius $r$ for the given disk is $r=f(h)$.  The volume is then the sum $(V=\int dV)$ given by
\[V=\int_a^b\pi r^2\,dh=\int_a^b\pi\left[f(h)\right]^2dh\]
If the symmetry axis is the $x$-axis (so $h=x$ and $r=y=f(x)$) then the formula for a solid of revolution about the $x$-axis becomes
\[V=\int_a^b\pi\left[f(x)\right]^2dx\]
This method of calculating volumes is called the \textbf{Disk Method} \index{disk method! for approximating volume} because we are approximating the volume with disks (i.e. short right cylinders).
\begin{example}
Find the volume of the solid obtained by revolving the following regions about the $x$-axis. 
\begin{enumerate}
\item The region bounded by the curves $\displaystyle y=x^{\frac{5}{3}}, x=5$, and $y=0$\,.
\item The region bounded by the curves $\displaystyle y=6x-x^2$ and $y=0$\,.
\end{enumerate}
\begin{solution*}
\begin{enumerate}
\item The region bounded by the curves $\displaystyle y=x^{\frac{5}{3}}, x=5$, and $y=0$\,.

A graph of the region of rotation is the following.
\begin{center}
\includegraphics[width=.3\textwidth]{graphics/integration_applications7}%
\end{center}
The vertical area element at $x$ being rotated has upper coordinate $(x,y)$.  The radius then is $r=y$ and the height of the disk $dh=dx$.  The disk volume element $dV$ is
\begin{align*}
  dV=\pi r^2dh=\pi y^2dx=\pi\left(x^{\frac{5}{3}}\right)^2dx\,.
\end{align*}
Here the variable $y$ must be written in terms of $x$, the variable found in the differential ($dx$).  Since the upper point $(x,y)$  sits on the curve $y=x^{\frac{5}{3}}$ this means it satisfies that equation allowing the replacement. In addition to replacing all variables in terms of those of the differential $dx$, the limits are also determined to be along the axis of the differential (again $x$), so $x=0$ to $x=5$\,.  The Disk Method volume using the above derivation or by formula $V=\int_a^b\pi\left[f(x)\right]^2dx$ is
\begin{align*}
V= \int_{0}^{5}\pi\left(x^{\frac{5}{3}}\right)^2\, dx
=\pi \int_{0}^{5}x^{\frac{10}{3}}\, dx 
  =\pi\left[\frac{3}{13}x^{\frac{13}{3}}\right|^{5}_{0}
  =\pi\left[\frac{3}{13}(5)^{\frac{13}{3}}-0\right]
=\frac{3\pi}{13}(5)^{\frac{13}{3}}\,\,\,\text{cubic units.}
\end{align*}
\item The region bounded by the curves $\displaystyle y=6x-x^2$ and $y=0$\,.

  The curves intersect when
  \begin{align*}
    6x-x^2=0\implies x(6-x)=0\implies x=0\ \text{or}\ x=6\,,
  \end{align*}
  so at points $(0,0)$ and $(6,0)$.  A graph of the region is the following.
  \begin{center}
    \includegraphics[width=.35\textwidth]{graphics/integration_applications8}%
  \end{center}
Labelling the upper end of the area element $(x,y)$ we see that
\begin{align*}
  dV=\pi r^2dh=\pi y^2dx=\pi\left(6x-x^2\right)^2dx\,.
\end{align*}
with limits of integration along the axis of the differential ($dx$) being $x=0$ and $x=6$. The volume using the above differential or the formula $V=\int_a^b\pi\left[f(x)\right]^2dx$ is
\begin{align*}
V&= \int_{0}^{6}\pi\left(6x-x^2\right)^2\, dx
=\pi \int_{0}^{6}(36x^2-12x^3+x^4)\, dx 
=\pi\left[12x^3-30x^4+\frac{1}{5}x^5\right|^{6}_{0}\\
&=\pi\left(12(6)^3-3(6)^4+\frac{1}{5}(6)^5-[0-0+0]\right)
=\frac{1296\pi}{5}\,\,\, \text{cubic units.}
\end{align*}
\end{enumerate}
As this example illustrates, rather than memorizing a formula and going straight to the integral, one may simply remember the intuitive formula for the differential volume $dV=\pi r^2 dh$ and systematically replace the variables and differential for the given situation.  Remember that the entire differential as well as the limits of the integral must ultimately be written in terms of the variable found in the differential~$dh$.
\end{solution*}

\begin{furtherquestions}
Find the volume of the solid obtained by revolving the following regions about the $x$-axis.
\begin{enumerate}
\item The region bounded by the curves $y=x^{\frac{3}{2}}$, $x=4$, and $y=0$\,.
\item The region bounded by the curve $y=2x-x^2$ and the line $y=0$\,.
\end{enumerate}
\end{furtherquestions}
\end{example}

The constant $\pi$ is defined to be the ratio of the circumferance to the diameter of any circle, $\pi=\frac{C}{D}$.  Archimedes (c. 287-212 BCE) discovered (long before the invention of calculus) that the constant $\pi$ surprisingly shows up in the formulae for the area of a circle ($A=\pi R^2$) as well as for a sphere's volume ($V=\frac{4}{3}\pi R^3$) and surface area ($S=4\pi R^2$).
\begin{example}
  Use calculus to show that the volume of a sphere of radius $R$ is $V=\frac{4}{3}\pi R^3$.
  
\begin{solution}
Consider a circle of radius $R$ with centre at the origin. The region bounded by this circle and the lines $x=0$ and $y=0$ lying in the first quadrant is shown.
\begin{center}
\includegraphics[width=.4\textwidth]{graphics/integration_applications9}%
\end{center}
Here $y=f(x)=\sqrt{R^2-x^2}$\,. If we rotate this region about the $x$-axis, we obtain half the volume of a sphere.  Using the Disk Method, the differential volume element is
\begin{align*}
  dV=\pi r^2dh=\pi y^2dx=\pi\left(R^2-x^2\right)dx\,.
\end{align*}
The limits along the differential axis ($x$) are $x=0$ and $x=R$ and the integral is then, using the differential or formula $V=\int_a^b\pi\left[f(x)\right]^2dx$,
\begin{align*}
V=\pi\int^{R}_{0}(R^2-x^2)\,dx 
=\pi\left[R^{2}x-\frac{1}{3}x^{3}\right|_{0}^{R}
=\pi\left(R^3-\frac{1}{3}R^3-\left[R^2(0)-\frac{1}{3}(0)^3\right]\right)=\frac{2}{3}\pi R^3\,.
\end{align*}
Therefore, the volume of the sphere is $\displaystyle 2\left(\frac{2}{3}\pi R^3\right)=\frac{4}{3}\pi R^3$.
\end{solution}

\begin{furtherquestion}
  Use calculus to show that the volume of a right circular cone with base radius $R$ and height $H$ is $V=\frac{1}{3}\pi R^2H$.
\end{furtherquestion}
\end{example}

We sometimes wish to revolve a region about the $y$-axis.  For a region bounded by the curve $x=f(y)$ and the lines $x=0$ (i.e. the $y$-axis), $y=c$, and $y=d$ we can use horizontal disks.  In this case $dh=dy$ (since the symmetry axis $h$ is now $y$) and $r=x=f(y)$ as shown in the following diagram.
\begin{center} 
\parbox[t]{.65\textwidth}{\vspace{0em}{\resizebox{.65\textwidth}{!}{\includegraphics{graphics/disk_method_vertical}}}}
\end{center}

The formula for the volume of such a region revolved about the $y$-axis is therefore
\[V=\int_c^d\pi\left[f(y)\right]^2dy\,.\]

\begin{example}\label{exam:disk1}%
Find the volume of the solid obtained by revolving the region bounded by $y=x^7$, $x=0$, and $y=2$ about the $y$-axis.
  
\begin{solution}
  The lines $y=x^7$ and $y=2$ intersect when $x^7=2\implies x=2^{\frac{1}{7}}$, and so at the point $(2^{\frac{1}{7}},2)$\,.
A graph of the region being rotated is as follows.
\begin{center}
\includegraphics[width=.45\textwidth]{graphics/integration_applications10}%
\end{center}
Consideration of the point $(x,y)$ at the end of the horizontal differential element shows that $r=x$ and $dh=dy$, so the volume element is
\begin{align*}
  dV=\pi r^2dh=\pi x^2dy=\pi\left(y^{\frac{1}{7}}\right)^2dy\,,
\end{align*}
where here we used that $(x,y)$ lies on the curve $y=x^7\implies x=f(y)=y^{\frac{1}{7}}$ to replace $x$ in terms of the differential variable $y$.  The limits in $y$ are $y=0$ and $y=2$ so the Disk Method volume is therefore, using the differential or the formula $V=\int_c^d\pi\left[f(y)\right]^2dy$, 
\begin{align*}
V=\int_{0}^{2}\pi\left(y^{\frac{1}{7}}\right)^2\, dy
=\pi \int_{0}^{2}y^{\frac{2}{7}}\, dy
= \pi \left[\frac{7}{9}y^{\frac{9}{7}}\right|^{2}_{0}
  =\frac{7\pi}{9}\left[(2)^{\frac{9}{7}}-(0)^{\frac{9}{7}}\right]
  =\frac{7\pi}{9}(2)^{\frac{9}{7}}\,\,\, \text{cubic units.}
\end{align*}
\end{solution}

\begin{furtherquestion}
  Find the volume of revolution of the solid obtained by revolving the region bounded by $y=x^2$, $x=0$, and $y=4$ about the $y$-axis.
\end{furtherquestion}
\end{example}

In general we may be interested in revolving an area about a horizontal or vertical axis which is neither the $x$ nor the $y$-axis.  In this case we must work from the differential volume $dV$ to determine the volume integral.  Follow the following steps which will work for any volume problem: \index{volume! steps for calculations}
\begin{enumerate}
\item Draw the region to be revolved labelling the curves appropriately as well as points of intersection. Draw the symmetry line about which the curve is to be revolved.
\item Decide on the differential volume element you will use and indicate a typical cross-section rectangle on your region.
\item Label the endpoints of rectangle with coordinates $(x,y)$, etc.  Note that for vertical elements the endpoint coordinates will have the same endpoint $x$ values while you will need to distinguish the $y$ values (i.e. $y_1$, $y_2$) as these values will differ.  If a value is constant (e.g. 0) for all potential cross-section rectangles label it as such instead of using a variable.  The opposite holds if you are using a horizontal element (i.e. same $y$ values at endpoints, different $x$'s.)
\item Label the differential width of the cross-section rectangle as $dx$ or $dy$ according to whether you are using a vertical or horizontal elements respectively.
\item Write down your integral, $V=\int dV$, and substitute your volume element for $dV$.
\item Next replace each variable in the integral in terms of the variables of your cross-section rectangle endpoint coordinate variables.  Some calculation may be required here, especially if you are not revolving about a coordinate axis.
\item From your diagram determine the limits of integration.  These must be along the same axis as found in your differential, i.e. $x$ for $dx$ or $y$ for $dy$.  Note the limits are for the rectangles only within the region being revolved.
\item All variables of the integrand must be in terms of the integration variable.  Replace any variables in your integral that are not the same as the differential variable in terms of the differential variable by considering the curve that the endpoint involving the variable lies upon.  You may need to solve for the variable using the curve equation to do so.
\item Calculate the integral.
\end{enumerate}
\begin{example}
  Find the volume of the solid generated by revolving the region bounded by $y=x^7$, $x=0$, and $y=2$ about the line $y=2$\,.
  
\begin{solution}
  The region being rotated is the same as in Example~\ref{exam:disk1}, but now we are rotating about the line $y=2$ and so require a vertical area element as shown in the following diagram.
\begin{center}
\includegraphics[width=.27\textwidth]{graphics/integration_applications11}%
\end{center}
Because the rotation is not about the Cartesian axes there is no formula that can be used directly.  Instead we build up the volume element $dV$ by first labelling the endpoints of the vertical element $(x,y_1)$ and $(x,y_2)$.  Note these points have the same $x$ coordinate but different $y$ coordinates which must be distinguished. (The opposite would be true for a horizontal area element.) From the diagram we see that the radius $r=y_2-y_1$, where here we have subtracted the lower $y$ coordinate from the upper one to ensure a positive radius.  (For horizontal elements we would subtract the value on the left from that on the right to ensure positivity.) Finally $dh=dx$ and we have for the volume element
\begin{align*}
  dV=\pi r^2dh=\pi(y_2-y_1)^2dx=\pi\left(2-x^7\right)^2dx\,.
\end{align*}
Here we replaced $y_2$ and $y_1$ to get them in terms of the differential variable $x$ by consideration of which curve the points lie on in the diagram. The limits in $x$ are $x=0$ on the left to $x=2^{\frac{1}{7}}$ on the right. The volume is then
\begin{align*}
V&=\int_{0}^{2^\frac{1}{7}} \pi (2-x^7)^2\, dx
= \pi \int^{2^\frac{1}{7}}_{0}(4-4x^7+x^{14})\, dx
=\pi \left[4x-\frac{1}{2}x^8+\frac{1}{15}x^{15}\right|_{0}^{2^\frac{1}{7}}\\
  &=\pi \left(4(2^\frac{1}{7})-\frac{1}{2}(2^\frac{1}{7})^8 + \frac{1}{15}(2^\frac{1}{7})^{15}-[0-0+0]\right) \\
  &=\pi \left(4(2^\frac{1}{7})-\frac{1}{2}(2^\frac{8}{7}) + \frac{1}{15}(2^\frac{15}{7})\right)=\pi (2^{\frac{1}{7}})\left(4-\frac{1}{2}(2)^{\frac{7}{7}} + \frac{1}{15}(2)^{\frac{14}{7}}\right)\\
  &=\pi (2^{\frac{1}{7}})\left(4-1+\frac{4}{15}\right)=\frac{49\pi (2^{\frac{1}{7}})}{15}\,\,\, \text{cubic units}.
\end{align*}
The reader is encouraged to visualize the volume generated in this problem compared to that of Example~\ref{exam:disk1}\,. The axis about which one rotates matters!
\end{solution}

\begin{furtherquestion}
  Find the volume of the solid generated by revolving the region bounded by $y=x^2$ and $y=4$ about the line $y=4$.
\end{furtherquestion}
\end{example}

\subsection{The Washer Method}
Suppose we generate a solid of revolution in which the lower boundary is not the axis of revolution but, more generally the curve $g(h)$ above it.  These regions are what we considered when we looked at areas between curves having an upper boundary determined by a function $f$ and a lower boundary determined by a function $g$.  If such a region is revolved about a chosen axis of symmetry $h$ then the solid of revolution generated will have a hole in it and the disk method will not, at first glance, work.
\begin{center} 
\parbox[t]{.65\textwidth}{\vspace{0em}{\resizebox{.65\textwidth}{!}{\includegraphics{graphics/solids_of_revolution_washer}}}}
\end{center}

The disk method can be used to calculate the volume of the outer solid and then again the inner solid with the actual volume being the difference of the two, but instead we introduce a new method which solves the problem with a single integral.  Consider a right circular cylinder of height $h$ and outer radius $r_{\s o}$ with an interior cylinder of inner radius $r_{\s i}$ removed:
\begin{center} 
\parbox[t]{.4\textwidth}{\vspace{0em}{\resizebox{.4\textwidth}{!}{\includegraphics{graphics/washer}}}}
\end{center}

The volume of the solid is just the difference of the volumes of the cylinders:
\[V=V_o-V_i=\pi r_o^2h-\pi r_i^2h\]
Factoring out the $\pi$ and $h$ gives:
\[V=\pi(r_o^2-r_i^2)h\]

If we imagine shrinking the height $h$ to a small height $\Delta h$ the shape looks like a metal \emph{washer}.  In the \textbf{Washer Method}\index{washer method} \index{solids of revolution! calculating volume using the washer method} we will slice solids of revolution with symmetrical holes in them into such washer-shaped volumes and sum them up to find the total volume of the solid.
\begin{center} 
\parbox[t]{.8\textwidth}{\vspace{0em}{\resizebox{.8\textwidth}{!}{\includegraphics{graphics/washer_method}}}}
\end{center}

In the Washer Method the height of the washer $\Delta h$ is replaced with the differential $dh$ so the differential element of volume is:
\[\boxed{dV=\pi(r_o^2-r_i^2)dh}\]
The total volume is just the sum $(V=\int dV)$ and we have:
\[V=\int_a^b\pi(r_o^2-r_i^2)dh=\int_a^b\pi\left([f(h)]^2-[g(h)]^2\right)dh\]
where we have replaced the outer and inner radii by their functional values at $h$ (see diagram).
If the symmetry axis is the $x$-axis (so $h=x$, $r_o=y_2=f(x)$, and $r_i=y_1=g(x)$) then the formula for a solid of revolution about the $x$-axis becomes
\[V=\int_a^b\pi\left([f(x)]^2-[g(x)]^2\right)dx\]

This may always be found starting with the differential volume element $dV=\pi(r_o^2-r_i^2)dh$ and following the general steps.\index{washer method! for approximating volume}

\textbook{\clearpage}
\begin{example}\label{exam:washer1}%
  Find the volume of the solid obtained by revolving the region bounded by the curves $y=2x$ and $y=x^4$ about the $x$-axis.
  
  \begin{solution}
    The curves $y=2x$ and $y=x^4$ intersect when
    \begin{align*}
      2x=x^4\implies 0=x^4-2x\implies 0=x(x^3-2)\implies x=0\ \text{or}\ x=2^{\frac{1}{3}}\,,
    \end{align*}
    and so at the points $(0,0)$ and $(0,2^{\frac{4}{3}})$.
A graph of the region is
\begin{center}
\includegraphics[width=.3\textwidth]{graphics/integration_applications12}%
\end{center}
Because the area to be revolved is not flush with the axis of revolution (the $x$-axis), we use the Washer Method that can accommodate the hole. As with the Disk Method, the area element for the Washer Method is perpendicular to the axis of symmetry as shown.  From the diagram the outer and inner radii are $r_o=y_2$ and $r_i=y_1$, and the differential height is $dh=dx$.  The volume element is
\begin{align*}
  dV=\pi(r_o^2-r_i^2)dh=\pi(y_2^2-y_1^2)dx=\pi\left((2x)^2-(x^4)^2\right)dx\,,
\end{align*}
where we write the $y$ coordinates in terms of the differential variable $x$ by consideration of the curves upon which their points lie.  The limits of that variable are $x=0$ and $x=2^{\frac{1}{3}}$\,.  The volume is therefore, using the above reasoning or directly by formula $V=\int_a^b\pi\left([f(x)]^2-[g(x)]^2\right)dx$,
\begin{align*}
V&=\int_{0}^{2^\frac{1}{3}} \pi\left((2x)^2-(x^4)^2\right)dx
= \pi \int^{2^\frac{1}{3}}_{0}\!\left(4x^2-x^8\right)dx
=\pi \left[\frac{4}{3}x^3-\frac{1}{9}x^9 \right|_{0}^{2^\frac{1}{3}}\\ 
&=\pi  \left(\frac{4}{3}(2^{\frac{1}{3}})^3-\frac{1}{9}(2^{\frac{1}{3}})^9-[0-0]\right)=\pi\left(\frac{4}{3}(2)-\frac{1}{9}(2)^3\right)
=\pi \left(\frac{8}{3}-\frac{8}{9}\right) \\
&= \frac{16\pi}{9}\,\,\, \text{cubic units.}
\end{align*}
\end{solution}

\begin{furtherquestion}
  Find the volume of the solid obtained by revolving the region bounded by the curves $y=x^2+1$ and $y=3-x^2$ about the $x$-axis.
\end{furtherquestion}
\end{example}

If the symmetry axis is the $y$-axis then $h=y$ and we can write the outer radii as $r_o=x_2=f(y)$ and the inner radii as $r_i=x_1=g(y)$.  The general formula for a solid of revolution about the $y$-axis becomes:
\[V=\int_c^d\pi\left([f(y)]^2-[g(y)]^2\right)dy\]
Once again we use our convention of using constants $c$ and $d$ for limits along the $y$-axis.

\begin{example}
Find the volume of the solid obtained by revolving the region bounded by $y=2x$ and $y=x^6$ about the $y$-axis.
  
\begin{solution}
The curves intersect when
\begin{align*}
2x = x^6 \implies
2x-x^6 =0 \implies
x(2-x^5) =0 \implies
x=0\ \text{or}\ x=2^{\frac{1}{5}}\,.
\end{align*}
Evaluating the $y$ coordinates for each give $y=0$ and $y=2^{\frac{6}{5}}$ respectively, so the intersection points are $(0,0)$ and $(2^{\frac{1}{5}},2^{\frac{6}{5}})$\,.  A graph of the region follows.
\begin{center}
\includegraphics[width=.25\textwidth]{graphics/integration_applications13}%
\end{center}
Since we are rotating about the $y$-axis we require a horizontal area element perpendicular to the axis to use the Washer Method.  We label the endpoints $(x_1,y)$ and $(x_2,y)$ since they share the same $y$ coordinate but different $x$ coordinates.  The differential volume element is
\begin{align*}
  dV=\pi(r_o^2-r_i^2)dh=\pi(x_2^2-x_1^2)dy=\pi\left(\left(y^{\frac{1}{6}}\right)^2-\left(\frac{1}{2}y\right)^2\right)dy\,,
\end{align*}
where here we had to solve our curves in terms of $y$ as
\begin{align*}
  y=x^6 & \implies x=f(y)=y^{\frac{1}{6}}\\
  y=2x & \implies x=g(y)=\frac{y}{2}\,,
\end{align*}
in order to write $x_1$ and $x_2$ in terms of the differential variable $y$.  The limits in that variable are $y=0$ and $y=2^{\frac{6}{5}}$.

The volume is, using the previous reasoning or by formula $V=\int_c^d\pi\left([f(y)]^2-[g(y)]^2\right)dy$,
\begin{align*}
V&=\pi\int^{2^\frac{6}{5}}_{0}\left(\left(y^{\frac{1}{6}}\right)^2-\left(\frac{1}{2}y\right)^2\right)dy
=\pi\int^{2^\frac{6}{5}}_{0}\left(y^{\frac{1}{3}}-\frac{1}{4}y^2\right)dy \\
&=\pi\left[\frac{3}{4}y^{\frac{4}{3}}-\frac{1}{12}y^3\right|_{0}^{2^\frac{6}{5}} 
=\pi\left(\frac{3}{4}(2^{\frac{6}{5}})^{\frac{4}{3}}-\frac{1}{12}(2^{\frac{6}{5}})^3-[0-0]\right) \\
  &=\pi \left(\frac{3}{4}(2^{\frac{8}{5}})-\frac{1}{12}(2^{\frac{18}{5}})\right)
  =\pi(2^{\frac{3}{5}})\left(\frac{3}{4}(2^{\frac{5}{5}})-\frac{1}{12}(2^{\frac{15}{5}})\right)\\
  &=\pi(2^{\frac{3}{5}})\left(\frac{3}{2}-\frac{2}{3}\right)
  =\frac{5\pi(2^{\frac{3}{5}})}{6} \, \, \, \text{cubic units.}
\end{align*}
\end{solution}

\begin{furtherquestion}
  Find the volume of the solid obtained by revolving the region bounded by $y=2x$ and $y=x^2$ about the $y$-axis.
\end{furtherquestion}
\end{example}

If one is revolving about an axis that is not a coordinate axis there is no immediate formula since the curve functions no longer represent distances from the axes of revolution.  Follow the steps outlined previously starting with the volume element and identifying the variables from your diagram.  In this case the radii will need to be calculated in terms of the variables.

\begin{example}
Find the volume of the solid obtained by revolving the region bounded by the curves $y=2x$ and $y=x^6$ about the line $x=5$\,.
  
\begin{solution}
This is the same region as in Example~\ref{exam:washer1} but now rotated about the line $y=5$\,.  A graph of the situation is
\begin{center}
\includegraphics[width=.6\textwidth]{graphics/integration_applications14}%
\end{center}
As we are not rotating about one of the Cartesian axes a formula will not work.  Consideration of the diagram shows that the inner and outer radii of the washer are
\begin{align*}
  r_i&=5-x_2=5-y^{\frac{1}{6}}\\
  r_o&=5 -x_1=5 -\frac{1}{2}y\,.
\end{align*}
Here we ensured the radii were positive by subtracting the left endpoint from the right endpoint.  We then replaced the $x$ coordinates in terms of the differential variable $y$ using the curves that the points lie upon as in Example~\ref{exam:washer1}\,.  The volume element is therefore
\begin{align*}
  dV=\pi(r_o^2-r_i^2)dh=\pi[(5-x_1)^2-(5-x_2)^2]dy=\pi\left[\left(5-\frac{1}{2}y\right)^2-\left(5-y^{\frac{1}{6}}\right)^2\right]dy\,,
\end{align*}
and the volume is
\begin{align*}
V&=\pi\int^{2^\frac{6}{5}}_{0}\left[\left(5-\frac{1}{2}y\right)^2-\left(5-y^{\frac{1}{6}}\right)^2\right]dy
=\pi \int_{0}^{2^\frac{6}{5}}\left[25-5y+\frac{1}{4}y^2-\left(25-10y^{\frac{1}{6}}+\left(y^{\frac{1}{6}}\right)^2\right)\right]\, dy \\
&=\pi \int_{0}^{2^\frac{6}{5}}\left[-5y + \frac{1}{4}y^2+10y^{\frac{1}{6}}-y^{\frac{1}{3}}\right]\, dy
=\pi\left[-\frac{5}{2}y^2+ \frac{1}{12}y^3+\frac{60}{7}y^{\frac{7}{6}}-\frac{3}{4}y^{\frac{4}{3}}\right|_{0}^{2^\frac{6}{5}}\\ 
&=\pi \left(-\frac{5}{2}(2^{\frac{6}{5}})^2+ \frac{1}{12}(2^{\frac{6}{5}})^3 +\frac{60}{7}(2^{\frac{6}{5}})^{\frac{7}{6}}-\frac{3}{4}(2^{\frac{6}{5}})^{\frac{4}{3}}\right)
  =\pi \left(-\frac{5}{2}(2^{\frac{12}{5}})+\frac{1}{12}(2^{\frac{18}{5}})+\frac{60}{7}(2^{\frac{7}{5}})-\frac{3}{4}(2^{\frac{8}{5}})\right)\\
  &=\pi \left(-\frac{5}{2}(2)^{\frac{10}{5}}(2^{\frac{2}{5}})+\frac{1}{12}(2^{\frac{15}{5}})(2^{\frac{3}{5}})+\frac{60}{7}(2^{\frac{5}{5}})(2^{\frac{2}{5}})-\frac{3}{4}(2^{\frac{5}{5}})(2^{\frac{3}{5}})\right)\\
  &=\pi \left(\left[-10+\frac{120}{7}\right](2^{\frac{2}{5}})+\left[\frac{2}{3}-\frac{3}{2}\right](2^{\frac{3}{5}})\right)
    =\frac{50\pi}{7}(2^{\frac{2}{5}})-\frac{5\pi}{6}(2^{\frac{3}{5}})
    \approx 25.6  \, \, \, \text{cubic units.}
\end{align*}
The reader should compare the volume generated in this example with that of Example~\ref{exam:washer1}\,.
\end{solution}

\begin{furtherquestion}
  Find the volume of the solid obtained by revolving the region bounded by the curves $y=\sqrt[3]{x}$ and $y=\frac{1}{4}x$ and lying in the first quadrant ($x\ge 0$ and $y\ge 0$) about the line $x=10$.
\end{furtherquestion}
\end{example}

\begin{exercise}
\input{INTEGRATION_APPLICATIONS/calculation-of-volume.tex}
\end{exercise}

\subsection{The Shell Method}\index{shell method}
Suppose one wished to calculate the volume of the solid generated by revolving the region bounded by the curve $y=\sin(x^2)$ and the line $y=0$ about the $y$-axis as shown.
\begin{center} 
\parbox[t]{.4\textwidth}{\vspace{0em}{\resizebox{.4\textwidth}{!}{\includegraphics{graphics/sinxsqr}}}}
\end{center}
While one can imagine dividing the solid into washer-shaped regiond perpendicular to the $y$-axis, there would be difficulties. 
\begin{center} 
\parbox[t]{.4\textwidth}{\vspace{0em}{\resizebox{.4\textwidth}{!}{\includegraphics{graphics/sinxsqr_washer}}}}
\end{center}
Ultimately we would need to solve for $x$ in terms of $y$ to find the inner and outer radii of the washer in terms of the position $y$ of the disk.  This could be tricky in part because of the trigonometric function.  Moreover in inverting the function we would somehow need to generate both an upper and lower function $x(y)$ corresponding to the two radii of the washer!

We now consider a different approach by the introduction of a new volume element.  Recall the cylinder with the inner cylinder removed.
\begin{center} 
\parbox[t]{.4\textwidth}{\vspace{0em}{\resizebox{.4\textwidth}{!}{\includegraphics{graphics/washer}}}}
\end{center}
We found before the formula for the volume of this solid was just the difference of the volume of the cylinders:
\[V=\pi\left(r_o^2-r_i^2\right)h\]
Notice that we can factor the difference of squares:
\[V=\pi\left(r_o+r_i\right)(r_o-r_i)h\]
Defining the average radius $r$ to be:
\[r=\frac{r_o+r_i}{2}\ ,\]
and the difference of the radii to be
\[\Delta r=r_o-r_i\ ,\]
we can rewrite the volume as
\[V=\pi(2r)(\Delta r)h\ ,\]
or equivalently
\[V=2\pi rh\,\Delta r\ .\]
Now suppose we make the two radii approximately equal.  Then the average radius $r$ approximately equals either of them with little error, while their difference $\Delta r$ will be small.  The solid will be like a cylindrical tin can (with the ends cut off) which we will refer to as a ``shell''.  The volume formula for the shell now becomes quite intuitive.  If we imagine cutting the shell along its length and unrolling it we would effectively have a rectangular sheet with top length equalling the circumference of the circle ($2\pi r$) and the other side length equalling the height.  The thin thickness will be $\Delta r$. \index{shell method! for approximating volume}
\begin{center} 
\parbox[t]{\textwidth}{\vspace{0em}{\resizebox{\textwidth}{!}{\includegraphics{graphics/shell_intuitive}}}}
\end{center}
So the volume of the rectangular sheet (the shell) is then:
\[V=(\mbox{circumference})(\mbox{height})(\mbox{depth})=(2\pi r)(h)(\Delta r)=2\pi rh\,\Delta r\]
as found before.    

Now in the Shell Method we imagine slicing the solid of revolution into concentric cylindrical shells and adding up the volumes. \index{solids of revolution! calculating volume using the shell method}
\begin{center} 
\parbox[t]{.65\textwidth}{\vspace{0em}{\resizebox{.65\textwidth}{!}{\includegraphics{graphics/shell_method}}}}
\end{center}
In the limit that the number of shells goes to infinity, so $\Delta r\rightarrow 0$, the limit of the Riemann Sum is an integral ($\int dV$) of differential volume
\[\boxed{dV=2\pi rh\,dr}\]
with, as usual the finite $V$ and $\Delta r$ replaced with $dV$ and $dr$ in our shell volume formula.  Specifically, for the region depicted above we have for the volume of the solid of revolution
\[V=\int_a^b 2\pi rh\,dr=\int_a^b 2\pi rf(r)\,dr\,.\]
Notice that in the Shell Method the differential $dr$ is now \emph{perpendicular} to the axis of symmetry.

If the axis of symmetry is the $y$-axis then, setting $h=y$ and $r=x$ in the above formula we get the following result for an a solid of revolution generated by revolving the region bounded by the curve $y=f(x)$, $y=0$, $x=a$, and $x=b$ about the $y$-axis
\[V=\int_a^b 2\pi xf(x)\,dx\,.\]

\begin{example}
  Find the volume of the solid generated by revolving the region bounded by the curves $y=\sin(5x)$ and  $y=0$ from $x=0$ to $x=\frac{\pi}{5}$ about the $y$-axis.
  
\begin{solution}
A graph of the region of interest is as follows.
\begin{center}
\includegraphics[width=.3\textwidth]{graphics/integration_applications15}%
\end{center}
We note that the area is not flush with the axis of symmetry so the Disk Method will not work.  To use the Shell Method we draw an area element that is parallel to the axis of symmetry as shown. From the diagram it follows that $dr=dx$, $r=x$, and $h=y$.  The differential volume element is therefore
\begin{align*}
  dV=2\pi rh\,dr=2\pi xy\,dx=2\pi x\sin(5x)dx\,,
\end{align*}
where here used that $(x,y)$ lies on the curve $y=\sin(5x)$ to write $y$ in terms of the differential variable $x$.  The limits in that variable are $x=0$ and $x=\frac{\pi}{5}$ so the volume becomes, using the above differential or formula $V=\int_a^b 2\pi xf(x)\,dx$,
\begin{align*}
V&=\int_{0}^{\frac{\pi}{5}} 2\pi x \sin(5x)\, dx
   =2\pi \int^{\frac{\pi}{5}}_{0} x\sin (5x) \, dx
   =2\pi\left.\left[x\left(-\frac{1}{5}\cos(5x)\right)\right|_0^{\frac{\pi}{5}}-\int_0^{\frac{\pi}{5}}\left(-\frac{1}{5}\cos(5x)\right)dx\right]\\
   &=2\pi \left[-\frac{1}{5}x\cos(5x) + \frac{1}{25}\sin(5x) \right|^{\frac{\pi}{5}}_{0}
=2\pi\left(-\frac{\pi}{25}\cos \pi + \frac{1}{25}\sin \pi -\left[-\frac{1}{5}(0)(1)+\frac{1}{25}(0)\right]\right)\\ 
&=2\pi \left(\frac{\pi}{25}\right) 
=\frac{2\pi^2}{25}\,\, \text{cubic units.}
\end{align*}
This problem could also be done using horizontal area elements and the Washer Method but it becomes much more difficult.  One needs first to solve for $y=\sin(5x)$ for $x$ to determine $x=g(y)=\frac{1}{5}\sin^{-1}\!y$ for the left of the hill and $x=f(y)=\frac{\pi}{5}-\frac{1}{5}\sin^{-1}\!y$ for the right of the hill.  The resulting Washer Method integral also is considerably more challenging.  As such when confronted with a volume of rotation it is worth considering which of the Shell or Washer Methods is easier to apply.
\end{solution}

\begin{furtherquestion}
  Find the volume of the solid generated by revolving the region bounded by the curves $y=\sin(x^2)$ and $y=0$ between $x=0$ and $x=\sqrt{\pi}$ about the $y$-axis.
\end{furtherquestion}
\end{example}

If the axis of symmetry is the $x$-axis then, setting $h=x$ and $r=y$ in the above formula we get the following formula for a solid of revolution generated by the region bounded by the curves $x=f(y)$, $x=0$, $y=c$, and $y=d$.
\[V=\int_c^d 2\pi yf(y)\,dy\]

More generally the lower bound of the region need not be along the radial axis: 
\begin{center} 
\parbox[t]{.5\textwidth}{\vspace{0em}{\resizebox{.5\textwidth}{!}{\includegraphics{graphics/shell_method_general_area}}}}
\end{center}

If the upper bound of the figure remains described by the function $h=f(r)$ while the lower bound is now the function $h=g(r)$ the the height of the shell is $h=f(r)-g(r)$.  The volume for the solid generated by revolving about the $h$-axis is then
\[V=\int_a^b2\pi rh\,dr=\int_a^b2\pi r\left[f(r)-g(r)\right]dr\]

One observes that, in the example depicted, the washer method would now be impossible in a single integral. For even should we be able to solve for $r(h)$ on the curves, there is no single washer generated toward the bottom of the region.  One also notes in this example that the region is generally bounded by four curves, two are vertical lines.   In general the shell method works well for solids with straight cylindrical boundaries parallel to the axis of revolution.

In the event we are revolving about the $y$-axis, so $h=y$ and $r=x$ in the above diagram, the solid generated by revolving the region bounded by $y=f(x)$, $y=g(x)$, $x=a$ and $x=b$ is
\[V=\int_a^b2\pi x\left[f(x)-g(x)\right]dx\]

On the other hand the volume of the solid generated by revolving about the $x$-axis the region bounded by $x=f(y)$, $x=g(y)$, and the horizontal lines $y=c$ and $y=d$ is, setting $h=x$, and $r=y$, 
\[V=\int_c^d 2\pi y\left[f(y)-g(y)\right]dy\]

These formulae may all, as usual be derived following the general steps starting with the differential volume element $dV=2\pi rh\,dr$. 
\begin{example}
  Find the volume of the solid generated by revolving the region bounded by the curves $y=2x^4$ and $y=2x$ about the $x$-axis.
  
\begin{solution}
  The curves intersect when
  \begin{align*}
    2x^4=2x\implies x^4-x=0\implies x(x^3-1)=0\implies x=0\ \text{or}\ x=1\,.
  \end{align*}
  Evaluating $y$ at these values shows the intersection points are $(0,0)$ and $(2,0)$\,.  A graph of the region follows.
\begin{center}
\includegraphics[width=.25\textwidth]{graphics/integration_applications16}%
\end{center}
Here we have drawn a horizontal area element parallel to the axis of symmetry as required by the Shell Method. The volume element is
\begin{align*}
  dV=2\pi rh\,dr=2\pi y(x_2-x_1)dy=2\pi y\left(\left(\frac{1}{2}y\right)^{\frac{1}{4}}-\left(\frac{1}{2}y\right)\right)dy\,,
\end{align*}
where here we needed to solve
\begin{align*}
  y&=2x^4\implies x=f(y)=\left(\frac{1}{2}y\right)^{\frac{1}{4}}\\
  y&=2x\implies x=g(y)=\frac{1}{2}y\,,
\end{align*}
in order to write $x_1$ and $x_2$ in terms of the differential variable $y$. The shells start at $y=0$ and end at $y=2$ which become the limits of the desired integral. The volume is therefore, using this reasoning or by formula $V=\int_c^d 2\pi y\left[f(y)-g(y)\right]dy$,
\begin{align*}
V&=\int_{0}^{2} 2\pi y\left(\left(\frac{1}{2}y\right)^{\frac{1}{4}}-\left(\frac{1}{2}y\right)\right)dy
=2\pi \int_{0}^{2} \left(\left(\frac{1}{2}\right)^{\frac{1}{4}}y^{\frac{5}{4}}-\left(\frac{1}{2}\right)y^2\right)dy\\
&=2\pi\left[\frac{1}{(2)^{\frac{1}{4}}}\left(\frac{4}{9}\right)y^{\frac{9}{4}}-\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)y^3\right|_{0}^{2}
=2\pi \left(\left(\frac{4}{9}\right)\frac{(2)^{\frac{9}{4}}}{(2)^{\frac{1}{4}}}-\left(\frac{1}{6}\right)(2)^3-[0-0]\right)\\
&=\frac{32}{9}\pi-\frac{8}{3}\pi=\frac{8\pi}{9} \,\, \text{cubic units}.
\end{align*}
\end{solution}

\begin{furtherquestion}
  Find the volume of the solid generated by revolving the region bounded by the curves $x^2=4y$ and $y=x$ about the $x$-axis.
\end{furtherquestion}
\end{example}

If one wishes to use the shell method to evaluate a solid generated by revolution of a region about a non-coordinate axis the differential volume element $dV=2\pi rh\,dh$ and the general steps must be used.

\begin{example}
  Find the volume of the solid generated by revolving the region bounded by the curves $y=4-2x-x^2$ and $y=x$ about the line $x=10$\,.
  
\begin{solution}
The curves intersect when
\begin{align*}
4-2x-x^2=x\implies 0=x^2+3x-4\implies 0=(x-1)(x+4)\implies x=1\ \text{or}\ x=-4\,, 
\end{align*}
and so at the points $(1,1)$ and $(-4,-4)$\,. A graph of the region is the following.
\begin{center}
\includegraphics[width=.5\textwidth]{graphics/integration_applications17}%
\end{center}
Since we are using the Shell Method a vertical area element parallel to the axis of symmetry has been drawn with endpoints $(x,y_1)$ and $(x,y_2)$ as shown.  No formula can be used in this case as we are not rotating about the $x$ or $y$-axis.  From the diagram we see the radius is, subtracting the left $x$ value from the right one, $r=10-x$\,. The differential $dr=dx$ and the differential volume is
\begin{align*}
  dV=2\pi rh\,dr=2\pi (10-x)(y_2-y_1)dx=2\pi(10-x)\left(\left(4-2x-x^2\right)-x\right)dx\,,
\end{align*}
where we used the curves on which our area endpoints lie to replace $y_1$ and $y_2$ in terms of the differential variable $x$. The limits, also in terms of $x$ are $x=-4$ to $x=1$.  The volume of the solid is
\begin{align*}
V&=\int_{-4}^{1} 2\pi (10-x)\left(4-2x-x^2-x\right)dx
=\int_{-4}^{1} 2\pi (10-x)(4-3x-x^2)dx \\
&=2\pi \int^{1}_{-4} (40-34x-7x^2+x^3)dx 
=2\pi \left[40x-\frac{34}{2}x^2-\frac{7}{3}x^3 + \frac{1}{4}x^4 \right|_{-4}^{1}\\
&  =2\pi \left(40-17-\frac{7}{3} + \frac{1}{4}-
\left[-160-272+\frac{448}{3}+ 64 \right]\right)\\
  &=2\pi \left(391-\frac{455}{3}+\frac{1}{4}\right)
 =\pi \left(782-\frac{910}{3}+\frac{1}{2}\right)
=\frac{2875 \pi}{6} \,\, \text{cubic units.}
\end{align*}
The reader may consider finding this volume using horizontal area elements and the Washer Method.  In that case the volume would need to be broken into two integrals as the right-most curve changes at $y=1$.  Additionally one would need to solve the quadratic $y=4-2x-x^2$ for $x$ to get the two functions of $y$ needed.
\end{solution}

\begin{furtherquestion}
  Find the volume of the solid generated by revolving the region bounded by the curves $y=x^2$ and $x+y=2$ about the line $x=3$.
\end{furtherquestion}
\end{example}

\begin{exercise}
\input{INTEGRATION_APPLICATIONS/the-shell-method.tex}
\end{exercise}

\subsection{Determining the Volume Method} \index{disk method} \index{washer method} \index{shell method} \index{volume! determining method for solids of revolution}
We have considered several methods for calculating volumes of solids of revolution.  Which to use in a particular instance can be determined by the following set of steps:
\begin{enumerate}
\item Draw the region to be revolved and identify the axis of revolution.
\item Next imagine a line \emph{perpendicular} to the axis of revolution.  Where this intersects the region represents (the cross-section of) a particular volume element.  Slide the line through the region keeping it perpendicular to the axis of revolution.  If all such lines intersect the boundary of the region two or fewer times then the \emph{Disk/Washer Method} will potentially work.  An exception to the intersection rule is when you first enter and leave the region; here one can intersect the boundary in a straight line.

As you pass the line through the region consider the endpoints of the intersection with the region.  If these do not always lie on the same curves then the region will need to be partitioned into two (or more) regions each requiring a separate integral to use this method.
\item Repeat the previous step but now with lines \emph{parallel} to the axis of revolution.  If the region passes this test then it is a candidate for the \emph{Shell Method}.
\item For any region that passes the \emph{Disk/Washer} test, if one of the boundary curves of the region is the axis of revolution then use the \emph{Disk Method} for that interval, otherwise the \emph{Washer Method} will be used.
\item To distinguish multiple candidate methods, next look at the curve equations.  Identify the variable of integration ($dx$ or $dy$).  It will be the variable of the coordinate axis \emph{parallel} to the axis of revolution in the case of the \emph{Disk/Washer Method} and \emph{perpendicular} to the axis of revolution for the \emph{Shell Method}.  The curves in the boundary region (aside from any straight lines at the beginning or end of the region for the method in question)  will have to have the non-differential variable written as a function of the differential variable.  If these curves cannot be solved for the non-differential variable, reject the method.
\item If several candidate methods still remain, set up the volume integral for one candidate.  If it is not easily solved, try the other candidate.
\item If no candidates lead to a solution, then the solid cannot be calculated with a single volume method.  Try partitioning the region into several areas and calculate the volume for each region revolved separately.  If these can be calculated the volume of the entire solid is their sum.
\end{enumerate}  

\textbook{\clearpage}
\begin{example}
For each of the following regions identify the simplest volume method(s) to use to calculate the solid of revolution found by rotating the region about the indicated axis.  For the method state whether to use vertical or horizontal differential volume elements. 
\begin{center} 
\parbox[t]{.95\textwidth}{\vspace{0em}{\resizebox{.95\textwidth}{!}{\includegraphics{graphics/volume_methods_example}}}}
\end{center}
\begin{solution*}
  \begin{enumerate}
  \item Passing a line perpendicular to the axis of the rotation (so horizontal) through the region shows the right and left endpoints remain on the same curves throughout the entire region.  This suggests using the Disk Method or Washer Method.  The Disk Method is sufficient because the region is flush with the axis of rotation. The volume element will be horizontal.  If we try passing a line parallel to the axis of rotation (so vertical) we see that the Shell Method could also be used with vertical volume elements.  This, however, would require solving the circular arc into an upper and lower boundary function for the shells.
  \item Passing a line perpendicular to the axis of rotation (so vertical) beginning at the left of the region shows it intersects the region in two distinct intervals.  Thus the region would need to be broken into several pieces to use the Washer Method.  Passing a line parallel to the axis of rotation (so horizontal) from top to bottom has the same curve on the left and right so the Shell Method with horizontal volume elements should be used instead.  Note that the horizontal lines at the top and bottom of the region also suggest that using a horizontal volume element is desirable here.
    \item Passing a line perpendicular to the axis of rotation (so horizontal) from top to bottom of the region shows the left and right endpoints of the intersection lie on the same curve over the entire region.  Since the region is not flush with the axis of rotation the Washer Method with horizontal volume elements could be used.  Passing a line parallel to the axis of rotation (so vertical) from left to right of the region shows the top and bottom endpoints of the intersection also lie on the same curves over the entire region.  Thus the Shell Method with vertical volume elements could also be used.  Which of the Washer or Shell Methods to use would likely be determined by whether the curves were written as functions of $x$ (use Shell Method) or $y$ (use Washer Method).  If the given functions were easily inverted both volume integrals could be written down and the one easier to integrate chosen.
  \end{enumerate}
\end{solution*}

\textbook{\clearpage}
\begin{furtherquestions}
For each of the following regions identify the simplest volume method(s) to use to calculate the solid of revolution found by rotating the region about the indicated axis.  For the method state whether to use vertical or horizontal differential volume elements. 
\begin{center} 
\parbox[t]{.95\textwidth}{\vspace{0em}{\resizebox{.95\textwidth}{!}{\includegraphics{graphics/volume_methods_exercise}}}}
\end{center}
\end{furtherquestions}
\end{example}

\begin{exercise}
\input{INTEGRATION_APPLICATIONS/determining-the-volume-method.tex}
\end{exercise}

\section{Arc Length}
From basic geometry the perimeter $P$ \index{perimeter} of a polygon or the circumference $C$ \index{circumference} of a circle are the distance around the given object.  They are the length one would measure if the boundary of the object were cut, straightened out, and measured with a ruler.
\begin{center}
\parbox[t]{4.6in}{\vspace{0em}{\resizebox{4.6in}{!}{\includegraphics{graphics/basic_arc_length}}}}
\end{center}

Intuitively the \textbf{arc length} \index{arc length} \index{arc length! of a curve} $\bm{s}$ of a curve between points $P$ and $Q$ is similarly the ``length of a piece of string'' that follows the curve joining those two points.
\begin{center}
\includegraphics{graphics/integration_applications3}%
\end{center}

Consider a curve defined by a function $y=f(x)$ with a continuous first derivative $f'(x)$ defined on interval $[a,b]$.  Such a function is called a \textbf{smooth function}\index{smooth function} and its graph $y=f(x)$ is said to be \textbf{smooth}.  An approximation to the arc length can be found by breaking the interval into $n$ subintervals of equal length $\Delta x=\frac{b-a}{n}$ with endpoints
\[a=x_0, x_1, x_2, \ldots, x_{n-1}, x_n=b\]
The points $P_k=\left(x_k,f(x_k)\right)$, $k=0,1,\ldots,n$, lie along the curve.  The length of the chord (straight line segment) joining $P_{k-1}$ to its neighbour $P_k$ will approximate the arc length between those two points.  Call that chord length $\Delta s_k$. It equals, by the Pythagorean Theorem, $\Delta s_k=\sqrt{(\Delta x)^2+(\Delta y_k)^2}$ where we define $\Delta y_k=y_k-y_{k-1}=f(x_k)-f(x_{k-1})$. 
\begin{center}
\includegraphics{graphics/integration_applications4}%
\end{center}
 By the Mean Value Theorem applied to the interval $[x_{k-1},x_k]$ there exists an $x_k^*$ in  $(x_{k-1},x_k)$ satisfying
\[\frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}}=f'(x_k^*)\]
Equivalently $\Delta y_k=f'(x_k^*)\Delta x$.  From this it follows
that we can write $\Delta s_k$ as:
\begin{eqnarray*}
\Delta s_k & = & \sqrt{(\Delta x)^2+(\Delta y_k)^2} \\
& = & \sqrt{(\Delta x)^2+\left[f'(x_k^*)\Delta x\right]^2} \\
& = & \sqrt{(\Delta x)^2+\left[f'(x_k^*)\right]^2(\Delta x)^2} \\
& = & \sqrt{\left(1+\left[f'(x_k^*)\right]^2\right)\cdot(\Delta x)^2} \\
& = & \sqrt{1+\left[f'(x_k^*)\right]^2}\cdot\sqrt{(\Delta x)^2} \\
& = & \sqrt{1+\left[f'(x_k^*)\right]^2}\Delta x
\end{eqnarray*}
An approximation to the arc length between the points $P(a,f(a))$ and $Q(b,f(b))$ is therefore the length of the polygonal curve we have constructed, namely the Riemann sum
\[\sum_{k=1}^n\Delta s_k=\sum_{k=1}^n\sqrt{1+\left[f'(x_k^*)\right]^2}\Delta x\]
Taking the limit as $n\rightarrow\infty$ (so $\Delta x\rightarrow 0$) we have the following:
\definition{If function $f$ has a continuous derivative $f'$ over the $x$-interval $[a,b]$ then the length of the graph over the interval (the \textbf{arc length}) is
\[s=\int_a^b\sqrt{1+[f'(x)]^2}\,dx\ \ \mbox{or}\ \ s=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\]
}

Note that if a curve is described by the equation $x=g(y)$ with continuous derivative $g'(y)$ over the $y$-interval $[c,d]$ then the arc length between the endpoints equals
\[s=\int_c^d\sqrt{1+[g'(y)]^2}\,dy\ \ \mbox{or}\ \ s=\int_c^d\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy\]

\begin{example}
  Find the  length of the following curves over the given intervals.
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle y=5x^{\frac{3}{2}} +2$;\ $0\le x\le 12$
    \item $\displaystyle x=\frac{2}{3}y^{\frac{3}{2}}$;\  $1\le y \le 9$
    \end{enumerate}
  \end{multicols}
\begin{solution*}
\begin{enumerate}
\item $\displaystyle y=5x^{\frac{3}{2}} +2$;\ $0\le x\le 12$

  The curve is between the points $(0,0)$ and $(0,5(12)^{\frac{3}{2}})\approx(12,207.8)$.
  \begin{center}
    \includegraphics[width=.35\textwidth]{graphics/integration_applications22}%
  \end{center}
The derivative is given by $\displaystyle \frac{dy}{dx}=\frac{15}{2}x^{\frac{1}{2}}$ and is continuous on $[0,12]$\,. The arc length of the curve over that interval is
  \begin{align*}
    s=\int_{0}^{12}\sqrt{1+ \left(\frac{dy}{dx}\right)^2}\, dx    
    &=\int_{0}^{12}\sqrt{1+ \left(\frac{15}{2}x^{\frac{1}{2}}\right)^2} \,dx 
      =\int_{0}^{12} \sqrt{1+ \frac{225}{4}x} \ dx\,.
  \end{align*}
  Substitute $u=1+\frac{225}{4}x$, so $du=\frac{225}{4}dx\implies dx=\frac{4}{225}du$.  The limits become
  \begin{align*}
    x&=12\implies u=676 \\
    x&=0\implies u=1,
  \end{align*}
  and so
    \begin{align*}
      s &=\frac{4}{225}\int_1^{676}u^{\frac{1}{2}}du=\frac{4}{225}\left[\frac{2}{3}u^{\frac{3}{2}}\right|_1^{676}
          =\frac{8}{675}\left[(676)^{\frac{3}{2}}-(1)^{\frac{3}{2}}\right]\\
        &=\frac{8(17575)}{675}=\frac{8(703)}{27}=\frac{5624}{27}\approx{208.296}\,\, \text{units.}  
    \end{align*}
    As a check note that the straight line distance between the endpoints is
    \begin{align*}
      \sqrt{(12)^2+\left(5(12)^{\frac{3}{2}}\right)^2}=\sqrt{43344}\approx 208.192\, \, \text{units,}
    \end{align*}
    which is just below the distance found, showing our curve is almost straight between the two points.
  \item $\displaystyle x=\frac{2}{3}y^{\frac{3}{2}}$;\ $1\le y \le 9$

    Since $x$ is written as a function of $y$ and our limits are in terms of $y$ we use the arc length formula with differential $dy$. The derivative is $\displaystyle \frac{dx}{dy}= y^{\frac{1}{2}}$ and is continuous on $[1,9]$. The arc length is
\begin{align*}
 s= \int_{1}^{9} \sqrt{1 + \left(\frac{dx}{dy}\right)^2}dy  
 &=\int_{1}^{9} \sqrt{1+ \left(y^{\frac{1}{2}}\right)^2}dy
 = \int_{1}^{9} \sqrt{1 + y}\, dy 
 =\left.\frac{2}{3}(1+ y)^{\frac{3}{2}} \right|_{1}^{9} \\
 &=\frac{2}{3}\left[(1+9)^{\frac{3}{2}}- (1+1)^{\frac{3}{2}}\right]
 =\frac{2}{3}\left[(10)^{\frac{3}{2}}-(2)^{\frac{3}{2}}\right]
 \approx 19.2 \,\, \text{units.}
\end{align*}
\end{enumerate}
Note that the integrand of the arc length formula typically results in integrals which are not integrable.  In that case numerical integration is used to find an approximate result.
\end{solution*}

\begin{furtherquestions}
  Find the lengths of the following curves over the given intervals.
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle f(x)=3x^{\frac{2}{3}}-10$;\ $8\le x\le 27$
    \item $\displaystyle y=\frac{x^4}{16}+\frac{1}{2x^2}$;\ $1\le x\le 2$
    \end{enumerate}
    \end{multicols}
\end{furtherquestions}
\end{example}

If we have a smooth function $f(x)$ on interval $[a,b]$ we can consider the arc length from the starting point $(a,f(a))$ up to the point $(x,f(x))$ for some value $x$ in $[a,b]$. This length is clearly dependent on the choice of $x$ and hence is a function $s(x)$.

\definition{Given a smooth curve described by the function $y=f(x)$ over the $x$-interval $[a,b]$, the arc length between the initial point $(a,f(a))$ and the point $(x,f(x))$ for $x$ in $[a,b]$ is given by the \textbf{arc length function}  \index{arc length! function of}
\[s(x)=\int_a^x\sqrt{1+[f'(t)]^2}\,dt\,.\]
}

Notes:
\begin{enumerate}
\item We changed the variable of integration to $t$ to distinguish it from the limit of integration~$x$.
\item The derivative of the arc length function $s(x)$ is, by the Fundamental Theorem of Calculus, $\frac{ds}{dx}=\sqrt{1+[f'(x)]^2}$~.  This in turn implies, since $ds=\frac{ds}{dx}\,dx$, that the \textbf{differential of the arc length} \index{arc length! differential of} is
  \[ds=\sqrt{1+[f'(x)]^2}dx\,,\]
  in agreement with our previous formula for arc length.
 Noting that $dy=f'(x)dx$ the differential of arc length is sometimes written
\[ds=\sqrt{dx^2+dy^2}~,\]
reflecting its Pythagorean origins.  The latter differential formula, along with the relation $dy=f'(x)dx$ for $x$-intervals $[a,b]$ or $dx=g'(y)dy$ for $y$-intervals $[c,d]$, and the idea of the integral as a sum ($s=\int ds$) make it easy to remember/derive the arc length formulae.
\end{enumerate}

\textbook{\clearpage}
\begin{example}\label{exam:arclength1}%
Find the arc length function $s(x)$ of 
\begin{enumerate}
    \item $\displaystyle y=\ln (\cos x)$ with starting point $P(0,0)$.
    \item $\displaystyle y=1-x^2$ with starting point $P(0,1)$.
\end{enumerate}
\begin{solution*}
\begin{enumerate}
\item $\displaystyle y=\ln (\cos x)$ with starting point $P(0,0)$.

  Since the argument of the logarithm must be positive we require $\cos x>0$\,.  Also the function must be continuous over the interval containing our starting point which has $x=0$ so our arc length function is restricted to $x$ in the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. A graph of the curve $y=f(x)=\ln(\cos x)$ with the arc length $s(x)$ from $(0,0)$ to $(x,f(x))$ is shown.
  \begin{center}
    \includegraphics[width=.4\textwidth]{graphics/integration_applications21}%
  \end{center}
  The needed derivative is given by 
  \begin{align*}
    f'(x)=\frac{1}{\cos x}(-\sin x)= -\frac{\sin x}{\cos x}= -\tan x\,.
  \end{align*}
  Therefore the arc length function is 
  \begin{align*}
    s(x)&= \int_{0}^{x} \sqrt{1 + (-\tan t)^2}\, dt
          =\int_{0}^{x} \sqrt{1 + \tan^{2}\!t} \, dt
          =\int_{0}^{x} \sqrt{ \sec^{2}\!t} \, dt
          =\int_{0}^{x} \sec t \, dt \\
        &=\left.\rule{0em}{1em}\ln|\sec t + \tan t|\right|^{x}_{0}
          =\ln|\sec x + \tan x|-\ln|\sec(0)+\tan(0)| \\
        & =\ln|\sec x+\tan x|-\ln|1+0|\\
        &=\ln|\sec x+\tan x|\,.
  \end{align*}
  The restriction of $x$ and therefore $t$ to $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ was used when simplifying $\sqrt{\sec^2t}=|\sec t|=\sec t$ in the integration since secant is positive over that interval.

  As an aside, note that our arc length formula is valid for $x$ in $[a,b)=[0,\frac{\pi}{2})$. If we consider a point to the left of our starting point with $x<0$ then the function $s(x)$ will actually be the negative of the arc length between the two points because then the differential $dt$ is effectively negative due to the upper limit being less than the lower limit.  From the symmetry of the original curve ($y=\ln(\cos x)$ is an even function) this means that $s(x)=\ln|\sec x+\tan x|$ is actually an odd function of $x$. This can be demonstrated by plotting the function.  This is somewhat surprising since secant is even and tangent is odd!  The reader is encouraged to prove the result. See Exercise~\ref{exer:arc-length} Problem~\ref{itm:arclengthodd}\,.

\item $\displaystyle y=1-x^2$ with starting point $P(0,1)$.
  
    The derivative of $y=f(x)=1-x^2$ is given by $f'(x)=-2x$\,.
    The arc length function is
    \begin{align*}
     s(x)&=\int_{0}^{x} \sqrt{1+(-2t)^2}dt
           =\int_{0}^{x} \sqrt{1+4t^2} \, dt=\int_{0}^{x} \sqrt{1+(2t)^2} \, dt\,.
    \end{align*}
    Using the trigonometric substitution $2t=\tan\theta\implies 2dt=\sec^2\!\theta\,d\theta$ gives
    \begin{align*}
      \int\sqrt{1+(2t)^2}\,dt&=\frac{1}{2}\int\sqrt{1+\tan^2\theta}\sec^2\theta\,d\theta=\frac{1}{2}\int\sqrt{\sec^2\!\theta}\sec^2\!\theta\,d\theta=\frac{1}{2}\int\sec^3\!\theta\,d\theta\,.
    \end{align*}
    Using Integration by Parts one has
    \begin{align*}
      \int\sec^3\!\theta\,d\theta&=\int\sec\theta\left(\sec^2\!\theta\,d\theta\right)=\sec\theta\tan\theta-\int\tan\theta(\sec\theta\tan\theta\,d\theta)\\
                                 &=\sec\theta\tan\theta-\int\sec\theta\tan^2\!\theta\,d\theta=\sec\theta\tan\theta-\int\sec\theta\tan^2\theta\,d\theta\\
                                 &=\sec\theta\tan\theta-\int\sec\theta\left(\sec^2\!\theta-1\right)d\theta=\sec\theta\tan\theta-\int\sec^3\!\theta\,d\theta+\int\sec\theta\,d\theta\\
      \implies \int\sec^3\!\theta\,d\theta &= \frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln|\sec\theta+\tan\theta|+C\,.
    \end{align*}
    \parbox{.7\textwidth}{To return to variable $t$ we draw a triangle with $\displaystyle\tan\theta=\frac{2t}{1}$ and hence hypotenuse $\sqrt{1+(2t)^2}=\sqrt{1+4t^2}$. The indefinite integral is then}\hfill\raisebox{-3em}{\includegraphics[width=.17\textwidth]{graphics/triangle6}}
    \begin{align*}
      \int\sqrt{1+(2t)^2}\,dt&=\frac{1}{4}\left(\frac{\sqrt{1+4t^2}}{1}\right)\left(\frac{2t}{1}\right)+\frac{1}{4}\ln\left|\frac{\sqrt{1+4t^2}}{1}+\frac{2t}{1}\right|+C\\
                             &=\frac{1}{2}t\sqrt{1+4t^2}+\frac{1}{4}\ln\left|2t+\sqrt{1+4t^2}\right|+C\,,
    \end{align*}
    and so the arc length function is
    \begin{align*}
     s(x) &=\left[\frac{1}{2}t\sqrt{1+4t^2} + \frac{1}{4}\ln (2t+ \sqrt{1+ 4t^2}) \right|_{0}^{x} \\
          &=\frac{1}{2}x\sqrt{1+4x^2} + \frac{1}{4}\ln\left(2x+ \sqrt{1+4x^2}\,\right)-\left[0+\frac{1}{4}\ln(1)\right]\\
          &=\frac{1}{2}x\sqrt{1+4x^2} + \frac{1}{4}\ln\left(2x+ \sqrt{1+4x^2}\,\right).
     \end{align*}
   \end{enumerate}
\end{solution*}

\begin{furtherquestions}
Find the arc length function $s(x)$ of
\begin{enumerate}
%The next one was on the assignment
%\item $\displaystyle y=2x^{\frac{3}{2}}$ with starting point $P(1,2)$.
\item $\displaystyle y=\sqrt{x-x^2}-\cos^{-1}\left(\sqrt{x}\right)$ with starting point $P(0,-\pi/2)$.
\item $\displaystyle y=e^x$ with starting point $P(0,1)$.
% Note $y=\ln x$ is also doable and ends up using a trigonometric substitution.
\end{enumerate}
\end{furtherquestions}
\end{example}

\textbook{\clearpage}
\begin{exercise}\label{exer:arc-length}
\input{INTEGRATION_APPLICATIONS/arc-length.tex}
\end{exercise}

\section{Areas of Surfaces of Revolution}

Just as volumes of revolution are constructed by revolving an area about a line, a \textbf{surface of revolution} \index{surface of revolution} is generated by revolving the graph of a continuous function about a line.
\begin{center}
\parbox[t]{.8\textwidth}{\vspace{0em}{\resizebox{.8\textwidth}{!}{\includegraphics{graphics/surfaces_of_revolution}}}}
\end{center}
Note that the surface generated in this manner does not produce any end surfaces such as the circular bottom of the cone or ends of the frustum of the cone.  (The spherical surface however is fully generated.)

We turn ourselves to the problem of calculating the area of such a surface of revolution.  Consider a simple object like the frustum of the cone above, with end radii of $r_1$ and $r_2$ and \textbf{slant height}\index{slant height} $s$.
\begin{center} 
\parbox[t]{.4\textwidth}{\vspace{0em}{\resizebox{.4\textwidth}{!}{\includegraphics{graphics/frustum}}}}
\end{center}
The \textbf{lateral} surface area of the frustum (i.e. the area of the side but not including the circular ends), can be shown to be 
\[S=2\pi rs\ ,\]
where $r=\frac{1}{2}\left(r_1+r_2\right)$ is the average of the two radii.

The approximate area of a surface of revolution may be found by slicing it perpendicular to its axis of revolution (which we will call $h$).  The intersection of these planes with the surface form the circular ends of conical frusta the net lateral surface area of which will approximate the area of the surface of revolution.

\begin{center}
\parbox[t]{.65\textwidth}{\vspace{0em}{\resizebox{.65\textwidth}{!}{\includegraphics{graphics/surface_method}}}}
\end{center}

A Riemann sum for the area is thereby formed:
\[S\approx\sum_{k=1}^n\Delta S_k~,\]
where $\Delta S_k$ is the lateral surface area of the $k^{\mbox{\tiny th}}$ conical frustum. Upon taking the limit of an infinite number ($n\rightarrow\infty$) of conical frusta one arrives at an exact expression for the surface area $S$ in terms of an integral $S=\int \!dS$ where the differential area of the conical frustum is given, analagously to the finite surface area:
\[dS=2\pi r\,ds\]
where here the finite slant height $s$ is replaced by the \textbf{differential arc length} \index{arc length! differential of} $ds=\sqrt{dx^2+dy^2}$ of the curve generating the surface.  Four different formulae are possible depending on whether the axis of symmetry $h$ is the $x$ or the $y$-axis and whether the curve-generating function is written as $y=f(x)$ over the $x$-interval $[a,b]$ or $x=g(y)$ over the $y$-interval $[c,d]$.  In all cases we assume the function is positive and has continuous first derivative (i.e. is smooth).

Considering first the case where the axis of symmetry $h$ is the $x$-axis and the curve revolved is described by the function $y=f(x)$, $a\le x\le b$, then our radius is $r=y=f(x)$ and we write the differential arc length in terms of the independent variable $x$ as $ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$ so that $S=\int dS=\int 2\pi r\,ds$ is given by the formula:
\[S=\int_a^b2\pi f(x)\sqrt{1+[f'(x)]^2}\,dx\ \ \ \mbox{or}\ \ \ S=\int_a^b2\pi f(x)\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\]
If we still consider the case where we are rotating about the $x$-axis but the curve is instead described by the function $x=g(y)$, $c\le y\le d$ then the differential arc length must be written in terms of the independent variable $y$ as $ds=\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy$. The radius $r=y$ still but as that is already expressed in terms of the variable of the differential ($dy$) variable our formula becomes (still using $S=\int dS=\int 2\pi r\,ds$):
\[S=\int_c^d2\pi y\sqrt{1+[g'(y)]^2}\,dy\ \ \ \mbox{or}\ \ \ S=\int_c^d2\pi y\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy\]

Consider next the cases where the surface is generated by revolving the curve around the $y$-axis.  In this case the radius $r=x$.  If the curve is described by the function $y=f(x)$, $a\le x\le b$ then the independent variable is $x$ so we use the differential $ds$ written in terms of $dx$ to get, since   $S=\int dS=\int 2\pi r\,ds$,
\[S=\int_a^b2\pi x\sqrt{1+[f'(x)]^2}\,dx\ \ \ \mbox{or}\ \ \ S=\int_a^b2\pi x\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\]
If instead the curve is written $x=g(y)$, $c\le y\le d$, then we must write the radius $x$ in terms of the differential ($dy$) as $g(y)$ and also select the appropriate form of the differential arc length in terms of $dy$:
\[S=\int_c^d2\pi g(y)\sqrt{1+[g'(y)]^2}\,dy\ \ \ \mbox{or}\ \ \ S=\int_c^d2\pi g(y)\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy\]

None of these formulae need to be memorized.  Just remember that we are integrating ($S=\int dS)$ the differential surface area $dS=2\pi r\,ds$.  Then look at the curve's function; the independent variable determines the form of $ds$ and the limits on the integral.  The radius $r$ is the variable perpendicular to the axis of revolution.  If that is not the independent variable then use the function for the curve to express it in terms of the independent (differential) variable.

\begin{example}
Find the area of the following surfaces of revolution generated by the given curves.
\begin{enumerate}
    \item $\displaystyle y=e^x$ from $x=1$ to $x=3$ revolved about the $x$-axis. 
    \item $\displaystyle y=\sqrt[3]{3x}$ from $x=9$  to $x=72$ revolved about the $y$-axis.
\end{enumerate}
\begin{solution*}
\begin{enumerate}
\item $\displaystyle y=e^x$ from $x=1$ to $x=3$ revolved about the $x$-axis.

A graph of the situation follows.
\begin{center}
\includegraphics[width=.45\textwidth]{graphics/integration_applications18}%
\end{center}
Since we have $y$ written as a function of $x$ it is easiest to write our differential area element in terms of $dx$. From the diagram
\begin{align*}
  dS=2\pi r\,ds=2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx=2\pi e^x\sqrt{1+(e^x)^2}\,dx
\end{align*}
where here the radius $r=y$ and the differential $ds$ were were written in terms of the differential variable $x$.  The limits in that variable are $x=1$ and $x=10$ so the total area is therefore, using this derivation or by formula $S=\int_a^b2\pi f(x)\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$,
\begin{align*}
  S&=\int_{1}^{3}2 \pi e^x \sqrt{1+(e^x)^2} \, dx
\end{align*}
Using substitution $u=e^x\implies du=e^x\,dx$ the indefinite integral is
\begin{align*}
  2\pi \int e^x \sqrt{1+ (e^{x})^2}\, dx&=2\pi\int\sqrt{1+u^2}\,du\,.
\end{align*}
Using the trigonometric substitution $u=\tan\theta\implies du=\sec^2\!\theta\,d\theta$ this equals
\begin{align*}
  2\pi\int\sqrt{1+\tan^2\theta}\,d\theta&=2\pi\int\sqrt{\sec^2\theta}\sec^2\!\theta\,d\theta=2\pi\int\sec^3\!\theta\,d\theta
  \\
  &= \pi\sec\theta\tan\theta+\pi\ln|\sec\theta+\tan\theta|+C\,,
\end{align*}
\parbox{.7\textwidth}{where we used the integral for $\sec^3\!\theta$ derived in Example~\ref{exam:arclength1}\,. Draw a triangle with $\displaystyle\tan\theta=\frac{u}{1}$ and thus hypotenuse $\sqrt{1+u^2}$. The indefinite integral is then}\hfill\raisebox{-3em}{\includegraphics[width=.17\textwidth]{graphics/triangle7}}

\begin{align*}
  2\pi \int e^x \sqrt{1+ e^{2x}}\, dx&= \pi\left(\frac{\sqrt{1+u^2}}{1}\right)\left(\frac{u}{1}\right)+\pi\ln\left|\frac{\sqrt{1+u^2}}{1}+\frac{u}{1}\right|+C\\
                                     &= \pi u\sqrt{1+u^2}+\pi\ln\left|u+\sqrt{1+u^2}\right|+C\\
                                     &=\pi e^x\sqrt{1+e^{2x}} + \pi\ln|e^x + \sqrt{1+e^{2x}}|+C\,.
\end{align*}
Therefore the surface area is
\begin{align*}
S  &=\pi\left[e^x\sqrt{1+e^{2x}} + \ln|e^x + \sqrt{1+e^{2x}}|\,\right|_{1}^{3}\\
   &=\pi \left[e^{3}\sqrt{1+e^{6}}+ \ln (e^{3}+ \sqrt{1 + e^{6}}\,)-e\sqrt{1+e^2}-\ln (e + \sqrt{1 + e^2}\,)\right] \\
   &\approx 1250.4 \,\, \text{square units}.
\end{align*}
Finally note that the area could also have been found using differential $dx$.  Since $y=f(x)=e^x$ it follows that $x=g(y)=\ln y$ and the differential area is
\begin{align*}
  dS=2\pi r\,ds=2\pi y\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy=2\pi y\sqrt{1+\left(\frac{1}{y}\right)^2}\,dy\,,
\end{align*}
which would need to be integrated from $y=f(1)=e$ to $y=f(3)=e^3$.  The reader is encouraged to solve this integral and confirm the above area.

\item $\displaystyle y=\sqrt[3]{3x}$ from $x=9$  to $x=72$ revolved about the $y$-axis.

    A graph of the situation follows.
\begin{center}
\includegraphics[width=.5\textwidth]{graphics/integration_applications19}%
\end{center}
Since $y=f(x)=\sqrt[3]{3x}=(3x)^{\frac{1}{3}}\implies \frac{dy}{dx}=\frac{1}{3}(3x)^{-\frac{2}{3}}(3)=(3x)^{-\frac{2}{3}}$ the differential area element written in terms of $dx$ is, from the diagram,
\begin{align*}
  dS=2\pi r\,ds=2\pi x\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx=2\pi x\sqrt{1+\left((3x)^{-\frac{2}{3}}\right)^2}\,dx=2\pi x\sqrt{1+\frac{1}{(3x)^{\frac{4}{3}}}}\,dx\,,
\end{align*}
which would need to be integrated from $x=8$ to $x=72$\,.  This integral looks challenging so instead consider writing $dS$ in terms of $dy$.  Solving $y=\sqrt[3]{3x}$ for $x$ gives $x=g(y)=\frac{1}{3}y^3$ with simple derivative $\frac{dy}{dx}=y^2$\,.  The differential area element in this case becomes
\begin{align*}
  dS=2\pi r\,ds=2\pi x\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy=2\pi \left(\frac{1}{3}y^3\right)\sqrt{1+(y^2)^2}\,dy\,,
\end{align*}
with the limits going between $y=f(9)=\sqrt[3]{3(9)}=3$ and $y=f(72)=\sqrt[3]{3(72)}=6$\,. The surface area is, using this differential or by formula $S=\int_c^d2\pi g(y)\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy$,
    \begin{align*}
     S &=2\pi \int_{3}^{6} \left(\frac{1}{3}y^3\right) \sqrt{1 +(y^2)^2} \,dy 
      =\frac{2\pi}{3}\int_{3}^{6} y^3\sqrt{1+ y^4} \,dy \ \ (\text{Let}\ u=1+y^4\implies du=4y^3dy)\\
     &=\frac{2\pi}{3}\left(\frac{1}{4}\right) \int_{82}^{1297} \sqrt{u} \,du   
         =\frac{\pi}{6}\left.\left[\frac{2}{3}u^{\frac{3}{2}}\right|_{82}^{1297}\right]
     =\frac{\pi}{9} \left[(1297)^{\frac{3}{2}}-(82)^{\frac{3}{2}}\right]\approx 5107.5 \,\, \text{square units}.
    \end{align*}
    While this integral was straightforward the one in terms of $dx$ can also be done and the reader is encouraged to try it.  (Hint: Write the term under the square root as a fraction and then pull the denominator $(3x)^{\frac{4}{3}}$ outside from under the root. Next simplify and use a substitution.) 
\end{enumerate}
\end{solution*}

\begin{furtherquestions}
Find the areas of the following surfaces of revolution generated by the given curves.
\begin{enumerate}
%\item $x=y^2$ from $(1,1)$ to $(4,2)$ revolved about the $x$-axis.
\item $y=x^3$ from $x=1$ to $x=2$ revolved about the $x$-axis.
\item $x=\frac{1}{8}y^3$ from $x=1$ to $x=8$ revolved about the $y$-axis.
\item $x=1+2y^2$, $1\le y\le2$ revolved about the $x$-axis.
\item $y=\sqrt[3]{x}$, $1\le x\le8$ revolved about the $y$-axis.
\item $y=e^{-x}$ from $x=0$ to $x=1$ revolved about the $x$-axis.
\end{enumerate}
\end{furtherquestions}
\end{example}

\begin{example}
  Use calculus to show that the lateral (side) surface area of a right circular cone of height $H$ and base radius $R$ is $S=\pi R\sqrt{R^2+H^2}$\,.
  
  \begin{solution}
    Represent the cone by a line of rise $R$ and run $H$ going through the origin rotated about the $x$-axis.  Then $y=f(x)=\frac{R}{H}x$ and the situation is as shown in the following diagram.
\begin{center}
\includegraphics[width=.55\textwidth]{graphics/integration_applications20}%
\end{center}
The differential area element is
\begin{align*}
 dS=2\pi r\,ds=2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx=2\pi \left(\frac{R}{H}x\right)\sqrt{1+\left(\frac{R}{H}\right)^2}\,dx\,.
\end{align*}
The limits in terms of the differential variable $x$ are $x=0$ and $x=H$.  The lateral surface area is, using this differential or by formula $S=\int_a^b2\pi f(x)\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$,
\begin{align*}
  S&=\int_0^H2\pi \left(\frac{R}{H}x\right)\sqrt{1+\left(\frac{R}{H}\right)^2}\,dx
   =\frac{2\pi R}{H}\sqrt{1+\frac{R^2}{H^2}}\int_0^Hx\,dx
   =\frac{2\pi R}{H}\sqrt{1+\frac{R^2}{H^2}}\left[\frac{1}{2}x^2\right|_0^H\\
   &=\frac{\pi R}{H}\sqrt{1+\frac{R^2}{H^2}}\left(H^2-(0)^2\right)
   =\pi RH\sqrt{1+\frac{R^2}{H^2}}
   =\pi R\sqrt{H^2\left(1+\frac{R^2}{H^2}\right)}\\
   &=\pi R\sqrt{R^2+H^2}
\end{align*}
For the total surface area of a cone one would need to add the area of the base ($\pi R^2$) to this result.

Note that our original derivation of the surface area integration formulae used the lateral area of a frustum as its starting point. Since a cone is just a frustum with upper radius of $r_1=0$, bottom radius of $r_2=R$,  and slant height $s=\sqrt{R^2+H^2}$ our formula above follows directly from the frustum lateral area formula $S=2\pi\left(\frac{r_1+r_2}{2}\right)s$\,.  The above derivation for the cone shows the surface area integral formula is self-consistent with this starting point.
\end{solution}

\begin{furtherquestion}
  Use calculus to show that the surface area of a sphere of radius $R$ is $S=4\pi R^2$.\\ (This result was originally discovered by Archimedes.)
\end{furtherquestion}
\end{example}

\textbook{\clearpage}
\begin{exercise}
\input{INTEGRATION_APPLICATIONS/areas-of-surface-of-revolution.tex}
\end{exercise}

\begin{reviewexercise}
\input{INTEGRATION_APPLICATIONS/integration-applications-review.tex}
\end{reviewexercise}

%As a first problem in a problem set we could say we wanted to make a cylinder as a measuring device.  If the radius at the bottom is such and such how high from the bottom should the 1 litre mark be placed.

% Another problem could be asking the question that while we used cylinders the volume wouldn't the slices be better estimated by the frustum of a cone.  Give the volume of that and get them to argue that in the limit of dh going to zero the two are the same.

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