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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Volume}

If you take a reasonable region in space, and slice it by planes perpendicular
to an axis (we'll think about the \( x \)- axis) with a machete, you might
be lucky and get something whose area is easy to describe. Now, if you think
about cutting the region with a chain saw, instead of a machete, so that you
actually cut out some bit \( dV \) of its volume, how much do you cut out?
It's the area \( A_{x} \) of the slice at the point \( x \) along the axis,
times the thickness \( dx \) of the saw. So, if you were to saw and saw until
there was no region left, you would have cut up 
\[
\sum _{i=1}^{n}A_{x_{i}}\, dx\]


volume of stuff. Well, approximately. Of course, you have to think about the
slope on the edges and all, but not for long. In the limit, as the thickness
of the saw blade goes to 0, the sum approaches a limit of 
\[
\int _{a}^{b}A_{x}dx,\]


where \( x=a \) is the first slice you need, and \( x=b \) is the last, to
cut through the whole region. 

Try to find integral formulas to find the volume of:  

\begin{enumerate}
\item A sphere of radius \( a \). \\
\textit{Solution:} If the sphere has center at the origin, and you slice the
sphere by planes perpendicular to the \( x \)-axis, the slice will be a circle,
but not one of radius \( a \), the radius will be \( \sqrt{a^{2}-x^{2}} \)
if the slice crosses the \( x \)-axis at \( x \). So, the area of the slice
will be \( A_{x}=\pi (a^{2}-x^{2}) \), so that \( dV=\pi (a^{2}-x^{2})dx \).
The sphere would be built from slices beginning at \( x=-a \) and continuing
until \( x=a \). So, the volume would be:
\begin{eqnarray*}
V & = & \int _{-a}^{a}\pi (a^{2}-x^{2})dx\\
 & = & \pi \left. \left( a^{2}x-\frac{x^{3}}{3}\right) \right| _{-a}^{a}\\
 & = & \frac{4\pi a^{3}}{3}.
\end{eqnarray*}

\item A cone, of base radius 2 and height 3. \\
\textit{Solution:} Here the slices would again be circles. If the cone is laid
out with its axis along the \( x \)-axis, point at the origin, then the radius
of the circle cut out by a plane cutting through perpendicularly, meeting the
axis at \( x \) would be \( 2x/3 \), because the ratio of radius to height
(distance from the apex to the slice) is \( 2/3 \) for the full height, and
would remain that ratio for the small cone that remains after being sliced.
Since the ``height'' is \( x \), the radius would be \( \frac{2}{3}x \).
Thus, the volume is 
\begin{eqnarray*}
V & = & \int _{0}^{3}\pi \left( \frac{2}{3}x\right) ^{2}dx\\
 & = & \left. \frac{4\pi }{27}x^{3}\right| _{0}^{3}\\
 & = & 4\pi .
\end{eqnarray*}

\item A pyramid, with a square base of \( 200\, \mathrm{cubits}\times 200\, \mathrm{cubits} \)
and a height of \( 150\, \mathrm{cubits} \).\\
Solution: Slice this one horizontally, a distance \( x \) cubits down from
the top. Again the ratios of corresponding sides is preserved, since the little
pyramid at the top is similar to the whole pyramid, so the ratio \( height\div base=150\div 200=3/4 \).
That means that the square base cut through at height \( x \) cubits down from
the top has side \( 4x/3 \), and the volume would be 
\begin{eqnarray*}
V & = & \int _{0}^{150}\left( \frac{4x}{3}\right) ^{2}dx\\
 & = & \left. \frac{16}{27}x^{3}\right| _{0}^{150}\\
 & = & 2,000,000\, cubic\, cubits.
\end{eqnarray*}

\end{enumerate}

\subsubsection{Volumes of revolution; disks and washers}

Many examples of this techniques compute volumes of regions which are rotationally
symmetric about some axis. These are called ``volumes of revolution'' for obvious
reasons. A volume of revolution can be described by ``sweeping out'' the region
by revolving a plane area about the axis of symmetry - so a sphere is swept
out by spinning a disk about a diameter, or a football by spinning an ellipse
about the major axis. If you describe the region that way, you can use that
planar description to find the areas of the slices \( A_{x} \). In this case,
 \textit{if} the region being spun about the \( x \)-axis is the region between
the curve \( y=f(x) \) and the \( x \)-axis, for \( x\in [a,b] \), then \( A_{x}=\pi r^{2}=\pi (f(x))^{2} \).
 

If the axis of rotation is \textit{not} the \( x \)-axis, then the radius function
has to be thought about. For example, find the volume of the region generated
when the semi-circle 
\[
y=\sqrt{1-x^{2}},\]
 for \( x\in [-1,1] \), is spun about the axis \( y=-1 \). Also, you can find
the volumes of more complicated regions the same way. If a region in the plane
\textit{between} two curves is revolved around an axis, then the slice is not
a disk, but an annulus, or ``washer''. The area is simply the difference between
the outer area and the inner, \( \pi R^{2}-\pi r^{2} \), so the volume really
should be thought of as: 
\[
Volume=\int _{a}^{b}(\pi R^{2}-\pi r^{2})dx,\]


where we assume that the axis of rotation is the \( x \)-axis.  

Here are a couple of examples: 

\noindent \textbf{1.} Find the volume of the region obtained when the area bounded
between the curves \( y=x \) and \( y=x^{2} \) is revolved about the \( x \)-axis.
Then, what is the volume when the same region is revolved about the \( y \)-axis
(switch variables in the formula). 

\noindent \textbf{2.} Find the volumes of the football-shaped region swept out
when the ellipse 
\[
\frac{x^{2}}{36}+\frac{y^{2}}{9}=1\]
 is revolved around the \( x \)-axis. 


\subsubsection{Shells}

The point of these volume problems is supposed to get you to see how integration
applications take shape -- you decompose the problem up into a bunch of little,
easy problems. Then you add them back together (integrate), and the combination
of the solutions to the easy problems solves the hard one. 

The shell method is just another example of that. You still compute volumes,
but you do it by decomposing the region into \textit{cylinders} rather than
slices. The direction of the \textbf{thickness} of the cylinder tells you the
variable to use. The area of the cylinders, (instead of the slices), is: \( A_{x}=2\pi rh \),
where \( r \) is the radius of the cylinder, and \( h \) is the height. The
volume, by this method, is: 
\[
\int _{a}^{b}2\pi rhdx,\]


if \( x \) gives the direction of the thickness of the cylinder. You have to
figure out what \( r \) and \( h \) are for each problem. 

You need to decompose the region into just enough cylinders to fill out the
region \textit{once}. While that's pretty obvious with slices, but not so with
shells -- because each one goes all the way around the axis. You don't have
to go from one edge to the opposite one, but from the center out.  

\begin{example}
Find the volume of a ``cored apple'', a sphere of radius 2 minus the stuff inside
a cylinder of radius 1 that passes through the center of the sphere. 
\end{example}

\subsubsection*{Solution }

Assume that things are arranged so that the axis of the ``corer'' is the \( y \)-axis.
Then, in the \( xy \)-plane, before you rotate the picture, the direction radially
away from the axis is the \( x \) direction, so that is what you use for the
variable. The height of a representative cylinder, expressed in terms of \( x \),
is 
\[
h=2y=2\sqrt{4-x^{2}}.\]
 Why the 2? You go up and down from the \( x \) axis to get to the top and
bottom of the cylinder. The radius \( r=x \), because we rotate around the
\( x \)-axis. So, the volume becomes: 
\[
Volume=\int _{a}^{b}dV=\int _{a}^{b}2\pi rhdx=\int _{1}^{2}2\pi 2\sqrt{4-x^{2}}x\, dx.\]


The limits of integration come from the fact that the \textit{innermost} slice
you need is at \( r=1 \), and the \textit{outermost} slice you need is at \( r=2 \).
I'll leave the final answer - or, rather, figuring it out, to you. The answer
is \( 4\, \sqrt{3}\pi  \), as my machine told me. 

\noindent Here are some more examples:  

\begin{example}
Find the volume obtained when the region bounded by the curves \( y=\sin (x) \),
\( y=-1 \), \( x=0 \), and \( x=\pi  \) is revolved about the \( y \)-axis. 
\end{example}
%

\begin{example}
Find the volume of the solid obtained by revolving the region bounded by the
curves \( y=2x-x^{2}+2 \), \( x=0 \), \( x=2 \), and \( y=x \) about the
\( y \)-axis. \textit{}
\end{example}
%

\begin{example}
A \textit{torus} is the doughnut-shaped region obtained by revolving a circle
about a line outside the circle. find the volume of the torus obtained by revolving
the circle \( (x-2)^{2}+y^{2}=1 \) about the \( y \)-axis. 

\end{example}
\end{document}
