\chapter[Falling Bodies; Volumes]{More on Falling Bodies; Volumes of Revolution}

If we know the \index{initial velocity} initial velocity
$v_0$ and the initial height $s_0$ of a
falling body at time $t=0$, then the formula
$$s(t)=-\frac{1}{2}gt^2+v_0t+s_0$$ 
tells us the height above the ground of the falling body at any time $t$. 
The formula for the velocity is  given by its derivative:
$$v(t) = -gt + v_0.$$ 

Note that we are neglecting air resistance and taking upward as the positive
direction. We usually use the traditional value $g=9.8$ m/sec${}^2$ for the 
gravitational acceleration.

But in some situations we do not know $v_0$ and/or
$s_0$, and must determine these values from other given information.
Generally speaking, any two bits of information can be used to determine
$v_0$ and $s_0$.


\begin{example}
%{\bf Example 1.}  
You see an object thrown downward from a building. It hits the ground
$5$ sec after it was thrown with an impact velocity of $60$ m/sec. What
was the height from which it was thrown, and with what initial velocity
was it thrown?  

\smallskip
We translate the information given into two equations. Since ``hits the
ground'' means that $s=0$, we have $s(5)=0$; we also know that $v(5)=-60$
(i.e., $60$ m/sec {\it downward}).  We plug $5$ into the $s(t)$ and $v(t)$
formulas to obtain two equations with the two unknowns $s_0$ and $v_0$:
\begin{align*} 
0 &= s(5)=-4.9\cdot 5^2+v_0\cdot 5+s_0=-122.5+5v_0+s_0;\\ -60 &=
v(5)=-9.8\cdot 5+v_0=-49+v_0.
\end{align*}
The second equation immediately gives $v_0=-60+49=-11$ m/sec.  Substituting
this value for $v_0$ in the first equation and solving for $s_0$, we
obtain: $$s_0=122.5-5v_0=122.5-5\cdot(-11)=177.5.$$ Thus, the object was
thrown downward at $11$ m/sec from a height of $177.5$ meters.
\end{example}


\section[Volume of Revolution]{Volume of a Solid of Revolution (Slice Method)} 

A \index{solid of revolution} solid of revolution
 is a solid which is symmetrical with respect to a
line called its {\it axis}.  Such a solid can be obtained by taking some
two-dimensional shape and revolving it around that axis.  Familiar examples
of solids of revolution include spheres, hemispheres, cylinders, and cones.

\smallskip

The basic situation to be dealt with first is the case when the axis of
symmetry of the solid is the $x$-axis and the shape being revolved around
the $x$-axis is the area under a curve $y=f(x)$ between $x=a$ and $x=b$.
The familiar examples listed above can be obtained in this way.

\smallskip

\begin{example} (Spheres and hemispheres)\

 
If we place the \index{sphere} sphere
 of radius
$r$ so that its center is at the origin of the $xy$-plane and take
$f(x)=\sqrt{r^2-x^2}$ (the upper semicircle of radius $r$), then we can get
the sphere by rotating $f(x)$ between $x=-r$ and $x=r$ around the $x$-axis.
To get a \index{hemisphere} hemisphere
 of radius $r$, we situate the hemisphere so that the
$x$-axis is its axis of symmetry, the $y$-axis between $r$ and $-r$ is a
diameter of its circular base, and the ``north pole'' of the hemisphere is
at the point $(r,0)$.  Then this hemisphere is the solid of revolution
obtained by revolving $f(x)=\sqrt{r^2-x^2}$ around the $x$-axis between
$x=0$ and $x=r$.

\begin{center}
%-figure BA -----------------------------
%
\begin{psfigure}{2in}{2.0in}{figBAa.ps}
%\putcrosshair{0,0}
\puttext{46, -10}{\small$x$}
\puttext{4.5, 35}{\small$y$}
\puttext{42, -17}{\small$r$}
\puttext{-42, -17}{\small$-r$}
\end{psfigure}
\begin{psfigure}{3in}{2.2in}{figBAb.ps}
%\putcrosshair{0,0}
\puttext{32, -5}{\small$x$}
\puttext{2, 45}{\small$y$}
\puttext{-50,-50}{Rotate around $x$-axis to get sphere of radius $r$.}
\end{psfigure}\\
\smallskip
\
%------------------------------
\end{center}
\end{example} 

\begin{example} (Right circular cones)\
%{\bf Example 2.  Right circular cones.}
 
By ``\index{right circular cone} right circular cone
'' we mean a
cone which is symmetric about an axis of symmetry, and for which any
cross-section perpendicular to this axis is a circle.  Such a cone is
determined if we know its height $h$ and its base radius $r$.  (In what
follows, we shall always use the term ``cone'' to mean a right circular
cone.)  To describe this cone as a solid of revolution, we first choose an
efficient way to situate it with respect to the $xy$-plane.  We place it so
that its vertex is at the origin and its axis of symmetry is the part of
the $x$-axis from 0 to $h$.  Then our cone is obtained by taking the line
joining the origin to the point $(h,r)$ and rotating it around the
$x$-axis.  The advantage in situating the cone in this way is that the
function we are revolving around the $x$-axis --- the line from $(0,0)$ to
$(h,r)$ --- has a very simple equation: $f(x)=mx$, where $m=r/h$ is the
slope of the line.

\begin{centering}
%-figure BB -----------------------------
%
%   NOTE: Make TWO figures here, showing how to actually GET the cone!
%

\begin{psfigure}{2in}{2.0in}{figBBa.ps}
%\putcrosshair{0,0}
\puttext{45.5, -15}{\small$x$}
\puttext{-25, 36.4}{\small$y$}
\puttext{32, 25}{$(h,r)$}
\end{psfigure}
\hspace{0.2in}
\begin{psfigure}{3in}{2.2in}{figBBb.ps}
%\putcrosshair{0,0}
\puttext{40.9, -13.6}{\small$x$}
\puttext{-25., 36.4}{\small$y$}
\puttext{-50,-50}{Rotate around $x$-axis to get cone of height $h$, base radius $r$.}
\end{psfigure}\\
%------------------------------
\smallskip
\
\end{centering}

\end{example}

\bigskip

We now return to the general situation of a solid of revolution obtained by
revolving $y=f(x)$ around the $x$-axis between $x=a$ and $x=b$.  We want to
find its volume.  To do this, we first take the region under the curve,
and divide it up into \index{Riemann sum rectangles} Riemann sum rectangles
 as we did earlier.
%in Chapter 17 of the Math 124 notes.  
Of course, the rectangles do not exactly fill out
the region: when the curve increases, the rectangles fall a little short,
and when the curve decreases, the rectangles give you a little too much
(see the second diagram below). But as the theory works out, the limit
of the Riemann sums gives the correct answer. 

Recall how the rectangles were constructed for the $n$-th left endpoint
Riemann sum.  First, the interval $[a,b]$ is divided into $n$ equal parts,
each of length $\Delta x=(b-a)/n$.  Next, above each subinterval we
draw the rectangle whose height is the distance up to the curve at the
left endpoint.  The heights of the rectangles are successively
$f(a)$, $f(a+\Delta x)$, $f(a+2\Delta x),\ldots$, $f(a+(n-1)\Delta x)$.

% Here there should be two diagrams, the first of which is the same as
% the first one below, but without the ``x [with dotted line above it].''
% The second diagram should show the same function with
% lightly shaded Riemann sum rectangles (say, with n=6 subdivisions).

\begin{centering}
%-figure BCa -----------------------------
\begin{psfigure}{1.7in}{2in}{figBCa.ps}
%\putcrosshair{0,0}
\puttext{10., 35.}{\small$y = f(x)$}
\puttext{-12.5, -25.}{\small$a$}
\puttext{42., -25.}{\small$b$}
\puttext{0,-45}{Graph of $y = f(x)$}
\end{psfigure}
\hspace{0.2in}
%-figure BCb -----------------------------
\begin{psfigure}{1.7in}{2in}{figBCb.ps}
%\putcrosshair{0,0}
\puttext{10., 35.}{\small$y = f(x)$}
\puttext{-12.5, -25.}{\small$a$}
\puttext{42., -25.}{\small$b$}
\puttext{0,-45}{Riemann sum rectangles}
\end{psfigure}\\
\end{centering}

\medskip
When we rotate our function $f(x)$ around the $x$-axis, what happens to
the rectangles?  As each narrow rectangle rotates through $360^\circ$, it
forms a sideways
cylinder whose ``height'' (actually, its distance across from left to
right) is $\Delta x$ and whose base radius is the height of the rectangle.
Since the volume of a cylinder is equal to its base area (which is
$\pi\cdot{\rm radius}^2$) times its height, it follows that the
volume of the first cylinder is $\pi f(a)^2\Delta x$; the volume of
the second is $\pi f(a+\Delta x)^2\Delta x$; the volume of the third is
$\pi f(a+2\Delta x)^2\Delta x$, and so on.  Thus, the total volume of
all of the cylindrical slices is
$$\Bigl(\pi f(a)^2+\pi f(a+\Delta x)^2+\pi f(a+2\Delta x)^2+\cdots+
\pi f(a+(n-1)\Delta x)^2\Bigr)\Delta x,$$ which can be written more
compactly using the \index{summation notation}summation notation, as follows:
$$\sum_{i=0}^{n-1}\pi f(a+i\Delta x)^2\Delta x.$$
If $\Delta x$ is small (i.e., if $n$ is large), this volume is a very 
close approximation to the true volume of the solid of revolution.
As $\Delta x\longrightarrow 0$ (i.e., as $n\longrightarrow\infty$), this
sum gets closer and closer to the volume of the solid we want.
As discussed in the Chapter on Riemann Sums,
%17 of the Math 124 notes, 
the limit of this sum will become the definite integral of
$\pi f(x)^2$ (rather than the definite integral of $f(x)$).  
We conclude that
$$
\text{Volume~of~solid~of~revolution}=\pi\int_a^bf(x)^2dx.
$$
(The constant $\pi$ that is next to $f(x)^2$ can be pulled out in front
of the integral sign.)

\smallskip

One way to think of this formula is to look at the contribution to
the volume that comes from the part of the region under $y=f(x)$ that
lies between $x$ and $x+\Delta x$, where we suppose that $\Delta x$ is
very small.  This is a thin slice whose thickness is $\Delta x$ and
whose circular base (at the left) has area $\pi f(x)^2$.  

Since $\Delta x$  
is small, the slice is very close to being a cylinder, and so its volume
is very close to $\pi f(x)^2\Delta x$.  Now to find the volume of the
entire solid of revolution we have to sum up the contributions as $x$ goes from
$a$ to $b$.  In the limit as $\Delta x$ goes to zero this sum becomes
an integral, and we write $dx$ instead of $\Delta x$.  That is, we get
$\int_a^b\bigl(\pi f(x)^2\bigr)dx$, as before.

{\bf More generally, we can think of this formula as follows. Suppose
that we have a solid caught between the planes $x=a$ and $x=b$ and
the \index{cross sectional area} cross sectional area corresponding to any value of $x$ is given 
by a function $F(x)$, then the total volume can be computed by the
integral 
$$\int_a^b F(x) \, dx $$

In our current situation the cross sectional area happend to be a circle.

This formula can also be understood by creating a variable ``volume''
function 
$$V(u) = \text{~volume~caught~between~} x=a \text{~and~}x=u$$
and deducing as before that $V'(x)=g(x)$. Then we use the Fundamental
Theorem of Calculus to deduce:
$$V(b) = \int_a^b g(x) \, dx.$$ 
}

%%%% To update! Add more suitable pictures.

% Now put in the 2nd and 3rd diagrams above (but in the reverse order,
% i.e., the single shaded slice comes first), and change the captions
% to ``A single slice of the volume'' and ``The full volume of revolution''

\begin{centering}

%-figure BCd -----------------------------
\begin{psfigure}{1.7in}{2in}{figBCd.ps}
%\putcrosshair{0,0}
%\puttext{15., 25.}{\small$\Delta x$}
\puttext{0,-45}{Volume of revolution.}
\end{psfigure}
%------------------------------
\hspace{0.2in}
%-figure BCc -----------------------------
\begin{psfigure}{1.7in}{2in}{figBCc.ps}
%\putcrosshair{0,0}
%\puttext{15., 25.}{\small$\Delta x$}
\puttext{0,-45}{Thin slice of solid.}
\end{psfigure}\\
%------------------------------

\end{centering}

\smallskip

\begin{example}
%{\bf Example 1 again.} 
Find the formula for the volume of a sphere of
radius $r$.

\smallskip
We saw that the \index{sphere} sphere is obtained by rotating the function
$f(x)=\sqrt{r^2-x^2}$ around the $x$-axis between $x=-r$ and $x=r$.
Actually, it turns out to be a little simpler just to rotate the part
between $x=0$ and $x=r$ --- thereby obtaining a hemisphere --- and then
doubling this volume.  That is, using the formula we just derived, we see
that the volume of the sphere is
\begin{align*}
2\pi\int_0^r\left(\sqrt{r^2-x^2}\right)^2\,dx
&=2\pi\int_0^r(r^2-x^2)\,dx=
2\pi\left(r^2x-\frac{x^3}{3}\right)\Bigm|_0^r\\
&=2\pi(r^3-r^3\!/3)=\frac{4}{3} \pi r^3.
\end{align*}

\end{example}

\begin{example}
%{\bf Example 2 again.} 
Find the formula for the volume of a cone of height
$h$ and base radius $r$.

\smallskip
According to our earlier discussion of how to get this cone as a solid of
revolution, we have 
$$
\text{Vol}=\pi\int_0^h\Bigl(\left(\frac{r}{h}\right)x\Bigr)^2dx=
\pi r^2\!/h^2\int_0^hx^2dx=\pi (r^2/h^2)(h^3/3)=\frac{1}{3}\pi r^2h.
$$

\smallskip
Expressed in words, this formula says that {\it the volume of a cone is one
third the volume of the cylinder of the same height and base radius}.  By
examining the computations in the last example, we see how the $1/3$ arose:
when we integrated $x^2$ to get $(x^3\!/3)|_0^h$.

\end{example}

\section{Some Variants} 

Suppose we want our solid to be ``right side up'' with the $y$-axis as its
axis of symmetry.  Then the curve that is rotated around the $y$-axis
should be viewed from the $y$-perspective as a function of the form
$x=g(y)$ (rather than $y=f(x)$).  Everything works the same with the roles
of $x$ and $y$ reversed.  Suppose that the solid is obtained by rotating
$x=g(y)$ around the $y$-axis between $y=c$ and $y=d$.  Then its volume is:
$$
\pi\int_c^d g(y)^2dy.
$$

\smallskip
As another variant, suppose we want to rotate around the line $y=c$ which
is parallel to the $x$-axis and a distance $c$ above it (if $c$ is
negative, a distance $|c|$ below it).  The only change is to replace $f(x)$
by $f(x)-c$, which is the distance from the axis of symmetry to the curve
at the position $x$ on the $x$-axis.  Thus, the volume is
$\pi\int_a^b(f(x)-c)^2dx$.  A similar remark applies to the case when we
take a function $x=g(y)$ and revolve it around the line $x=c$ rather than
around the $y$-axis.

\begin{example}
%{\bf Example 3}.  
Let $S$ be the shape in the $xy$-plane that is bounded by
the curve $y=x^2$, the line $y=4$ and the line $x=1$, and which is to the
right of the line $x=1$.  Find the volume of the shape obtained by
revolving $S$ around (a) the $x$-axis, (b) the line $y=4$, (c) the
$y$-axis, (d) the line $x=1$.

\smallskip

(a)~Here we obtain a ``hollowed out'' shape.  That is, to get our solid we
first take the cylinder that is obtained by rotating the line $y=4$ around
the $x$-axis between $x=1$ and $x=2$, and from this cylinder we then carve
out the solid obtained by rotating $y=x^2$ between $x=1$ and $x=2$.  Thus,
$$
\text{Vol}=\pi\cdot 4^2\cdot 1-\pi\int_1^2(x^2)^2dx=
16\pi-\pi\int_1^2x^4dx=(16-(32-1)/5)\pi=9.8\pi.
$$

\smallskip

(b)~Here we are taking the full solid of revolution obtained by rotating
$y=x^2$ around the line $y=4$.  The result is
$$
\text{Vol}=\pi\int_1^2(x^2-4)^2dx=\pi\int_1^2(x^4-8x^2+16)dx=53\pi/15.
$$

\smallskip

(c)~Here we start with the solid of revolution obtained by rotating
$x=\sqrt{y}$ around the $y$-axis between $y=1$ and $y=4$, and then hollow
out a cylinder of height 3 and base radius 1 (obtained by rotating $x=1$
around the $y$-axis between $y=1$ and $y=4$).  The result is:
$$
\text{Vol}=\pi\int_1^4(\sqrt{y})^2dy-\pi\cdot 1^2\cdot 3=
\pi\int_1^4y\,dy-3\pi=4.5\pi.
$$

\smallskip

(d)~Here we have the full solid of revolution where we rotate $x=\sqrt{y}$
around the line $x=1$.  Thus,
$$
\text{Vol}=\pi\int_1^4(\sqrt{y}-1)^2dy=\pi\int_1^4(y-2\sqrt{y}+1)dy=7/6\pi.
$$

\end{example} 

\begin{example}
% Example 2.7
Suppose that a solid has \index{{\bf square} cross-sections} {\bf square} cross-sections.  The centers of the
squares are on the $x$-axis, and the tops of the squares run through the
line $y=f(x)=4-x^2$ between $x=-2$ and $x=2$.  (See the diagram below.)
Find the volume of this solid.

\medskip


\begin{centering}
\begin{psfigure}{3in}{3in}{figBD.ps}
%\putcrosshair{0,0}
\puttext{-10,5}{\small$x$}
\puttext{-50,25}{\small$y$}
\puttext{-30,20}{\small$y = f(x)$}
\puttext{-30,-20}{\small$y=-f(x)$}
\puttext{-50,-40}{\small Side view}
\puttext{50,-40}{\small Perspective view}
\end{psfigure}\\
\end{centering}

\bigskip

Here the cross-section at $x$ is a square of side $2f(x)$.  So the
cross-sectional area is $\bigl(2f(x)\bigr)^2=4(4-x^2)^2$.  We integrate
this from $-2$ to $2$.  (Because of the symmetry --- i.e., the fact that
$f(x)=f(-x)$ --- we can take the integral from $0$ to $2$ and then
double the result.) Thus, the volume is
$$2\int_0^2 4(4-x^2)^2dx=8\int_0^2(16-8x^2+x^4)dx=136\frac{8}{15}.$$ 

\end{example}

% 
