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\begin{center}

{\Large {\bf 
Mathematics 1210-2,3 \hspace{5ex}  Prof. Ken Golden\hspace{5ex} \\
PRACTICE FINAL EXAM ANSWER KEY
\hspace{5ex}Fall 2009 \\}
Problems From Varberg, Purcell, Rigdon (Chapter.Section.Number)
\\
}
%\vspace{2ex}\\
%{\Large {\bf ANSWER KEY}}

\end{center}
\vspace{2ex}



\begin{enumerate}

\item[1.] 
(5.1.6.) The boundary curves intersect when $x^2-2=x+4$ or $x^2-x-6=(x-3)(x+2)=0$
so at $x=-2$ and $x=3$. We integrate the vertical slices whose height at
$x$ is
the (positive) distance from the lower boundary to the upper boundary
there,
$(x+4)-(x^2-2)$ and whose width is $dx$.

$$\int_{-2}^3 -x^2 +x +6 dx
= |_{-2}^3 -\frac{x^3}{3}  + \frac{x^2}{2} +6x
=    \frac{27}{2}  -  ( -\frac{22}{3} )  =  \frac{103}{6} = 17
\frac{1}{6} .$$



\item[2.] 
(5.2.17.) The upper and lower boundary curves intersect when $y=0$ and
$\frac{x^2}{a^2} +  \frac{y^2}{b^2} = 1$, so when
$x=\pm a$.


If we slice vertically, the
long direction of the
slices is
perpendicular
to the axis of revolution and they sweep out circular discs.
The volume of the disc at $x$ is its thickness, $dx$, times its
area  $\pi r^2$. The radius is just
$y=  b ( 1- \frac{x^2}{a^2} )^{1/2}$,
the height of the region
being revolved about its lower boundary curve.

So we integrate

$$V = \int_{-a}^a \pi  b ^2 ( 1- \frac{x^2}{a^2} ) dx
= \pi  b ^2 \int_{-a}^a 1- \frac{x^2}{a^2} dx
= \pi  b ^2 | _{-a}^a   x - \frac{x^3}{3a^2}
=\frac{4}{3} \pi a b^2.$$

The prolate spheroid has a circular cross section
with radius $b$ in the $y-z-$plane, and extends
a distance $a$ from the origin in either direction
in the $x-$direction. When $a=b=r$, it becomes a
sphere of radius $r$ and the formula gives its
volume, $V=\frac{4}{3} \pi r^3$.


\item[3.] 
(5.3.2.) The upper and lower boundary curves intersect when $x^2=0$, so when $x=0$
so the region extends from $x=0$ to $x=1$. If we slice vertically, the
long direction of the
slices is parallel to the axis of revolution and they sweep out
cylindrical
shells. The volume of the slice at $x$ is its thickness, $dx$ times its
height, the distance from the lower to the upper boundary $x^2-0$, times
its circumference,
$2\pi$ times $x$, the `radius' of that shell, given by the distance of the
generating slice from the axis of revolution.
So we integrate

$$V = \int_0^1 2\pi x (x^2) dx
= |_0^1 2\pi \frac{x^4}{4} = \frac{\pi}{2}  .$$

If we slice horizontally, the long direction of each slice is
perpendicular
to the axis of revolution and they sweep out circular annuli or `washers'.
The volume of the washer at height $y$ is its thickness, $dy$, times the
difference of the area of the outer disc and the cut-out inner disc,
$\pi r_o^2 - \pi r_i^2$. The height of the slices extends from the lowest
point of the lower boundary, $0$ everywhere, to the highest point of the
upper boundary, $y=1$ at $x=1$. At height $y$, the inner radius is
the $x$ corresponding to $y=x^2$ which is both the upper {\it and}
inner boundary, so $x=y^{1/2}$, and the outer radius is $1$ at
every $y$. So we integrate

$$V = \int_0^1 \pi (1^2 - (\sqrt y)^2  dy
= \pi \int_0^1  1 - y   dy
= \pi _0^1  y  - \frac{y^2}{2}
= \frac{\pi}{2}.$$

You can also see from the graph of $1-y$ that the final integral is
just the area of a triangle with base and height $1$, so the
area is $1/2$, recalling something we did at the very beginning
of the class.



\item[4.] 
(5.4.10.) The arclength of a section of a parametric curve, $(x(t),y(t))$
is computed by integrating the infinitesimal hypotenuse lengths,

$$\sqrt{dx^2+dy^2}=\sqrt{\frac{dx}{dt}^2+\frac{dy}{dt}^2} dt$$
between the $t$-values of interest.

\bigskip

$$x=\sqrt 5 \sin {2t} -2 \quad  \quad y=\sqrt 5 \cos {2t} - \sqrt 3 ;
\quad \quad 0 \leq t \leq \frac{\pi}{4} ,$$
In this case, using $\frac{dx}{dt} = 2 \sqrt 5 \cos {2t}$ and
$\frac{dy}{dt} = -2\sqrt 5   \sin {2t}$,

$$S=\int_0^\frac{\pi}{4} \sqrt{\frac{dx}{dt}^2+\frac{dy}{dt}^2} dt
= \int_0^\frac{\pi}{4} \sqrt{ 20 (\cos^2 {2t} +  \sin^2 {2t} } dt =
= 2 \sqrt 5 \int_0^\frac{\pi}{4} dt = \frac{\sqrt 5 \pi}{2}.
$$


\item[5.] 
(5.4.21a.) The arclength of a section of a graph, $y=f(x)$
is computed by integrating the infinitesimal hypotenuse lengths,

$$\sqrt{dx^2+dy^2}=\sqrt{1+\frac{dy}{dx}^2} dx =\sqrt{1+f'(x)^2} dx$$
between the $x$-values of interest.

In this case, since $f$ itself is given as an integral between $1$ and
$x$,
we use the Fundamental Theorem of Calculus to compute
$f'(x)= \sqrt{ x^3 -1 }$ so that $\sqrt{ 1+f'(x)^2 } = x^{3/2}$ and
$$S=\int_1^2 \sqrt{1+f'(x)^2} dx = \int_1^2 x^{3/2} dx =|_1^2 \frac{2}{5}
x^{5/2} =
\frac{2}{5} (2^{5/2} -1).$$



\item[6.] 
(5.5.6.) Since $F=ks^{4/3}$ and $F=2$ pounds when $s=8$ inches,
$k={1 \over 8}=0.125$ when units of $F$ are in pounds and
$s$ are in inches.
Integrating the force of the spring
times the infinitesimal displacement,
$F(s)ds$, while stretching it
from $s=0$ (equilibrium) to
$27$  inches from equilibrium,

$$W = \int_0^{27} \frac{1}{8} s^{4/3} ds 
= |_0^{27} \frac{3}{56} s^{7/3} = \frac{3^8}{56}
\sim 117.16$$
where the answer is in inch-pounds, and can be
converted to the more common foot-pounds by dividing
by $12$.


\item[7.] 
(5.5.22.) Coulomb's Law may be written $F = K \frac{q_1 q_2}{r^2}$
where here $q_1=q_2=q$ and $10$ dynes = $K\frac{q^2}{4}$.
The work done bringing the charges from a distance $r_1=5$
centimeters apart to a distance $r_2=1$ centimeter apart
is

$$W = -\int_{r_1}^{r_2} F(r) dr = -\int_{5}^{1} K q^2 r^{-2} dr,$$

where the minus sign is because the force you must apply to move the
charge in closer is equal and opposite at each point to the repulsive
force which is in the positive direction. 
Doing the integration,  we get
$$W = -Kq^2 |_5^1 (-r^{-1}) = Kq^2 (1-\frac{1}{5}).$$

Finally, using $Kq^2 = 40$ in dyne-cm units,
$W= 32$ dynes.


\item[8.] 
(5.6.25.) Using the method of cylindrical shells

\bigskip

Recall that slices parallel to
the axis of revolution give cylindrical shells, slices
perpendicular to the axis of revolution give discs or
washers. Also notice that since the slices are rectangles,
every slice is really both parallel AND perpendicular, so
that the difference between a cylindrical shell and a
washer is really just a matter of aspect ratio.

Each $dx$ slice parallel to the $y$-axis at a particular
value of $x$
extends from the lower boundary line $y=0$ to
the upper boundary at $y=x^3$, and the $x$ values
extend from $x=0$ where the upper and lower boundary
curves intersect (at the axis of revolution) to $x=1$.

\bigskip
The cylindrical shell generated by revolving such a
slice about the $y$-axis is its circumference times
its thickness times its height. (This is just length
times width times height if the cylindrical shell is
unrolled like a `Ding-dong' into a rectangular slab.)
The circumference is $2\pi$ times the radius from the
axis to the slice, in this case just $x$, the height
is the distance from the lower boundary to the upper
boundary at $x$, $x^3-0$, and the thickness is $dx$.
So the volume is given by
$$\int_{x=0}^{x=1} 2 \pi x (x^3) dx = 2 \pi |_0^1 \frac{x^5}{5} = \frac{2
\pi}{5}.$$


\bigskip

The centroid of the region is obtained by averaging
the value of $x$ and $y$ in the region, weighting
each by the fraction of the region having that
$x$ or $y$ value.
$$ \bar{x} = \int_a^b x \frac{f(x)}{\int_a^b f(x) dx }  dx$$

so in this case

$$ \bar{x} = \frac{1}{\int_0^1 x^3 dx } \int_0^1 x^4   dx =
\frac{1/5}{1/4} = \frac{4}{5}.$$

For $\bar{y}$ we can either slice along constant $y$-values
and use
$$\bar{y} = \frac{1}{\int_0^1 1-y^{1/3} dy }\int_0^1 y ( 1-y^{1/3} )dy =
\frac{2}{7}$$
and notice that the normalization factor is the same area of the region,
$1/4$, computed
with $dy$ slices instead of $dx$ as before, so we really didn't need to do
it twice.

The other way to compute the $y$-component of the centroid is to average
the
average $y$ value in each $dx$ slice, $\frac{1}{2} y$, weighted by
the the fraction of the region having that average $y$ value.

The area of the region is $\frac{1}{4}$. The distance swept out by
the centroid is $2 \pi$ times 
$\frac{4}{5}$,
its 
distance 
from the axis of revolution. 
Multiplying these according to Pappus' volume theorem (there
is also an area theorem) gives
$\frac{1}{4}(2 \pi \frac{4}{5} = \frac{2 \pi}{5}$,
confirming Pappus' Theorem in this case.

The alternate derivation of the $y$-coordinate of the centroid is
$$ \bar{y} = \int_a^b \frac{1}{2} y \frac{f(x)}{\int_a^b f(x) dx }  dx$$

$$ \bar{y} = 4  \int_0^1 \frac{1}{2} x^6  dx = \frac{2}{7}. $$


\item[9.]
To find $T(x)$ we integrate the differential equation twice to get:

\vspace{.1 in}

\begin{center}
  $\ds \frac{dT}{dx} = C$

  $T(x) = Cx + D$
\end{center}

\vspace{.1 in}

Based on the initial conditions we get $T(0) = D = 4$ and $T(1) = 2 = 2C+4$ and so $C = -1$. Therefore, the temperature function is:

\vspace{.1 in}

\begin{center}
  $\ds T(x) = -2x+4$
\end{center}
% \begin{enumerate}

%\item[a.] 

% \end{enumerate}

%\eject


%\begin{figure}  %[htp]
%\centering
%\includegraphics{prac_2.eps}
%\caption{Graph of $f(x)$ for \#2.}        
%\end{figure}

\end{enumerate}

\end{document}

