\def\whichexam{Exam III}
\def\examdate{April 20, 2004}

\long\def\newp #1.{
\par\noindent\vtop\bgroup\noindent{\bf #1.}
}
\long\def\problembreak{\par\egroup\vfill\eject\vtop\bgroup}
\def\endp{\par\noindent\vrule width\hsize depth 0pt height .1pt\egroup\par\bigskip}

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\font\sm=cmr7
\magnification1200
\centerline{\whichexam}
\centerline{\examdate}
\def\Int{\displaystyle\int}
\def\Lim{\displaystyle\lim}
\def\frac#1#2{{{#1}\over{#2}}}
\def\smfrac#1#2{{{\scriptstyle#1}\over{\scriptstyle#2}}}
\bigskip
\def\frac#1#2{{{\textstyle#1}\over{\textstyle#2}}}
\def\mbox#1{\hbox{\rm#1}}
% #1 - path to pdf file
% #2 - scale factor
% #3 #4  - size of box pdf is centered in this box. It can overflow box
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}

%%%%%
%:12
\newp 12.
Use the Integral Test to show that the series
$\sum_{n=3}^{\infty} \frac{\ln(n)}{n}$ diverges.

\noindent{\bf Remark:} Be sure to check that the Integral Test can be applied.

\noindent{\bf Solution:} Seven points were allocated for the correct
verification of the hypotheses in the statement of the Integral Test;
eight points were given for successfully applying the theorem.

The appropriate function to use in the Integral Test is clearly
$f(x) = \ln(x)/x$ since $f(n) = a_n$ for integer $n$.  The hypotheses 
for using this test are:

\item{} (1 point) {\bf Continuity}: $\ln(x)$ and $1/x$ are well--known to
be continuous for all $x>0$.  Clearly their product is continuous
on $[3,\infty)$.
\item{} (1 point) {\bf Positivity}: $\ln(x)$ is positive for $x>1$, while
$1/x$ is positive for $x>0$.  Clearly their product is positive
for $[3,\infty)$.
\item{} (5 points) {\bf Decreasing}:  The derivative of $f$ is
$$
f'(x) = \frac{ (1/x) x - \ln(x) 1 }{x^2} = \frac{1-\ln(x)}{x^2}.
$$
Since the denominator is always positive for $x \neq 0$, $f'$ is
negative where $1-\ln(x)<0$, i.e.\ for $x>e$.  Since $e<3$,
$f'$ is negative on $[3,\infty)$ and $f$ is decreasing for this
interval.

\bigskip\noindent
The Integral Test states that if $\int_3^{\infty} f(x) \; dx$ diverges
then so does $\sum_{n=3}^{\infty} f(n)$, which is $\sum_{n=3}^{\infty} a_n$.
If we let $u = \ln(x)$ then $du = (1/x) dx$ and
$$
\int_3^{\infty} \frac{\ln(x)}{x} \; dx
  = \int u \; du
  = \left. \frac{1}{2} (\ln(x))^2 \right|_{x=3}^{x=\infty}
$$
which diverges since $\ln(x) \rightarrow \infty$ as $x \rightarrow \infty$.
\endp

%:13
\newp 13.
Set up an integral which computes the shaded area. 
The polar equation of the region is $r = 1 -2\cos\theta$.
\par\bigskip\noindent{\bf Remarks:} $\cos\left(\frac{\pi}{3}\right) = \frac 12$.
The answer is the difference of two areas.

\hbox{%
\vbox{\LoadPDF{./areaPartial.pdf}{.7}{5in}{4in}}\hskip -2in\raise .5in
\vbox{\hsize=2 in \hfuzz=10pt%
This was a complicated problem to grade because there are so many correct answers.
Many people computed the shaded area above the $x${--}axis and doubled it, which got
full marks if done correctly.
Others computed the area directly.
The crux of the problem is to work out intervals for $\theta$ which sweep out the various pieces.

From the first remark, $\theta = \frac{\pi}{3}$ is one of the angles where the curve goes through
the pole (the origin).
As $\theta$ increases, it gets to $\frac{\pi}{2}$ which turns out to be the point $(0,1)$ in Cartesian
coordinates.
When $\theta = \pi$ we are at the point $(-3,0)$.
By the time $\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$, we have traced the outside curve
once and hence
}}
\vskip-40pt
\noindent the integral $
\frac{1}{2}\displaystyle \int^{\smfrac{5\pi}{3}}_{\smfrac{\pi}{3}}\ (1 - 2\cos \theta )^2\; d\theta$ 
computes the total area {\sl inside\/} the outer curve.

As $\theta$ continues to increase past $\frac{5\pi}{3}$, we begin to trace the inside loop.
At $\theta = 2\pi$ we are at $(-1,0)$ and by $\theta = 2\pi +\frac{\pi}{3} = \frac{7\pi}{3}$ we have
traced out the entire inner loop so one answer is\
$$\frac{1}{2}\displaystyle \int^{\smfrac{5\pi}{3}}_{\smfrac{\pi}{3}}\ (1 - 2\cos \theta )^2\; d\theta
- \frac{1}{2}\displaystyle \int^{\smfrac{7\pi}{3}}_{\smfrac{5\pi}{3}}\ (1 - 2\cos \theta )^2\; d\theta$$

Because the curve is periodic with period $2\pi$ the limits cab be changed: a popular choice
was to write the area of the inner loop as
$\frac{1}{2}\displaystyle \int^{\smfrac{\pi}{3}}_{\smfrac{-\pi}{3}}\ (1 - 2\cos \theta )^2\; d\theta$.

Another approach is to start with
$\frac{1}{2}\displaystyle \int^{2\pi}_{0}\ (1 - 2\cos \theta )^2\; d\theta$.
This is the area inside the outer curve plus the area inside the inner curve so
$$\frac{1}{2}\displaystyle \int^{2\pi}_{0}\ (1 - 2\cos \theta )^2\; d\theta
-2 \cdot  \frac{1}{2} \int^{\smfrac{7\pi}{3}}_{\smfrac{5\pi}{3}}\ (1 - 2\cos \theta )^2\; d\theta$$
also computes the requested area.
\endp


Computing the area above the $x${--}axis works out as follows.
From $\frac{\pi}{3}$ to $\pi$ the top half of the outer curve is traced: from
$0$ to $\frac{\pi}{3}$ the bottom loop of the inner curve is traced.
Hence another answer is
$$2\left(
\frac{1}{2}\displaystyle \int^{\pi}_{\smfrac{\pi}{3}}\ (1 - 2\cos \theta )^2\; d\theta
- \frac{1}{2}\displaystyle \int^{\smfrac{\pi}{3}}_{0}\ (1 - 2\cos \theta )^2\; d\theta\right)$$
One can also do
$$2\left(
\frac{1}{2}\displaystyle \int^{\pi}_{0}\ (1 - 2\cos \theta )^2\; d\theta
- 2 \cdot \frac{1}{2}\displaystyle \int^{\smfrac{\pi}{3}}_{0}\ (1 - 2\cos \theta )^2\; d\theta\right)$$


%:14
\newp 14.
For revolution about the $y$-axis, surface area is
$$A \; = \; \int 2 \pi x ds \; = \; \int_0^{2 \pi} 2 \pi x(t)
\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2   } \; dt\ .$$

$\frac{dx}{dt} = \frac{d}{dt}(5 + 4 \cos t) = -4\sin t$ and $\frac{dy}{dt}
= \frac{d}{dt}(3\sin t) = 3  \cos t$.  Thus
$$\eqalign{%
A =&  \int_0^{2 \pi} 2 \pi (5 + 4 \cos t) \sqrt{(-4 \sin t )^2 + (3 \cos t)^2   } \; dt \cr
=&  \int_0^{2 \pi} 2 \pi (5 + 4 \cos t) \sqrt{16 \sin^2 t +  9\cos^2 t   } \; dt .\cr}$$
\endp

\end

