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\hskip36pc \= Math.\ 302-01\\
\>Spring, 2005\\
\>R. B. Kearfott
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\bf Fifth Examination \\
\it Thursday, March 24, 2005
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\noindent {\small {\bf Instructions:} This exam should be done on
your own paper. Your name should be on each sheet and on the back
of the last sheet; the answers should appear written carefully and
in order. If in doubt, show intermediate steps: Full credit may
not be given, even for correct answers, unless work is arranged
clearly and explained. This exam is closed book. You may leave
after handing in your exam paper, but be sure to check your
answers carefully. You may keep this exam sheet.  Each part of
each is worth 30 points, and 10 points are ``free".}

\Large
\begin{enumerate}
   \item Suppose the Earth's atmosphere extends from radius $\rho =
         6,400$ kilometers  to radius $\rho = 6,700$ kilometers,
         and the density of the atmosphere at height $\rho$ kilometers
         is approximately
         $$D(\rho) \approx 1.4\times10^9 e^{-0.105(\rho-6400)}$$
         kilograms per cubic kilometer.  From this, compute the
         approximate weight of the earth's atmosphere.\newline
         {\it Hint: Integration by parts gives
         $$\int u^2 e^{-c(u-a)}\/du = -\frac{1}{c} u^2 e^{-c(u-a)}
           - \frac{2}{c^2} u e^{-c(u-a)}
           - \frac{2}{c^3}e^{-c(u-a)}.
         $$
         Also, you may take the upper limit of the Earth's atmosphere
         to be at $\rho = \infty$, if it is easier.}
   \item Consider
         $$I = \int\int_{\mathcal{A}}\left( x^2 + y^2\right)\/ dA,$$
         where $\mathcal{A}$ is the triangle with vertices
         $(0,0)$, $(1,0)$, and $(1,1)$.
         \begin{enumerate}
            \item Compute $I$ using rectangular coordinates.
            \item Compute $I$ using polar coordinates.\newline
                  \textit{Hint: the upper bound on $r$ is a
                  function of $\theta$, that can be obtained by
                  considering a trigonometric function, as in an
                  example worked in class.  Also,
                  $$\int \sec^4(\theta)\/d\theta =
                     \frac{2}{3}\tan(\theta)
                     + \frac{1}{3}\sec^2(\theta) \tan(\theta).
                  $$}
         \end{enumerate}
\end{enumerate}
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