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\begin{document}
\centerline{Mathematics 136, section 2 -- Advanced Placement Calculus}
\centerline{Solutions for Exam 3 -- December 2, 2009}
\vskip 10pt
\begin{enumerate}
\item[I.]  Let $R$ be the region in the plane bounded by $y = e^x$, $y = x/2$,
$x = 0$, and $x = 1$.
\begin{enumerate}
\item[A)] (5)  Sketch the region $R$.  
\item[] {\it Solution.}  See Figure 1.
\item[B)] (5)  Find the {\it area} of $R$.
\item[] {\it Solution:}  We have 
$$A = \int_0^1 e^x - x/2\ dx = \left. e^x - x^2/4\right|_0^1 = e - 1/4 - 1 = e - 5/4
\doteq 1.47.$$
\item[C)] (10) Find the {\it volume} of the solid generated by rotating $R$ about the
$x$-axis.
\item[] {\it Solution:}  The cross-sections are washers with outer radius
$e^x$ and inner radius $x/2$, so
\begin{align*}
V &= \int_0^1 \pi (e^x)^2 - (x/2)^2\ dx\\
 &= \pi \int_0^1 e^{2x} - x^2/4\ dx\\
 &= \pi \left(\left. e^{2x}/2 - x^3/12\right|_0^1\right)\\
 &= \pi (e^2/2 - 1/12 - 1/2)\\
 &= \pi (e^2/2 - 7/12).
\end{align*}
\begin{figure}
\begin{center}
\includegraphics[width=0.4\textwidth]{I.ps}
\end{center}
\caption{The region $R$ in question I.}
\end{figure}
\item[D)] (5)  Set up, but do not evaluate, an integral or integrals to compute the 
{\it volume} of the solid generated by rotating $R$ about the $y$-axis.  
\item[] {\it Solution:}  The cross-sections are different for different ranges of $y$:
disks with radius $x = 2y$ for $0 \le y \le 1/2$, disks with radius 1 for $1/2 \le y \le 1$, 
and washers with outer radius $1$ and inner radius $x = \ln(y)$ for $1 \le y \le e$. So the 
volume would be computed as the sum:
$$V = \int_0^{1/2} \pi (2y)^2\ dy + \int_{1/2}^1 \pi (1)^2\ dy + \int_1^e \pi((1)^2 - (\ln(y))^2)\ dy.$$

\end{enumerate}
\vskip 10pt
\item[II.] 
\begin{enumerate}
\item[A)] (10) Integrate by parts:  $\int x e^{-3x}\ dx$. 
\item[] {\it Solution:}  Let $u = x$ and $dv = e^{-3x}\ dx$.  Then $du = dx$
and $v = \frac{-1}{3} e^{-3x}$.  Applying the parts formula,
$$\int x e^{-3x}\ dx = \frac{-xe^{-3x}}{3} + \frac{1}{3} \int e^{-3x}\ dx = 
\frac{-xe^{-3x}}{3} - \frac{e^{-3x}}{9} + C.$$
\item[B)] (10) Does the integral $\int_0^\infty xe^{-3x}\ dx$ converge?
If so, find its value.  
\item[] {\it Solution:}  The integral does converge to the value $\frac{1}{9}$:
\begin{align*}
\int_0^\infty xe^{-3x}\ dx &= \lim_{b\to\infty} \int_0^b xe^{-3x}\ dx\\
&= \lim_{b\to\infty} \left(\left. \frac{-xe^{-3x}}{3} - \frac{e^{-3x}}{9} \right|_0^b\right)\\
&= \lim_{b\to\infty} \left(\frac{-b e^{-3b}}{3} - \frac{e^{-3b}}{9} + 0 + \frac{1}{9}\right).
\end{align*}
Now, 
$$\lim_{b\to\infty} \frac{e^{-3b}}{9} = \lim_{b\to\infty} \frac{1}{9e^{3b}} = 0.$$
Moreover, by L'Hopital's Rule,
\begin{align*}
\lim_{b\to\infty} \frac{-b e^{-3b}}{3} &= \lim_{b\to\infty} \frac{-b}{3e^{3b}} \qquad \text{(an $\infty/\infty$ form)}\\
&= \lim_{b\to\infty} \frac{-1}{9e^{3b}}\\
&=0.
\end{align*}
Hence the integral converges to $\frac{1}{9}$.  
\end{enumerate} 
\vskip 10pt
\item[III.] (15)   A water tank has the shape of an inverted cone
(that is, the ``point'' of the cone is at the bottom) with
height $10$ meters and radius $3$ meters.  If the tank is
full of water (density 1000kg per cubic meter), find the {\it work} done in
pumping all the water out the top of the tank.  The acceleration of 
gravity is $9.8$ meters per second squared.  
\item[] {\it Solution:}  Consider the slice of the water, thickness $\Delta y$,
originally at height $y$ from the bottom -- approximately a cylinder with 
radius $3y/10$.  The work to lift the slice
is 
$$W_{slice} = \pi (3y/10)^2 \Delta y \cdot 1000 \cdot 9.8 \cdot (10 - y).$$
Summing over the slices and taking a limit:
$$W = 9800\pi \int_0^{10} \frac{9y^2}{10} - \frac{9y^3}{100}\ dy = 
9800\pi \left( \left. \frac{3y^3}{10} - \frac{9y^4}{400}\right|_0^{10}\right) = 735000\pi.$$
 
\vskip 10pt
\item[IV.]  (15)  Set up and evaluate the integral to compute the {\it arclength} of the 
parabola $y = 4x^2$, $0 \le x \le 1$.  (Use the table to evaluate.)
\item[] {\it Solution:}  $y' = 8x$, so
$$L = \int_0^1 \sqrt{1 + (8x)^2}\ dx.$$
To evaluate, we use the substitution $u = 8x$ and \# 21 in the table of integrals:
$$=\frac{1}{8}\left(\left. \frac{8x}{2}\sqrt{1 + (8x)^2} + \frac{1}{2}\ln(8x + \sqrt{1 + (8x)^2}\right|_0^1\right) = \frac{\sqrt{65}}{2} + \frac{1}{16}\ln(8 + \sqrt{65}).$$
\vskip 10pt
\item[V.]  All parts of this question refer to the differential equation
\[
\frac{dy}{dx} = (y - 1)(y - 3)/4
\]
\begin{enumerate}
\item[A)] (10)  Sketch the {\it slope field} of this equation, showing 
the slopes at points on the lines $y = 0,1,2,3,4,5$, and $-3 \le x \le 3$.
\item[] {\it Solution:}  See Figure 2.  
\item[B)]  (5) On your slope field, sketch the graph of the 
{\it solution} of the equation with $y(0) = 2$.
\item[] {\it Solution:}  See Figure 2.  
\begin{figure}
\begin{center}
\includegraphics[width=0.4\textwidth]{V.ps}
\end{center}
\caption{The slope field and solution in V.}
\end{figure}
\item[C)]  (10) This is a separable equation; find an explicit formula
for the {\it solution} satisfying the initial condition $y(0) = 2$. 
\item[] {\it Solution:}  We separate variables and integrate using partial fractions
for the $y$-integral:
\begin{align*}
\int \frac{dy}{(y - 1)(y - 3)} &= \int \frac{1}{4}\ dx\\
\int \frac{-1/2}{y - 1} + \frac{1/2}{y - 3} &= \frac{x}{4} + c\\
-\frac{1}{2}\ln|y - 1| + \frac{1}{2}\ln|y-3| &= \frac{x}{4} + c\\
\ln\left|\frac{y-3}{y-1}\right| &= \frac{x}{2} + c\\
\frac{y-3}{y-1} &= k e^{-x/2} \quad \text{where $k = \pm e^c$}\\
y - 3 &= k(y-1)e^{-x/2}\\
y &= \frac{-k e^{x/2} + 3}{ -k e^{x/2} + 1}.
\end{align*}
To get $y(0) = 2$, we must have $2 = \frac{3-k}{1-k}$ so $2 - 2k = 3 - k$, or
$k = -1$.  The solution can be written as 
$$y = \frac{e^{x/2} + 3}{e^{x/2} + 1}.$$
\end{enumerate}
\vskip 10pt
\item[] {\it Extra Credit} (10)  Set up the integrals to 
find the coordinates of the {\it centroid}
of the region in the first quadrant inside the ellipse $\frac{x^2}{4} + y^2 = 1$.  
You do not need to evaluate.
\item[] The region is bounded by $y = \sqrt{1 - x^2/4}$, $x = 0$, $x = 2$, and the 
$x$-axis.  The coordinates of the centroid would be computed by 
$$\overline{x} = \frac{\int_0^2 x\sqrt{1 - x^2/4}\ dx}{\int_0^2 \sqrt{1 - x^2/4}\ dx}$$
and 
$$\overline{y} = \frac{\int_0^2 \frac{1}{2} (\sqrt{1 - x^2/4})^2 \ dx}{\int_0^2 \sqrt{1 - x^2/4}\ dx}.$$
(These are approximately $(\overline{x},\overline{y}) \doteq (.85,.42)$.  See Figure 3.)
\begin{figure}
\begin{center}
\includegraphics[width=0.4\textwidth]{EC.ps}
\end{center}
\caption{The region and the approximate location of the centroid.}
\end{figure}
\end{enumerate}
\end{document}
