3. Let $R$ be the region in the first quadrant enclosed by the hyperbola $x ^ { 2 } - y ^ { 2 } - 9$, the $x$-axis, and the line $x - 5$.\\
(a) Find the volume of the solid generated by revolving $R$ about the x-axis.\\
(b) Set up, but do not integrate, an integral expression in terms of a single variable for the volume of the solid generated when $R$ is revolved about the line $\mathrm { x } = - 1$.

\section*{Solution}
(a) Discs:

$$\begin{aligned}
V & = \pi \int _ { 3 } ^ { 5 } \left( x ^ { 2 } - 9 \right) d x \\
& = \pi \left[ \frac { 1 } { 3 } x ^ { 3 } - 9 x \right] _ { 3 } ^ { 5 } \\
& = \pi \left[ \left( \frac { 125 } { 3 } - 45 \right) - ( 9 - 27 ) \right] = \frac { 44 } { 3 } \pi
\end{aligned}$$

or\\
Shells:

$$\begin{aligned}
V & = 2 \pi \int _ { 0 } ^ { 4 } \left( 5 - \sqrt { 9 + y ^ { 2 } } \right) y d y \\
& = 2 \pi \left[ \frac { 5 } { 2 } y ^ { 2 } - \frac { 1 } { 3 } \left( 9 + y ^ { 2 } \right) ^ { \frac { 3 } { 2 } } \right] _ { 0 } ^ { 4 } \\
& = 2 \pi \left( 40 - \frac { 125 } { 3 } + \frac { 27 } { 3 } \right) = \frac { 44 } { 3 } \pi
\end{aligned}$$

(b) Shells:

$$\begin{aligned}
V & = 2 \pi \int _ { 3 } ^ { 5 } ( x + 1 ) y d x \\
& = 2 \pi \int _ { 3 } ^ { 5 } ( x + 1 ) \sqrt { x ^ { 2 } - 9 } d x
\end{aligned}$$

or\\
Washers:

$$\begin{aligned}
V & = \pi \int _ { 0 } ^ { 4 } \left[ 36 - ( x + 1 ) ^ { 2 } \right] d y \\
& = \pi \int _ { 0 } ^ { 4 } \left[ 36 - \left( \sqrt { 9 + y ^ { 2 } } + 1 \right) ^ { 2 } \right] d y
\end{aligned}$$

\section*{Distribution of Points}
(a)

$$5 : \begin{cases} 2 : & \text { for a correct integrand } \\ 1 : & \text { for appropriate limits } \\ & \text { and } k \pi \\ 1 : & \text { for correct antiderivative } \\ 1 : & \text { for substitution and/or } \\ & \text { evaluation } \end{cases}$$

(b)

$$4 : \left\{ \begin{array} { l } 
3 : \text { for a correct integrand } \\
1 : \text { for appropriate limits } \\
\text { and } k \pi
\end{array} \right.$$

\begin{enumerate}
  \setcounter{enumi}{3}
  \item Let $f$ be the function defined by $f ( x ) = 2 x e ^ { - x }$ for all real numbers $x$.\\
(a) Write an equation of the horizontal asymptote for the graph of $f$.\\
(b) Find the $x$-coordinate of each critical point of $f$. For each such $x$, determine whether $f ( x )$ is a relative maximum, a relative minimum, or neither.\\
(c) For what values of $x$ is the graph of $f$ concave down?\\
(d) Using the results found in parts (a), (b), and (c), sketch the graph of $y = f ( x )$ in the $x y$-plane provided below.
\end{enumerate}

\section*{Solution}
(a) $y = 0$\\
(b) $f ^ { \prime } ( x ) = 2 \left( - x e ^ { - x } + e ^ { - x } \right)$

$$= 2 e ^ { - x } ( 1 - x )$$

critical point at $x = 1$\\
relative maximum at $x = 1$\\
(c) $f ^ { \prime \prime } ( x ) = 2 e ^ { - x } ( - 1 ) + \left( - 2 e ^ { - x } \right) ( 1 - x )$

$$= 2 e ^ { - x } ( x - 2 )$$

Concave down when

$$\begin{aligned}
2 e ^ { - x } ( x - 2 ) & < 0 \\
( x - 2 ) & < 0 \\
x & < 2
\end{aligned}$$

\begin{center}
\includegraphics[max width=\textwidth, alt={}]{0e352c1f-977a-4194-a08a-58c028816852-58_696_928_1702_112}
\end{center}

\section*{Distribution of Points}
\section*{(a) 1: for correct equation}
(b)

$$3 : \left\{ \begin{array} { l } 
1 : \text { for correct derivative } \\
1 : \text { for critical value for } f ^ { \prime } \\
1 : \text { for identifying critical } \\
\text { point as relative maximum }
\end{array} \right.$$

(c)

$$2 : \begin{cases} 1 : & \text { for correct } f ^ { \prime \prime } ( x ) \text { for } f ^ { \prime } ( x ) \\ & \text { found in } ( b ) \\ 1 : & \text { for correct interval } \end{cases}$$

(d) 3: for graph consistent with information found in (a), (b), and (c)\\