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{\Large {\bf 
Mathematics 1210-2,3 \hspace{5ex}  Prof. Ken Golden\hspace{5ex} \\
PRACTICE FINAL EXAM ANSWER KEY
\hspace{5ex}Fall 2009 \\}
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%{\Large {\bf ANSWER KEY}}

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\begin{enumerate}

\item[1.] 
$\frac{103}{6} = 17 \frac{1}{6} $


\item[2.] 
$V=\frac{4}{3} \pi ab^2$

\item[3.] $V = \frac{\pi}{2} $ 


\item[4.] $\frac{\sqrt 5 \pi}{2}$


\item[5.] $\frac{2}{5} (2^{5/2} -1)$



\item[6.] $W = \frac{3^8}{56} \sim 117.16$ inch-pounds.


\item[7.] $W= 32$ dynes.


\item[8.] 
$\int_{x=0}^{x=1} 2 \pi x (x^3) dx = 2 \pi |_0^1 \frac{x^5}{5} = \frac{2
\pi}{5}$

\bigskip

The centroid of the region is 
$ (\bar{x} , \bar{y} ) = ( \frac{4}{5}, \frac{2}{7})$.
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.


\item[9.] $\ds T(x) = -2x+4$


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\end{enumerate}

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