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\documentclass[12pt,thmsa]{article}
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\usepackage{sw20jart}

%TCIDATA{TCIstyle=article/art4.lat,jart,sw20jart}

%TCIDATA{<META NAME="ViewSettings" CONTENT="17">}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{Created=Sat Nov 16 20:56:01 1996}
%TCIDATA{LastRevised=Thu Feb 13 11:05:09 1997}
%TCIDATA{Language=American English}
%TCIDATA{CSTFile=Exam.cst}

\input{tcilatex}
\begin{document}


\section{Quiz}

\subsection{Text}

\section{Calculus, 2nd Semester}

Your Name: 
%TCIMACRO{\HTML{<Input TYPE="Text" NAME="STUDENTNAME" VALUE=" " SIZE=0>}}
%BeginExpansion
%%
%EndExpansion

\subsection{\protect\vspace{1pt}Centroids}

The \textit{centroid} of a region $R$ is a generalization of the \textit{%
center of mass }of \ a finite collection of objects. If $R$ is a region in
the plane with area $A$ then the coordinates of the centroid are given by $%
\left( \bar{x},\bar{y}\right) $ where

\begin{eqnarray*}
\bar{x} &=&\frac{1}{A}\int_{R}xdA \\
\bar{y} &=&\frac{1}{A}\int_{R}ydA
\end{eqnarray*}

\hrulefill \bigskip

\subsubsection{\protect\vspace{1pt}Setup}

Choice Space: 2

Choices: No Break, Radio buttons, Permute

Title:Centroids

\section{Question}

\subsection{Setup}

$a:=\limfunc{rand}(1,4)$

\subsection{Statement}

Find the centroid of the region in the first quadrant bounded by the axes
and the curve $y=%
%TCIMACRO{\FORMULA{a^{2}}{a^{2}}{evaluate}}
%BeginExpansion
a^{2}%
%EndExpansion
-x^{2}$.

\subsection{Choices}

\begin{itemize}
\item  $\left( 
%TCIMACRO{\FORMULA{\frac{3a}{8}}{\frac{3}{8}a}{evaluate}}
%BeginExpansion
\frac{3}{8}a%
%EndExpansion
,%
%TCIMACRO{\FORMULA{\frac{2a}{5}}{\frac{2}{5}a}{evaluate}}
%BeginExpansion
\frac{2}{5}a%
%EndExpansion
\right) \correctchoice{}$

\item  $\left( 
%TCIMACRO{\FORMULA{\frac{a}{2}}{\frac{1}{2}a}{evaluate}}
%BeginExpansion
\frac{1}{2}a%
%EndExpansion
,%
%TCIMACRO{\FORMULA{\frac{a^{2}}{2}}{\frac{1}{2}a^{2}}{evaluate}}
%BeginExpansion
\frac{1}{2}a^{2}%
%EndExpansion
\right) $

\item  $\left( 
%TCIMACRO{\FORMULA{\frac{a}{2}}{\frac{1}{2}a}{evaluate}}
%BeginExpansion
\frac{1}{2}a%
%EndExpansion
,%
%TCIMACRO{\FORMULA{\frac{2a}{5}}{\frac{2}{5}a}{evaluate}}
%BeginExpansion
\frac{2}{5}a%
%EndExpansion
\right) $

\item  $\left( 
%TCIMACRO{\FORMULA{\frac{3a}{8}}{\frac{3}{8}a}{evaluate}}
%BeginExpansion
\frac{3}{8}a%
%EndExpansion
,%
%TCIMACRO{\FORMULA{\frac{a^{2}}{2}}{\frac{1}{2}a^{2}}{evaluate}}
%BeginExpansion
\frac{1}{2}a^{2}%
%EndExpansion
\right) $
\end{itemize}

\subsection{Answer}

$\left( 
%TCIMACRO{\FORMULA{\frac{3a}{8}}{\frac{3}{8}a}{evaluate}}
%BeginExpansion
\frac{3}{8}a%
%EndExpansion
,%
%TCIMACRO{\FORMULA{\frac{2a}{5}}{\frac{2}{5}a}{evaluate}}
%BeginExpansion
\frac{2}{5}a%
%EndExpansion
\right) $

\subsection{Solution}

\begin{enumerate}
\item  The curve intersects the axes at $\left( 0,%
%TCIMACRO{\FORMULA{a^{2}}{a^{2}}{evaluate}}
%BeginExpansion
a^{2}%
%EndExpansion
\right) $and $\left( 
%TCIMACRO{\FORMULA{a}{a}{evaluate}}
%BeginExpansion
a%
%EndExpansion
,0\right) $.

\item  The area of the enclosed region is $A=\int_{0}^{%
%TCIMACRO{\FORMULA{a}{a}{evaluate}}
%BeginExpansion
a%
%EndExpansion
}%
%TCIMACRO{\FORMULA{a^{2}}{a^{2}}{evaluate}}
%BeginExpansion
a^{2}%
%EndExpansion
-x^{2}dx=%
%TCIMACRO{
%\FORMULA{\int_{0}^{a}\left( a^{2}-x^{2}\right) dx}{\frac{2}{3}a^{3}}{evaluate}}
%BeginExpansion
\frac{2}{3}a^{3}%
%EndExpansion
$

\item  Then $\bar{x}=\frac{\int xdA}{A}=\frac{\int_{0}^{%
%TCIMACRO{\FORMULA{a}{a}{evaluate}}
%BeginExpansion
a%
%EndExpansion
}x\left( 
%TCIMACRO{\FORMULA{a^{2}}{a^{2}}{evaluate}}
%BeginExpansion
a^{2}%
%EndExpansion
-x^{2}\right) dx}{%
%TCIMACRO{
%\FORMULA{\int_{0}^{a}\left( a^{2}-x^{2}\right) dx}{\frac{2}{3}a^{3}}{evaluate}}
%BeginExpansion
\frac{2}{3}a^{3}%
%EndExpansion
}=\frac{%
%TCIMACRO{
%\FORMULA{\int_{0}^{a}x\left( a^{2}-x^{2}\right) dx}{\frac{1}{4}a^{4}}{evaluate}}
%BeginExpansion
\frac{1}{4}a^{4}%
%EndExpansion
}{%
%TCIMACRO{
%\FORMULA{\int_{0}^{a}\left( a^{2}-x^{2}\right) dx}{\frac{2}{3}a^{3}}{evaluate}}
%BeginExpansion
\frac{2}{3}a^{3}%
%EndExpansion
}=%
%TCIMACRO{\FORMULA{\frac{3a}{8}}{\frac{3}{8}a}{evaluate}}
%BeginExpansion
\frac{3}{8}a%
%EndExpansion
$

\item  And $\bar{y}=\frac{\int ydA}{A}=\frac{\int_{0}^{%
%TCIMACRO{\FORMULA{a^{2}}{a^{2}}{evaluate}}
%BeginExpansion
a^{2}%
%EndExpansion
}y\sqrt{%
%TCIMACRO{\FORMULA{a^{2}}{a^{2}}{evaluate}}
%BeginExpansion
a^{2}%
%EndExpansion
-y}dy}{%
%TCIMACRO{
%\FORMULA{\int_{0}^{a}\left( a^{2}-x^{2}\right) dx}{\frac{2}{3}a^{3}}{evaluate}}
%BeginExpansion
\frac{2}{3}a^{3}%
%EndExpansion
}=\frac{%
%TCIMACRO{
%\FORMULA{\int_{0}^{a^{2}}y\sqrt{a^{2}-y}dy}{\frac{4}{15}\func{csgn}\left( a\right) a^{5}}{evaluate}}
%BeginExpansion
\frac{4}{15}\func{csgn}\left( a\right) a^{5}%
%EndExpansion
}{%
%TCIMACRO{
%\FORMULA{\int_{0}^{a}\left( a^{2}-x^{2}\right) dx}{\frac{2}{3}a^{3}}{evaluate}}
%BeginExpansion
\frac{2}{3}a^{3}%
%EndExpansion
}=%
%TCIMACRO{\FORMULA{\frac{2a^{2}}{5}}{\frac{2}{5}a^{2}}{evaluate}}
%BeginExpansion
\frac{2}{5}a^{2}%
%EndExpansion
$
\end{enumerate}

\section{Question}

\subsection{Setup}

$a:=\limfunc{rand}(1,4)$

\subsection{Statement}

Find the centroid of the region in the first quadrant bounded by the curve $%
y=x^{3}$ and the line $y=%
%TCIMACRO{\FORMULA{a^{2}x}{a^{2}x}{evaluate}}
%BeginExpansion
a^{2}x%
%EndExpansion
$.

\subsection{Choices}

\begin{itemize}
\item  $\left( 
%TCIMACRO{\FORMULA{\frac{8a}{15}}{\frac{8}{15}a}{evaluate}}
%BeginExpansion
\frac{8}{15}a%
%EndExpansion
,%
%TCIMACRO{\FORMULA{\frac{a^{3}}{2}}{\frac{1}{2}a^{3}}{evaluate}}
%BeginExpansion
\frac{1}{2}a^{3}%
%EndExpansion
\right) $

\item  $\left( 
%TCIMACRO{\FORMULA{\frac{a}{2}}{\frac{1}{2}a}{evaluate}}
%BeginExpansion
\frac{1}{2}a%
%EndExpansion
,%
%TCIMACRO{\FORMULA{\frac{a^{3}}{2}}{\frac{1}{2}a^{3}}{evaluate}}
%BeginExpansion
\frac{1}{2}a^{3}%
%EndExpansion
\right) $

\item  $\left( 
%TCIMACRO{\FORMULA{\frac{a}{2}}{\frac{1}{2}a}{evaluate}}
%BeginExpansion
\frac{1}{2}a%
%EndExpansion
,%
%TCIMACRO{\FORMULA{\frac{2a^{3}}{21}}{\frac{2}{21}a^{3}}{evaluate}}
%BeginExpansion
\frac{2}{21}a^{3}%
%EndExpansion
\right) $

\item  $\left( 
%TCIMACRO{\FORMULA{\frac{8a}{15}}{\frac{8}{15}a}{evaluate}}
%BeginExpansion
\frac{8}{15}a%
%EndExpansion
,%
%TCIMACRO{\FORMULA{\frac{2a^{3}}{21}}{\frac{2}{21}a^{3}}{evaluate}}
%BeginExpansion
\frac{2}{21}a^{3}%
%EndExpansion
\right) \correctchoice{}$
\end{itemize}

\subsection{Answer}

$\left( 
%TCIMACRO{\FORMULA{\frac{8a}{15}}{\frac{8}{15}a}{evaluate}}
%BeginExpansion
\frac{8}{15}a%
%EndExpansion
,%
%TCIMACRO{\FORMULA{\frac{2a^{3}}{21}}{\frac{2}{21}a^{3}}{evaluate}}
%BeginExpansion
\frac{2}{21}a^{3}%
%EndExpansion
\right) $

\subsection{Solution}

\begin{enumerate}
\item  The curves intersect at $\left( 0,0\right) $ and $\left( 
%TCIMACRO{\FORMULA{a}{a}{evaluate}}
%BeginExpansion
a%
%EndExpansion
,%
%TCIMACRO{\FORMULA{a^{3}}{a^{3}}{evaluate}}
%BeginExpansion
a^{3}%
%EndExpansion
\right) $.

\item  The area of the enclosed region is $A=\int_{0}^{%
%TCIMACRO{\FORMULA{a}{a}{evaluate}}
%BeginExpansion
a%
%EndExpansion
}\left( 
%TCIMACRO{\FORMULA{a^{2}x}{a^{2}x}{evaluate}}
%BeginExpansion
a^{2}x%
%EndExpansion
-x^{3}\right) dx=\allowbreak \frac{1}{4}a^{4}$

\item  Then $x=\frac{\int xdA}{A}=\frac{\int_{0}^{%
%TCIMACRO{\FORMULA{a}{a}{evaluate}}
%BeginExpansion
a%
%EndExpansion
}x\left( 
%TCIMACRO{\FORMULA{a^{2}x}{a^{2}x}{evaluate}}
%BeginExpansion
a^{2}x%
%EndExpansion
-x^{3}\right) dx}{%
%TCIMACRO{
%\FORMULA{\int_{0}^{\FORMULA{a}{a}{evaluate}}\left( \FORMULA{a^{2}x}{a^{2}x}{evaluate}-x^{3}\right) dx}{\frac{1}{4}a^{4}}{evaluate}}
%BeginExpansion
\frac{1}{4}a^{4}%
%EndExpansion
}=\frac{%
%TCIMACRO{
%\FORMULA{\int_{0}^{a}x\left( a^{2}x-x^{3}\right) dx}{\frac{2}{15}a^{5}}{evaluate}}
%BeginExpansion
\frac{2}{15}a^{5}%
%EndExpansion
}{%
%TCIMACRO{
%\FORMULA{\int_{0}^{\FORMULA{a}{a}{evaluate}}\left( \FORMULA{a^{2}x}{a^{2}x}{evaluate}-x^{3}\right) dx}{\frac{1}{4}a^{4}}{evaluate}}
%BeginExpansion
\frac{1}{4}a^{4}%
%EndExpansion
}=%
%TCIMACRO{\FORMULA{\frac{8a}{15}}{\frac{8}{15}a}{evaluate}}
%BeginExpansion
\frac{8}{15}a%
%EndExpansion
$

\item  And $\bar{y}=\frac{\int ydA}{A}=\frac{\int_{0}^{%
%TCIMACRO{\FORMULA{a^{3}}{a^{3}}{evaluate}}
%BeginExpansion
a^{3}%
%EndExpansion
}y\left( y^{\frac{1}{3}}-\frac{y}{a^{2}}\right) dy}{%
%TCIMACRO{
%\FORMULA{\int_{0}^{\FORMULA{a}{a}{evaluate}}\left( \FORMULA{a^{2}x}{a^{2}x}{evaluate}-x^{3}\right) dx}{\frac{1}{4}a^{4}}{evaluate}}
%BeginExpansion
\frac{1}{4}a^{4}%
%EndExpansion
}=\frac{%
%TCIMACRO{
%\FORMULA{\int_{0}^{a^{3}}y\left( y^{\frac{1}{3}}-\frac{y}{a^{2}}\right) dy}{\frac{3}{7}a^{6}\sqrt[3]{\left( a^{3}\right) }-\frac{1}{3}a^{7}}{evaluate}}
%BeginExpansion
\frac{3}{7}a^{6}\sqrt[3]{\left( a^{3}\right) }-\frac{1}{3}a^{7}%
%EndExpansion
}{%
%TCIMACRO{
%\FORMULA{\int_{0}^{\FORMULA{a}{a}{evaluate}}\left( \FORMULA{a^{2}x}{a^{2}x}{evaluate}-x^{3}\right) dx}{\frac{1}{4}a^{4}}{evaluate}}
%BeginExpansion
\frac{1}{4}a^{4}%
%EndExpansion
}=%
%TCIMACRO{\FORMULA{\frac{2a^{3}}{21}}{\frac{2}{21}a^{3}}{evaluate}}
%BeginExpansion
\frac{2}{21}a^{3}%
%EndExpansion
$
\end{enumerate}

\end{document}
