\thepage } %} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \subsection*{Ma 227 \hspace{2.25in} Exam IIB \ \ Solutions\hfill 4/5/04} \subsection*{{\protect\normalsize Name: \protect\underline{\hspace{2.5in}} \hfill ID: \protect\underline{\hspace{2.0in}}}} \subsection*{{\protect\normalsize Lecture Section: \_\_\_\_\_\_\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\ \ \ \ \ \ \ \ \ \ \ Recitation Section: \_\_\_\_\hspace{2.5in}\hfill }} \hfill {\small \textit{I pledge my honor that I have abided by the Stevens Honor System.}\hspace{2pt} \ \ \ \ \ \ \ \ \ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_% \_\_\_\_} \hfill \paragraph{You may not use a calculator, cell phone, or computer while taking this exam. All work must be shown to obtain full credit. Credit will not be given for work not reasonably supported. When you finish, be sure to sign the pledge.} \vspace{1pt} Score on Problem \ \#1 \_\_\_\_\_\_\_\_\vspace{0.2cm} \qquad \qquad \qquad \qquad\ \ \ \#2 \_\_\_\_\_\_\_\_\vspace{0.2cm} \qquad \qquad \qquad \qquad\ \ \ \#3 \_\_\_\_\_\_\_\_\vspace{0.2cm} \qquad \qquad \qquad \qquad\ \ \ \#4 \_\_\_\_\_\_\_\_ \qquad \qquad \qquad \qquad\ \ \qquad \qquad \qquad \qquad\ \ \qquad \qquad \qquad \qquad\ \ \qquad \qquad \qquad \qquad\ \ \qquad \qquad \qquad \qquad\ \ Total Score \ \ \ \ \ \ \ \ \ \ \ \ \ \ \_\_\_\_\_\_\_\_\bigskip \pagebreak \begin{description} \item[{1 [$25$ pts.]}] Give two expressions with two different orders of integration for \begin{equation*} \diint\limits_{R}xydA \end{equation*} \end{description} where $R$ is the region bounded by the lines $x=0,y=2x,$ and $y=4.$ Be sure to sketch $R.$ Evaluate \textit{one} of these expressions. Solution: $R$ is the triangular region shown below. $x$\FRAME{dtbpFX}{4.4996in}{3in}{0pt}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 4.4996in;height 3in;depth 0pt;display "USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine "Maple";xmin "0";xmax "2";xviewmin "-0.04";xviewmax "2.0408";yviewmin "-0.08";yviewmax "4.0816";plottype 4;constrained TRUE;numpoints 100;plotstyle "patch";axesstyle "normal";xis \TEXUX{x};yis \TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};function \TEXUX{$2x$};linecolor "black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";var1range "0,2";num-x-gridlines 100;curveColor "[flat::RGB:0000000000]";curveStyle "Line";rangeset"X";function \TEXUX{$\left( 0,4,2,4\right) $};linecolor "black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";curveColor "[flat::RGB:0000000000]";curveStyle "Line";function \TEXUX{$\left( 0,0,0,4\right) $};linecolor "black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";curveColor "[flat::RGB:0000000000]";curveStyle "Line";}} The lines $y=2x$ and $y=4$ intersect when $2x=4$, i.e. $x=2.$Therefore, we have% \begin{equation*} \int_{0}^{4}\int_{0}^{\frac{y}{2}}xydxdy=\int_{0}^{4}\left. \frac{x^{2}}{2}% \right] _{0}^{\frac{y}{2}}ydy=\int_{0}^{4}\left( \frac{y^{3}}{8}\right) dy=\left. \frac{y^{4}}{32}\right] _{0}^{4}=8 \end{equation*} or% \begin{equation*} \int_{0}^{2}\int_{2x}^{4}xydydx=\int_{0}^{2}\left. \frac{y^{2}}{2}\right] _{2x}^{4}xdx=\int_{0}^{2}\left( 8x-2x^{3}\right) dx=\left. 4x^{2}-\frac{x^{4}% }{2}\right] _{0}^{2}=\left( 16-8\right) =8 \end{equation*} \pagebreak \begin{description} \item[{2 a $\left[ 20\text{ pts.}\right] $}] Give two different triple integral expressions for the volume of the region beneath the graph with equation $z=\frac{1}{2}x^{2}+\frac{1}{3}y^{2}$ and above the triangular region $R$ in the $x,y-$ plane with vertices $\left( 0,0,0\right) ,\left( 1,1,0\right) ,\left( 2,0,0\right) .$ Be sure to sketch $R.$ Do not evaluate this expression\textit{.} \end{description} Solution: The region $R$ is shown below. $\left( 0,0,1,1,1,1,2,0,0,0\right) $\FRAME{dtbpFX}{4.0145in}{3.0113in}{0pt}{% }{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 4.0145in;height 3.0113in;depth 0pt;display "USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine "Maple";xmin "-5";xmax "5";xviewmin "-0.04";xviewmax "2.0408";yviewmin "-0.02";yviewmax "1.0204";plottype 4;numpoints 49;plotstyle "patch";axesstyle "normal";xis \TEXUX{x};var1name \TEXUX{$x$};function \TEXUX{$\left( 0,0,1,1,1,1,2,0,0,0\right) $};linecolor "black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";curveColor "[flat::RGB:0000000000]";curveStyle "Line";}} We need the equations of the lines that bound $R.$ The line through $\left( 0,0\right) $ and $\left( 1,1\right) $ is $y=x.$ The line through $\left( 2,0\right) $ and $\left( 1,1\right) $ is $y=-x+2.$ Therefore% \begin{eqnarray*} \text{volume} &=&\int_{0}^{1}\int_{y}^{-y+2}\int_{0}^{\frac{1}{2}x^{2}+\frac{% 1}{3}y^{2}}dzdxdy \\ &=&\int_{0}^{1}\int_{0}^{x}\int_{0}^{\frac{1}{2}x^{2}+\frac{1}{3}% y^{2}}dzdydx+\int_{1}^{2}\int_{0}^{-x+2}\int_{0}^{\frac{1}{2}x^{2}+\frac{1}{3% }y^{2}}dzdydx \end{eqnarray*} \vspace{1pt} \textbf{Finished to here!Finished to here!Finished to here!Finished to here!Finished to here!Finished to here!Finished to here!Finished to here!} \textbf{\vspace{1pt}} \begin{description} \item[{2 b $\left[ 20\text{ pts.}\right] $}] Consider the integral% \begin{equation*} \int_{0}^{2}\int_{0}^{\sqrt{2y-y^{2}}}y\left( x^{2}+y^{2}\right) ^{\frac{1}{2% }}dxdy \end{equation*} \end{description} Sketch the region of integration and then convert the integral to polar coordinates. \textit{Do} \textit{not evaluate.} Solution: Note that $x$ goes from $0$ to $\sqrt{2y-y^{2}}.$ Thus $x=\sqrt{% 2y-y^{2}}$ or $x^{2}=2y-y^{2}$ so we have the upper half of the circle% \begin{equation*} x^{2}+y^{2}-2y=0 \end{equation*} or% \begin{equation*} x^{2}+(y-1)^{2}=1 \end{equation*} which is the right half of the circle of radius $1$ with center at $\left( 0,1\right) $. This region is shown below. \ \ \vspace{1pt} \FRAME{dtbpFX}{3.9998in}{3.9998in}{0pt}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 3.9998in;height 3.9998in;depth 0pt;display "USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine "Maple";xmin "0";xmax "1.5708";xviewmin "-0.020007469406702E0";xviewmax "1.02039892887099";yviewmin "-0.039999999999460E0";yviewmax "2.04079999997247";plottype 8;constrained TRUE;numpoints 49;plotstyle "patch";axesstyle "normal";xis \TEXUX{v952};var1name \TEXUX{$\theta $};function \TEXUX{$2\sin \theta $};linecolor "black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";coordinateSystem "polar";var1range "0,1.5708";num-x-gridlines 49;curveColor "[flat::RGB:0000000000]";curveStyle "Line";rangeset"X";}} \vspace{1pt} In polar coordinates the equation of the circle is% \begin{equation*} x^{2}+y^{2}=r^{2}=2y=2r\sin \theta \end{equation*} or% \begin{equation*} r=2\cos \theta \end{equation*} Thus% \begin{equation*} \int_{0}^{2}\int_{0}^{\sqrt{2x-x^{2}}}y\left( x^{2}+y^{2}\right) ^{\frac{1}{2% }}dydx=\int_{0}^{\frac{\pi }{2}}\int_{0}^{2\sin \theta }r\sin \theta \left( r\right) rdrd\theta =\int_{0}^{\frac{\pi }{2}}\int_{0}^{2\sin \theta }r^{3}\sin \theta drd\theta \end{equation*} \vspace{1pt} \begin{description} \item[{3 $\left[ 20\text{ pts.}\right] $}] Let $V$ be the volume in the first octant that is bounded by the cylinder $x^{2}+y^{2}=2x,$ the paraboloid $% z=x^{2}+y^{2}$ and \end{description} the $x,y-$plane. Give a triple integral expression in \textbf{cylindrical} coordinates for the volume of $V.$ Sketch $V.$ \textit{Do not evaluate the integral you give.} Solution: The equation of the paraboloid is $z=r^{2},$ and the equation of the circle is $r=2\cos \theta .$ Thus the graph is below. (Everything in front of the y-z plane is shown, althought the region of interest is just in the first octant.) $\left( r,\theta ,r\right) $\FRAME{dtbpFX}{4.4996in}{3in}{0pt}{}{}{Plot}{% \special{language "Scientific Word";type "MAPLEPLOT";width 4.4996in;height 3in;depth 0pt;display "USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine "Maple";xmin "0";xmax "2";ymin "-1.6";ymax "1.6";xviewmin "-0.099567025494629E0";xviewmax "2.04199134050989";yviewmin "-2.07911309432633";yviewmax "2.0807124120912";zviewmin "-0.08";zviewmax "4.0816";phi 66;theta 25;plottype 14;constrained TRUE;num-x-gridlines 25;num-y-gridlines 25;plotstyle "patch";axesstyle "normal";plotshading "XYZ";xis \TEXUX{z};yis \TEXUX{v952};var1name \TEXUX{$z$};var2name \TEXUX{$\theta $};function \TEXUX{$\left( r,\theta ,r^{2}\right) $};linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";coordinateSystem "cylindrical";var1range "0,2";var2range "-1.6,1.6";surfaceColor "[linear:XYZ:RGB:0x00ff0000:0x000000ff]";surfaceStyle "Color Patch";num-x-gridlines 25;num-y-gridlines 25;surfaceMesh "Mesh";rangeset"XY";function \TEXUX{$\left( 2\cos \theta ,\theta ,z\right) $};linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";coordinateSystem "cylindrical";var1range "0,4";var2range "0,4";surfaceColor "[linear:XYZ:RGB:0x00ff0000:0x000000ff]";surfaceStyle "Color Patch";num-x-gridlines 25;num-y-gridlines 25;surfaceMesh "Mesh";rangeset"XY";}} Thus \begin{equation*} \text{volume}=\diiint\limits_{V}dV=\int_{0}^{\frac{\pi }{2}}\int_{0}^{2\cos \theta }\int_{0}^{r^{2}}rdzdrd\theta \end{equation*} \begin{description} \item[{4 [$15$ pts.]}] Let S be a sphere of radius $a$ centered at the origin. Give an integral expression in \textbf{spherical} coordinates for the volume of half of S$.$ \textit{Do not evaluate this\ expression. }Which hemisphere have you chosen? \ (For example, "above x-y plane, bounded by y-z plane with x negative, etc.)\textit{.} \end{description} Solution: In spherical coordinates% \begin{equation*} \text{volume}=\diiint\limits_{V}\rho ^{2}\sin \phi dV \end{equation*}% so% \begin{equation*} \int_{0}^{2\pi }\int_{0}^{\frac{\pi }{2}}\int_{0}^{a}\rho ^{2}\sin \phi d\rho d\phi d\theta \end{equation*} describes the volume of the "top hal\'{f}", i.e ., the protion abouve the x-y plane. \ Another choice might be the "right half", that is the portion bounded by the x-z plane with y positive, described by \begin{equation*} \int_{0}^{\pi }\int_{0}^{\pi }\int_{0}^{a}\rho ^{2}\sin \phi d\rho d\phi d\theta \end{equation*} \vspace{1pt} \end{document} %%%%%%%%%%%%%% End /document/ma227_exam_2b_04s_v2_sol.tex %%%%%%%%%%%%%