\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 IIA Solutions\hfill 4/4/05} \subsection*{{\protect\normalsize Name: \protect\underline{\hspace{2.5in}} \hfill }} \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\ \ Total Score \ \ \ \ \ \ \ \ \ \ \ \ \ \ \_\_\_\_\_\_\_\_\bigskip \pagebreak \begin{description} \item[{1 [$25$ pts.]}] Give two double integral expressions for the area of the region $R$ bounded by the parabola $x=-y^{2}$ and the line \end{description} $y=x+2.$ Do not evaluate these integrals. Be sure to sketch $R.$ Solution: $x=-y^{2}$\FRAME{dtbpFX}{4.4998in}{2.9999in}{0pt}{}{}{Plot}{\special% {language "Scientific Word";type "MAPLEPLOT";width 4.4998in;height 2.9999in;depth 0pt;display "USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine "Maple";xmin "-5";xmax "0";ymin "-5";ymax "5";xviewmin "-5.1";xviewmax "0.102";yviewmin "-3.1047138047138";yviewmax "2.34249831649832";plottype 12;num-x-gridlines 24;num-y-gridlines 24;plotstyle "patch";axesstyle "normal";xis \TEXUX{x};yis \TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};function \TEXUX{$x=-y^{2}$};linecolor "black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";var1range "-5,0";var2range "-5,5";num-x-gridlines 73;num-y-gridlines 73;curveColor "[flat::RGB:0000000000]";curveStyle "Line";rangeset"X";function \TEXUX{$y=x+2$};linecolor "black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";var1range "-5,0";var2range "-5,5";num-x-gridlines 24;num-y-gridlines 24;curveColor "[flat::RGB:0000000000]";curveStyle "Line";rangeset"X";}} The curves intersect when $y=-y^{2}+2$ or $y^{2}+y-2=\left( y+2\right) \left( y-1\right) =0,$ that is at $y=-2$ and $y=1.$ Hence at the points $% \left( -4,-2\right) $ and $\left( -1,1\right) $ \vspace{1pt} \begin{eqnarray*} \text{Area} &=&\int_{-4}^{-1}\int_{-\sqrt{-x}}^{x+2}dydx+\int_{-1}^{0}\int_{-% \sqrt{-x}}^{\sqrt{-x}}dydx \\ &=&\int_{-2}^{1}\int_{y-2}^{-y^{2}}dxdy=\frac{9}{2} \end{eqnarray*} \begin{description} \item[{2 a $\left[ 20\text{ pts.}\right] $}] Find the area of the ellipse cut from the plane $z=2x+2y+1$ by the cylinder $x^{2}+y^{2}=4.$ \end{description} Solution:% \begin{equation*} \text{Surface Area }=\diint\limits_{R}\sqrt{1+z_{x}^{2}+z_{y}^{2}}dA \end{equation*} $R$ is the circle centered at the origin of radius $2.$ Now $z_{x}=2$ and $% z_{y}=2$ so% \begin{equation*} \text{Surface Area}=\diint\limits_{x^{2}+y^{2}\leq 4}\sqrt{1+4+4}% dA=3\diint\limits_{x^{2}+y^{2}\leq 4}dA=3\pi \left( 2^{2}\right) =12\pi \end{equation*} \begin{description} \item[{2 b $\left[ 20\text{ pts.}\right] $}] Give a double integral in polar coordinates for the area enclosed by one petal of the rose \begin{equation*} r=\sin 3\theta \end{equation*} \end{description} Sketch the part of the petal in the first quadrant. \ Do \textit{not} evaluate this integral. \vspace{1pt}Solution: $\frac{\pi }{3}=\allowbreak 1.\,\allowbreak 047\,197\,551\,\allowbreak 197$ $% \ \ \ \ \ \ \ \ \ \ \ \ \sin 3\theta $ \FRAME{dtbpFX}{4.4998in}{2.9999in}{0pt}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 4.4998in;height 2.9999in;depth 0pt;display "USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine "Maple";xmin "0";xmax "1.05";xviewmin "-0.021868487075261E0";xviewmax "0.898119996639583";yviewmin "-0.018688226783418E0";yviewmax "0.574110112919307";plottype 8;numpoints 100;plotstyle "patch";axesstyle "normal";xis \TEXUX{v58130};var1name \TEXUX{$\theta $};function \TEXUX{$\sin 3\theta $};linecolor "black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";coordinateSystem "polar";var1range "0,1.05";num-x-gridlines 100;curveColor "[flat::RGB:0000000000]";curveStyle "Line";rangeset"X";}} Now $\sin 3\theta =0$ implies that $3\theta =\pi $ or $\theta =\pi /3.$ Thus% \begin{equation*} \text{Area}=\int \int dA=\int_{0}^{\frac{\pi }{3}}\int_{0}^{\sin 3\theta }rdrd\theta \end{equation*} \begin{description} \item[{3 $\left[ 20\text{ pts.}\right] $}] Let $V$ be the volume bounded by the paraboloids $z=x^{2}+y^{2}$ and $z=4-x^{2}-y^{2}.$Give a triple integral expression \end{description} in \textbf{cylindrical} coordinates for the volume of $V.$ Sketch $V.$ \textit{Do not evaluate the integral you give}$.$ Solution: $r^{2}$\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 "MuPAD";xmin "-3.1416";xmax "3.1416";ymin "0";ymax "1.5";xviewmin "-1.50299999995948";xviewmax "1.50299999999996";yviewmin "-1.50299999998986";yviewmax "1.50299999998986";zviewmin "-0.004";zviewmax "4.004";phi 45;theta 45;plottype 14;num-x-gridlines 25;num-y-gridlines 25;plotstyle "patch";axesstyle "normal";plotshading "XYZ";xis \TEXUX{v58130};yis \TEXUX{z};zis \TEXUX{r};var1name \TEXUX{$\theta $};var2name \TEXUX{$z$};var3name \TEXUX{$r$};function \TEXUX{$r^{2}$};linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";coordinateSystem "cylindrical";var1range "-3.1416,3.1416";var2range "0,1.5";surfaceColor "[linear:XYZ:RGB:0x00ff0000:0x000000ff]";surfaceStyle "Color Patch";num-x-gridlines 25;num-y-gridlines 25;surfaceMesh "Mesh";rangeset"Y";function \TEXUX{$4-r^{2}$};linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";coordinateSystem "cylindrical";var1range "-3.1416,3.1416";var2range "0,1.5";surfaceColor "[linear:XYZ:RGB:0x00ffff80:0x0000ffff]";surfaceStyle "Color Patch";num-x-gridlines 25;num-y-gridlines 25;surfaceMesh "Mesh";rangeset"Y";}% } The paraboloids intersect when $x^{2}+y^{2}=4-x^{2}-y^{2},$ that is the circle $x^{2}+y^{2}=2$ Thus since $z$ goes from the paraboloid $z=x^{2}+y^{2}=r^{2}$ to the paraboloid $z=4-\left( x^{2}+y^{2}\right) =4-r^{2}$ \vspace{1pt} \begin{equation*} \text{Volume}=\diiint\limits_{V}dV=\int_{0}^{2\pi }\int_{0}^{\sqrt{2}% }\int_{r^{2}}^{4-r^{2}}rdzdrd\theta =4\pi \end{equation*} \begin{description} \item[{4 [$15$ pts.]}] Give an integral expression in \textbf{spherical} coordinates for the volume of the solid that lies above the cone \end{description} $\phi =\frac{\pi }{6}$ and below the sphere $\rho =2a\cos \phi .$ Sketch the volume. \textit{Do not evaluate this expression.} \vspace{1pt}Solution: \vspace{1pt}$\left( \rho ,\theta ,\frac{\pi }{6}\right) $\FRAME{dtbpFX}{% 4.4998in}{2.9999in}{0pt}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 4.4998in;height 2.9999in;depth 0pt;display "USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine "Maple";xmin "-3.1416";xmax "3.1416";ymin "0";ymax "3.1416";xviewmin "-1.63363199995676";xviewmax "1.63488863999914";yviewmin "-1.63363199998898";yviewmax "1.63488863998897";zviewmin "-0.054414108170584E0";zviewmax "2.77620779886319";phi 82;theta 31;plottype 15;constrained TRUE;num-x-gridlines 25;num-y-gridlines 25;plotstyle "wireframe";axesstyle "normal";plotshading "NONE";xis \TEXUX{v58130};yis \TEXUX{v58144};var1name \TEXUX{$\theta $};var2name \TEXUX{$\phi $};function \TEXUX{$\left( \rho ,\theta ,\frac{\pi }{6}\right) $};linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";coordinateSystem "spherical";var1range "-3.1416,3.1416";var2range "0,3.1416";surfaceColor "[flat::RGB:0x00ff0000:0x000000ff]";surfaceStyle "Wire Frame";num-x-gridlines 25;num-y-gridlines 25;surfaceMesh "Mesh";rangeset"Y";function \TEXUX{$2\cos \phi $};linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";coordinateSystem "spherical";var1range "-3.1416,3.1416";var2range "0,3.1416";surfaceColor "[flat::RGB:0x00ff0000:0x000000ff]";surfaceStyle "Wire Frame";num-x-gridlines 25;num-y-gridlines 25;surfaceMesh "Mesh";}} The volume is given by% \begin{equation*} \text{Volume}=\diiint\limits_{V}dV=\int_{0}^{2\pi }\int_{0}^{\frac{\pi }{6}% }\int_{0}^{2a\cos \phi }\rho ^{2}\sin \phi d\rho d\phi d\theta \end{equation*} \pagebreak \section{\protect\vspace{1pt}Table of Integrals} \begin{center} \vspace{1pt} \begin{tabular}{|l|} \hline $\int \sin ^{2}axdx=\allowbreak \frac{1}{a}\left( -\frac{1}{2}\cos ax\sin ax+% \frac{1}{2}ax\right) +C$ \\ \hline $\int \cos ^{2}axdx=\allowbreak \frac{1}{a}\left( \frac{1}{2}\cos ax\sin ax+% \frac{1}{2}ax\right) +C$ \\ \hline $\int \sin ^{3}axdx=\allowbreak \frac{1}{a}\left( -\frac{1}{3}\sin ^{2}ax\cos ax-\frac{2}{3}\cos ax\right) +C$ \\ \hline $\int \cos ^{3}axdx=\allowbreak \frac{1}{a}\left( \frac{1}{3}\cos ^{2}ax\sin ax+\frac{2}{3}\sin ax\right) +C$ \\ \hline \\ \hline \end{tabular} \end{center} \end{document} %%%%%%%%%%%%%% End /document/ma227_exam_2a_05s_sol_v3.tex %%%%%%%%%%%%%