\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 II \ \ \ Solutions\hfill 4/2/03} \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 {\small \textbf{SHOW ALL WORK! You may not use a calculator on this exam.}} \begin{description} \item[{1 [25 pts.]}] Find a \textit{particular} solution of \begin{equation*} x^{\prime }\left( t\right) =\left[ \begin{array}{cc} 6 & 1 \\ 4 & 3% \end{array}% \right] x\left( t\right) +\left[ \begin{array}{c} -11 \\ -5% \end{array}% \right] \end{equation*} \end{description} Solution: \textbf{Note: Since the problem asks for a particular solution only, there is no need to consider the homogeneous solution of the system. } For a particular solution we assume \begin{equation*} x_{p}\left( t\right) =t\vec{a}+\vec{b}=t\left[ \begin{array}{c} a_{1} \\ a_{2}% \end{array}% \right] +\left[ \begin{array}{c} b_{1} \\ b_{2}% \end{array}% \right] \end{equation*} The System implies% \begin{equation*} \left[ \begin{array}{c} a_{1} \\ a_{2}% \end{array}% \right] =\left[ \begin{array}{cc} 6 & 1 \\ 4 & 3% \end{array}% \right] \left( t\left[ \begin{array}{c} a_{1} \\ a_{2}% \end{array}% \right] +\left[ \begin{array}{c} b_{1} \\ b_{2}% \end{array}% \right] \right) +\left[ \begin{array}{c} -11 \\ -5% \end{array}% \right] \end{equation*} : $\allowbreak $ or% \begin{equation*} \left[ \begin{array}{c} a_{1} \\ a_{2}% \end{array}% \right] =\left[ \begin{array}{c} 6ta_{1}+6b_{1}+ta_{2}+b_{2} \\ 4ta_{1}+4b_{1}+3ta_{2}+3b_{2}% \end{array}% \right] +\left[ \begin{array}{c} -11 \\ -5% \end{array}% \right] \end{equation*} Thus from the terms times $t$% \begin{eqnarray*} 6a_{1}+a_{2} &=&0 \\ 4a_{1}+3a_{2} &=&0 \end{eqnarray*} Therefore $a_{1}=a_{2}=0.$ We now have% \begin{equation*} \left[ \begin{array}{c} 0 \\ 0% \end{array}% \right] =\left[ \begin{array}{c} 6b_{1}+b_{2} \\ 4b_{1}+3b_{2}% \end{array}% \right] +\left[ \begin{array}{c} -11 \\ -5% \end{array}% \right] \end{equation*} or% \begin{eqnarray*} 6b_{1}+b_{2} &=&11 \\ 4b_{1}+3b_{2} &=&5 \end{eqnarray*} Multiplying the first equation by $-3$ and adding to the second equation gives $-14b_{1}=-28,$ so $b_{1}=2.$ Hence $b_{2}=-1.$ \begin{equation*} x_{p}\left( t\right) =\left[ \begin{array}{c} 2 \\ -1% \end{array}% \right] \end{equation*} \begin{description} \item[{2 a $\left[ 15\text{ pts.}\right] $}] Evaluate the integral \begin{equation*} \diint\limits_{R}\frac{\sin x}{x}dA \end{equation*} \end{description} where $R$ is the triangle in the $x,y-$plane bounded by the $x-$axis, the line $y=x,$ and the line $x=1.$ Sketch $R.$ Solution: The region $R$ is shown below. $\left( 0,0,1,0,1,1,0,0\right) $\FRAME{dtbpFX}{4.4996in}{3in}{0pt}{}{}{Plot}{% \special{language "Scientific Word";type "MAPLEPLOT";width 4.4996in;height 3in;depth 0pt;display "USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine "Maple";xmin "-5";xmax "5";xviewmin "-0.02";xviewmax "1.0204";yviewmin "-0.02";yviewmax "1.0204";plottype 4;numpoints 100;plotstyle "patch";axesstyle "normal";xis \TEXUX{x};var1name \TEXUX{$x$};function \TEXUX{$\left( 0,0,1,0,1,1,0,0\right) $};linecolor "black";linestyle 1;pointstyle "point";linethickness 2;lineAttributes "Solid";curveColor "[flat::RGB:0000000000]";curveStyle "Line";}} We can write the given integral in two ways:% \begin{equation*} \int_{0}^{1}\int_{y}^{1}\frac{\sin x}{x}dxdy=\int_{0}^{1}\int_{0}^{x}\frac{% \sin x}{x}dydx \end{equation*} The first integral cannot be evaluated, but the second one can.% \begin{equation*} \int_{0}^{1}\int_{0}^{x}\frac{\sin x}{x}dydx=\int_{0}^{1}\left. \frac{\sin x% }{x}y\right] _{0}^{x}dx=\int_{0}^{1}\sin xdx=-\cos 1+1 \end{equation*} \begin{description} \item[{2 b $\left[ 15\text{ pts.}\right] $}] Use polar coordinates to evaluate% \begin{equation*} \int_{-1}^{1}\int_{0}^{\sqrt{1-x^{2}}}\left( x^{2}+y^{2}\right) ^{\frac{3}{2}% }dydx \end{equation*} \end{description} Sketch the region of integration. \vspace{1pt}Solution: The region of integration is the upper half of the circle of radius $1$ centered at the origin, and is shown below $\sqrt{1-x^{2}}$\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 "Maple";xmin "-1.1";xmax "1.1";xviewmin "-1.02762038012";xviewmax "1.0295375880824";yviewmin "0.129496874766036";yviewmax "1.01734012879368";plottype 4;constrained TRUE;numpoints 100;plotstyle "patch";axesstyle "normal";xis \TEXUX{x};var1name \TEXUX{$x$};function \TEXUX{$\sqrt{1-x^{2}}$};linecolor "black";linestyle 1;pointstyle "point";linethickness 2;lineAttributes "Solid";var1range "-1.1,1.1";num-x-gridlines 100;curveColor "[flat::RGB:0000000000]";curveStyle "Line";rangeset"X";}} Thus in polar coordinates the integral becomes since $r=\left( x^{2}+y^{2}\right) ^{\frac{1}{2}}$% \begin{eqnarray*} \int_{-1}^{1}\int_{0}^{\sqrt{1-x^{2}}}\left( x^{2}+y^{2}\right) ^{\frac{3}{2}% }dydx &=&\int_{0}^{\pi }\int_{0}^{1}\left( r^{3}\right) rdrd\theta \\ &=&\int_{0}^{\pi }\frac{1}{5}d\theta =\frac{\pi }{5} \end{eqnarray*} \begin{description} \item[{3 $\left[ 20\text{ pts.}\right] $}] Let $V$ be the volume that is bounded above by the hemisphere $z=\sqrt{25-x^{2}-y^{2}},$ below by the \end{description} $x,y-$plane and laterally by the cylinder $x^{2}+y^{2}=9.$ Given an integral expression in cylindrical coordinates for the volume of $V.$ Be sure to sketch $V.$ (Do \textit{not} evaluate the integral you give.) Solution: In cylindrical coordinates the equation of the hemisphere is $z=% \sqrt{25-r^{2}},z\geq 0$ and the equation of the cylinder is $r=3.$ The volume is shown below. \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 "-3.1416";xmax "3.1416";ymin "0";ymax "5.0";xviewmin "-5.19999999986238";xviewmax "5.20399999999725";yviewmin "-5.19999999996492";yviewmax "5.20399999996489";zviewmin "-0.1";zviewmax "5.102";phi 83;theta 20;plottype 14;constrained TRUE;num-x-gridlines 25;num-y-gridlines 25;plotstyle "wireframe";axesstyle "normal";plotshading "NONE";xis \TEXUX{v952};yis \TEXUX{z};var1name \TEXUX{$\theta $};var2name \TEXUX{$z$};function \TEXUX{$\left( r,\theta ,\sqrt{25-r^{2}}\right) $};linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";coordinateSystem "cylindrical";var1range "-3.1416,3.1416";var2range "0,5.0";surfaceColor "[flat::RGB:0x00ff0000:0x000000ff]";surfaceStyle "Wire Frame";num-x-gridlines 25;num-y-gridlines 25;surfaceMesh "Contour";rangeset"Y";function \TEXUX{$\left( 3,\theta ,z\right) $};linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";coordinateSystem "cylindrical";var1range "-3.1416,3.1416";var2range "0,5";surfaceColor "[linear:XYZ:RGB:0x00ff0000:0x000000ff]";surfaceStyle "Wire Frame";num-x-gridlines 25;num-y-gridlines 25;surfaceMesh "Contour";}} \begin{equation*} Vol=\int_{0}^{2\pi }\int_{0}^{3}\int_{0}^{\sqrt{25-r^{2}}}rdzdrd\theta \end{equation*} \begin{description} \item[{4 [15 pts.]}] Give an integral expression in \textbf{spherical} coordinates for the volume of the solid $D$ that lies above the cone \end{description} $z^{2}=\frac{1}{3}\left( x^{2}+y^{2}\right) ,$ $z\geq 0$ and below the sphere $x^{2}+y^{2}+z^{2}=z.$ Sketch $D.$ (Note: $\ \tan 30^{\circ }=\allowbreak \frac{1}{3}\sqrt{3},$ \ $\tan 45^{\circ }=\allowbreak 1,$ \ $\tan 60^{\circ }=\allowbreak \sqrt{3}$) Solution: (This problem is similar to example 4 on page 897 of Stewart.) We may write the equation of the sphere as \begin{equation*} x^{2}+y^{2}+\left( z-1\right) ^{2}=1 \end{equation*}% Hence the sphere has radius $1$ and is centered at $\left( 0,0,1\right) .$ Since% \begin{equation*} x=\rho \cos \theta \sin \phi ,\text{ \ }y=\rho \sin \theta \sin \phi ,\text{ \ \ }z=\rho \cos \phi \end{equation*} the equation of the sphere in spherical coordinates is \begin{equation*} \rho ^{2}=\rho \cos \phi \end{equation*} or% \begin{equation*} \rho =\cos \phi \end{equation*} For the cone we have% \begin{equation*} \rho ^{2}\cos ^{2}\phi =\frac{1}{3}\left( \rho ^{2}\cos ^{2}\theta \sin ^{2}\phi +\rho ^{2}\sin ^{2}\theta \sin ^{2}\phi \right) =\frac{1}{3}\rho ^{2}\sin ^{2}\phi \end{equation*} Thus% \begin{equation*} \tan ^{2}\phi =3 \end{equation*} and hence the equation of the cone is% \begin{equation*} \phi =\frac{\pi }{3} \end{equation*} The solid is given in the following diagram \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 "-3.1416";xmax "3.1416";ymin "0";ymax "1.0472";xviewmin "-0.943177874931824";xviewmax "0.943903396398562";yviewmin "-0.943177874950424";yviewmax "0.943903396392694";zviewmin "-0.02";zviewmax "1.0204";phi 81;theta 27;plottype 15;num-x-gridlines 25;num-y-gridlines 25;plotstyle "wireframe";axesstyle "none";plotshading "ZHUE";xis \TEXUX{v952};yis \TEXUX{v961};var1name \TEXUX{$\theta $};var2name \TEXUX{$\rho $};function \TEXUX{$\left( \cos \phi ,\theta ,\phi \right) $};linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";coordinateSystem "spherical";var1range "-3.1416,3.1416";var2range "0,1.0472";surfaceColor "[rainbow:Z:RGB:0x00ff0000:0000000000]";surfaceStyle "Wire Frame";num-x-gridlines 25;num-y-gridlines 25;surfaceMesh "Contour";rangeset"Y";function \TEXUX{$\left( \rho ,\theta ,\frac{\pi }{3}\right) $};linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";coordinateSystem "spherical";var1range "-3.1416,3.1416";var2range "0,1.0472";surfaceColor "[linear:XYZ:RGB:0x00ff0000:0x000000ff]";surfaceStyle "Color Patch";num-x-gridlines 25;num-y-gridlines 25;surfaceMesh "Mesh";rangeset"Y";}% } Therefore% \begin{equation*} Vol=\int_{0}^{2\pi }\int_{0}^{\frac{\pi }{3}}\int_{0}^{\cos \phi }\rho ^{2}\sin \phi d\rho d\phi d\theta \end{equation*} \vspace{1pt} \end{document} %%%%%%%%%%%%%%%% End /document/ma227_exam_2_03s_sol.tex %%%%%%%%%%%%%%%