\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 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Exam III Solutions\hfill 4/28/03} \subsection*{{\protect\normalsize Name: \protect\underline{\hspace{2.5in}} \hfill ID: \protect\underline{\hspace{2.0in}}}} \subsection*{{\protect\normalsize Lecture Section: \protect\underline{% \hspace{2.5in}} \hfill \hspace{2in}}} \hfill {\small \textit{I pledge my honor that I have abided by the Stevens Honor System.}\hspace{2pt}\hfill\underline{\hspace{2.0in}}} \hfill {\small \textbf{SHOW ALL WORK! You may not use a calculator on this exam.}} \begin{description} \item[{1 [$20$ pts.]}] Let $r=\left\vert \vec{r}\right\vert ,$ where $\vec{r}% =x\vec{i}+y\vec{j}+z\vec{k}$ and $\left\vert \vec{r}\right\vert $ denotes the magnitude of $\vec{r}.$ Show that \begin{equation*} gradr=\vec{\nabla}r=\frac{\vec{r}}{r} \end{equation*} \end{description} Solution:% \begin{equation*} r=\left\vert \vec{r}\right\vert =\left( x^{2}+y^{2}+z^{2}\right) ^{\frac{1}{2% }} \end{equation*} \begin{eqnarray*} \vec{\nabla}r &=&\vec{i}\frac{\partial r}{\partial x}+\vec{j}\frac{\partial r% }{\partial y}+\vec{k}\frac{\partial r}{\partial z} \\ &=&\frac{x}{\left( x^{2}+y^{2}+z^{2}\right) ^{\frac{1}{2}}}\vec{i}+\frac{y}{% \left( x^{2}+y^{2}+z^{2}\right) ^{\frac{1}{2}}}\vec{j}+\frac{z}{\left( x^{2}+y^{2}+z^{2}\right) ^{\frac{1}{2}}}\vec{k} \\ &=&\frac{\vec{r}}{r} \end{eqnarray*} \vspace{1pt} \begin{description} \item[{2a $\left[ 20\text{ pts.}\right] $}] \end{description} Evaluate% \begin{equation*} \int_{C}x^{2}ydx+xydy \end{equation*} where $C$ consists is the path $y=\sqrt{1-x^{2}}$ connecting $\left( 0,1\right) $ to $\left( -1,0\right) $. Sketch $C.$ Note: $\int \cos ^{2}u\sin ^{2}udu=\allowbreak -\frac{1}{4}\sin u\cos ^{3}u+% \frac{1}{8}\cos u\sin u+\frac{1}{8}u+C$ Solution: $\sqrt{1-x^{2}}$\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";xmax "0";xviewmin "-1.02";xviewmax "0.020400000000001E0";yviewmin "-0.02";yviewmax "1.0204";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 "-1,0";num-x-gridlines 100;curveColor "[flat::RGB:0000000000]";curveStyle "Line";rangeset"X";}} We parametrize $C$ as \begin{equation*} x=\cos t\text{ \ \ }y=\sin t\text{ \ }\frac{\pi }{2}\leq t\leq \pi \end{equation*} Thus% \begin{eqnarray*} \vec{r}\left( t\right) &=&\cos t\vec{i}+\sin t\vec{j} \\ \vec{r}^{\prime }\left( t\right) &=&-\sin t\vec{i}+\cos t\vec{j} \end{eqnarray*} \begin{equation*} \vec{F}\left( t\right) =\cos ^{2}t\sin t\vec{i}+\cos t\sin t\vec{j} \end{equation*} \begin{equation*} \vec{F}\left( t\right) \cdot \vec{r}^{\prime }\left( t\right) =-\cos ^{2}t\sin ^{2}t+\cos ^{2}t\sin t \end{equation*} Therefore% \begin{eqnarray*} \int_{C}x^{2}ydx+xydy &=&\int_{\frac{\pi }{2}}^{\pi }\left( -\cos ^{2}t\sin ^{2}t+\cos ^{2}t\sin t\right) dt \\ &=&-\left[ -\frac{1}{4}\sin u\cos ^{3}u+\frac{1}{8}\cos u\sin u+\frac{1}{8}u% \right] _{\frac{\pi }{2}}^{\pi }+\left[ -\frac{1}{3}\cos ^{3}t\right] _{% \frac{\pi }{2}}^{\pi } \\ &=&-\frac{\pi }{8}+\frac{\pi }{16}+\frac{1}{3}=-\frac{1}{16}\pi +\frac{1}{3} \end{eqnarray*} \begin{description} \item[{2b [$20$ pts.]}] Evaluate% \begin{equation*} \oint_{C}\left( 1+10xy+y^{2}\right) dx+\left( 6xy+5x^{2}\right) dy \end{equation*} \end{description} where $C$ is the square with vertices $\left( 0,0\right) ,\left( a,0\right) ,\left( a,a\right) ,\left( 0,a\right) $ with counterclockwise orientation$.$ Solution: We use Green's Theorem since the path is closed. Now% \begin{equation*} P\left( x,y\right) =1+10xy+y^{2}\text{ and }Q\left( x,y\right) =6xy+5x^{2} \end{equation*} Thus% \begin{equation*} P_{y}=10x+2y\text{ \ and \ }Q_{x}=6y+10x \end{equation*} \begin{equation*} Q_{x}-P_{y}=4y \end{equation*} By Green's Theorem% \begin{eqnarray*} \oint_{C}\left( 1+10xy+y^{2}\right) dx+\left( 6xy+5x^{2}\right) dy &=&\diint\limits_{R}\left( Q_{x}-P_{y}\right) dA \\ &=&\int_{0}^{a}\int_{0}^{a}4ydxdy=2a^{3} \end{eqnarray*} \begin{description} \item[3] Let $S$ be the surface of the unit sphere $x^{2}+y^{2}+z^{2}=1$. \end{description} Note: See Example 4 on page 967 of the text. \begin{description} \item[{3 a $\left[ 12\text{ pts.}\right] $}] Use spherical coordinates to give a vector parametrization of $S.$ \end{description} Solution: Here $\rho =1$ so \begin{equation*} x=\cos \theta \sin \phi \text{ \ \ }y=\sin \theta \sin \phi \text{ \ }z=\cos \phi \text{ \ \ \ \ \ }0\leq \theta \leq 2\pi \text{ \ \ }0\leq \phi \leq \pi \end{equation*} so \begin{equation*} \vec{r}\left( \theta ,\phi \right) =\cos \theta \sin \phi \vec{i}+\sin \theta \sin \phi \vec{j}+\cos \phi \vec{k}\text{ \ \ \ \ \ }0\leq \theta \leq 2\pi \text{ \ \ }0\leq \phi \leq \pi \end{equation*} \vspace{1pt} \begin{description} \item[{3 b $[15$ pts.$]$}] Find a vector normal to this sphere $S$. \end{description} Solution: $\vec{r}_{\theta }\times \vec{r}_{\phi }$ is normal to the surface.% \begin{eqnarray*} \vec{r}_{\theta } &=&-\sin \theta \sin \phi \vec{i}+\cos \theta \sin \phi \vec{j} \\ \vec{r}_{\phi } &=&\cos \theta \cos \phi \vec{i}+\sin \theta \cos \phi \vec{j% }-\sin \phi \vec{k} \end{eqnarray*} Thus \ \begin{eqnarray*} \vec{r}_{\theta }\times \vec{r}_{\phi } &=&\left\vert \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ -\sin \theta \sin \phi & \cos \theta \sin \phi & 0 \\ \cos \theta \cos \phi & \sin \theta \cos \phi & -\sin \phi% \end{array}% \right\vert \\ &=&-\cos \theta \sin ^{2}\phi \vec{i}-\left( \sin \theta \sin ^{2}\phi \right) \vec{j}-\cos \phi \sin \phi \vec{k} \end{eqnarray*} \vspace{1pt} \begin{description} \item[{3 c $\left[ 13\text{ pts.}\right] $ }] Let \begin{equation*} \vec{F}\left( x,y,z\right) =z\vec{i}+y\vec{j}+x\vec{k} \end{equation*} \end{description} Give an expression for \begin{equation*} \int \int_{S}\overrightarrow{F}\cdot \vec{n}ds \end{equation*} where $\vec{n}$ is a vector normal to surface of the unit sphere $S$. Do \textbf{not }evaluate this expression. Solution:% \begin{equation*} \vec{n}=\pm \left( \vec{r}_{\theta }\times \vec{r}_{\phi }\right) \end{equation*} \begin{equation*} \int \int_{S}\overrightarrow{F}\cdot \vec{n}ds=\int \int_{D}\left[ \vec{F}% \left( \vec{r}\left( \theta ,\phi \right) \right) \cdot \left( \pm \right) \left( \vec{r}_{\theta }\times \vec{r}_{\phi }\right) \right] dA \end{equation*}% \begin{equation*} \vec{F}\left( \vec{r}\left( \theta ,\phi \right) \right) =\cos \phi \vec{i}% +\sin \theta \sin \phi \vec{j}+\cos \theta \sin \phi \vec{k} \end{equation*} \begin{equation*} \vec{F}\left( \vec{r}\left( \theta ,\phi \right) \right) \cdot \left( \pm \right) \left( r_{\theta }\times r_{\phi }\right) =\pm \left( -2\cos \theta \cos \phi \sin ^{2}\phi -\sin ^{2}\theta \sin ^{3}\phi \right) \end{equation*} Thus \begin{equation*} \int \int_{S}\overrightarrow{F}\cdot \vec{n}ds=\pm \int_{0}^{2\pi }\int_{0}^{\pi }\left( -2\cos \theta \cos \phi \sin ^{2}\phi -\sin ^{2}\theta \sin ^{3}\phi \right) d\phi d\theta \end{equation*} \end{document} %%%%%%%%%%%%%%%% End /document/ma227_exam_3_sol_03s.tex %%%%%%%%%%%%%%%