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\hfill \thepage} %} \input{tcilatex} \begin{document} \title{Mathematical Scratchpad } \subsection{\protect\Large Ma 227\qquad \qquad \qquad \qquad Exam III\qquad \qquad \qquad Spring 1997} \vspace{1pt} \subsubsection{PROBLEM\ 1. (20 points)} Compute $\nabla \phi $ where $\phi =2xy-3z^{2}.$ Verify explicitly that $% \nabla \times (\nabla \varphi )=0.$ \paragraph{\protect\vspace{1pt}} \paragraph{SOLUTION} \vspace{1pt}$\nabla \phi =\frac{\partial \phi }{\partial x}\overrightarrow{i}% +\frac{\partial \phi }{\partial y}\overrightarrow{j}+\frac{\partial \phi }{% \partial z}\overrightarrow{k}=2y\overrightarrow{i}+2x\overrightarrow{j}-6z% \overrightarrow{k}.$ $\nabla \times (\nabla \phi )=\det \left( \begin{array}{lll} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ 2y & 2x & -6z \end{array} \right) =2\overrightarrow{k}-2\overrightarrow{k}=\overrightarrow{0},$ QED. \vspace{1pt} \subsubsection{PROBLEM 2. (20 points)} $\smallskip $Let $\mathbf{\vec{F}=\vec{i}-}y\mathbf{\vec{j}}+xyz\mathbf{\vec{% k}}.$ Calculate the line integral $\int_{C}^{{}}\mathbf{\vec{F}\cdot d\vec{r}% }$ where $C$ is the curve parametrized by \begin{enumerate} \item \begin{center} $x(t)=t,\qquad y(t)=t^{2},\qquad z(t)=t,\qquad 0\leq t\leq 1.$ \end{center} \end{enumerate} \begin{center} \vspace{1pt} \end{center} \paragraph{SOLUTION} Introduce the parameter into $\overrightarrow{F}$: $\overrightarrow{F}=\overrightarrow{i}\mathbf{-}t^{2}\overrightarrow{j}+t^{4}% \overrightarrow{k}.$ Furthermore, $\overrightarrow{r}=x\overrightarrow{i}\mathbf{+}y% \overrightarrow{j}+z\overrightarrow{k}=t\overrightarrow{i}\mathbf{+}t^{2}% \overrightarrow{j}+t\overrightarrow{k}.$ Then \ $d\overrightarrow{r}=\overrightarrow{i}\mathbf{+}2t\overrightarrow{j}+% \overrightarrow{k}.$ \vspace{1pt} Then the wanted integral becomes $\int_{C}\overrightarrow{F}\cdot d\overrightarrow{r}% =\int_{0}^{1}(1-2t^{3}+t^{4})dt=\allowbreak \frac{7}{10}.$ \begin{center} \vspace{1pt} \end{center} \subsubsection{\protect\vspace{1pt}PROBLEM 3. (20 points)} Evaluate the surface integral $\int \int_{S}^{{}}z^{2}dS,$ where $S$ is the part of the plane $z=x$ over $0\leq x\leq 2,$ $0\leq y\leq 4.$ \vspace{1pt} \paragraph{SOLUTION} Use the parametrization $x=x$ $y=y$ $z=x$ \vspace{1pt} The surface in question is a plane that is at an angle of $\frac{\pi }{4}$ to the $xy$-plane and crosses it along the $y$-axis. \vspace{1pt} Then $dS=\sqrt{1+z_{x}^{2}+z_{y}^{2}}dA=\sqrt{2}dA,$ \ \ because \ $z_{x}=1$ \ \ and \ \ $z_{y}=0$ The wanted integral is $\int \int_{S}z^{2}dS=\int \int_{A}x^{2}\sqrt{2}dA=\int_{0}^{4}\int_{0}^{2}% \sqrt{2}x^{2}dxdy=\allowbreak \frac{32}{3}\sqrt{2}$ \subsubsection{\protect\vspace{1pt}Problem 4. (20 points)} Sketch the region $M$ over which the integral $\ $ $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \int_{0}^{3}\int_{x^{2}}^{9}\int_{0}^{9-y}f(x,y,z)dzdydx$ is carried. Express the same integral with the order of integration changed to integrate with respect to $x,$then $y$ then \ $z.$\pagebreak \paragraph{SOLUTION} $x\,$goes from $0$ to $3,$ OR from $0$ to $\sqrt{y}$ $z$ goes from $0$ (the $xy$-plane) to the plane $9-y$, which is parallel to the $x$-axis and crosses the \qquad $xy$-plane along the line $y=9$. OR $z$ goes from $0$ to $9$. $y$ goes from the parabola $x^{2}$ to the line $y=9$; the parabola in the $% xy $-plane is formed by \qquad \qquad intersection with the parabolic cylinder $y=x^{2}$. OR $y$ goes from $% 0$ to the plane $9-z.$ \vspace{1pt} Then, with changed order of integration, $\int_{0}^{3}\int_{x^{2}}^{9}% \int_{0}^{9-y}f(x,y,z)dzdydx=\allowbreak \int_{0}^{9}\int_{0}^{9-z}\int_{0}^{% \sqrt{y}}f(x,y,z)dxdydz.$ \vspace{1pt}\FRAME{dtbpF}{2.9603in}{2.9603in}{0pt}{}{}{solid1.jpg}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.9603in;height 2.9603in;depth 0pt;original-width 210.8125pt;original-height 210.8125pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'solid1.jpg';file-properties "XNPEU";}} Sketch: \subsubsection{\protect\vspace{1pt}PROBLEM\ 5. (20 points)} Rewrite the integral $\qquad \qquad \qquad \qquad \qquad \qquad I=\int_{0}^{2}\int_{0}^{\sqrt{% 4-x^{2}}}\int_{0}^{8}yzdzdydx=\allowbreak \frac{256}{3}$ \begin{enumerate} \item \begin{center} in cylindrical coordinates. DO NOT EVALUATE THE INTEGRAL. \end{center} \end{enumerate} \vspace{1pt} \paragraph{SOLUTION} Introduce cylindrical coordinates: $x=r\cos \theta $ $y=r\sin \theta $ $z=z.$ \vspace{1pt} Integrand becomes: $yz=zr\sin \theta ;$ \ $dV=rdrd\theta dz$ (in the appropriate order) Limits of integration: Originally, \vspace{1pt}$x\,$goes from $0$ to $2$ $z$ goes from $0$ (the $xy$-plane) to the plane $z=8$, which is parallel to the $xy$-plane $y$ goes from $0$ to the surface $\sqrt{4-x^{2}}$, which is, in fact, the cylinder $x^{2}+y^{2}=4$ \vspace{1pt} In cylindrical coordinates, the cylinder becomes $r=2;$ so $r$ will go from $% 0$ to $2.$ The original $y$ and $x$ both start at $0$, so we deal only with a quarter of the cylinder: $0\leq \theta \leq \frac{\pi }{2}$ \vspace{1pt}$z$ stays the same. \vspace{1pt} So, we have: $\int_{0}^{2}\int_{0}^{\sqrt{4-x^{2}}}\int_{0}^{8}yzdzdydx=% \int_{0}^{8}\int_{0}^{\frac{\pi }{2}}\int_{0}^{2}zr^{2}\sin \theta drd\theta dz$. \vspace{1pt} \end{document} 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