\hfill \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} \section{Ma 227 Review of Gradient, Curl, Divergence, and Line Integrals} \vspace{1pt} \subsection{Vector Fields, Gradient, Divergence, Curl} \begin{definition} Let $D$ denote a subset of the plane. \ A \textbf{vector field} on $D$ is a function $\mathbf{F}$ that assigns to each point $(x,y)$ in $D$ a two-dimensional vector $\mathbf{F}(x,y)$. \ In terms of its component functions, the vector field $\mathbf{F}$ is given by \begin{equation*} \mathbf{F}(x,y)=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j}=\left\langle P\left( x,y\right) ,Q\left( x,y\right) \right\rangle \end{equation*}% or, for short, \begin{equation*} \mathbf{F}=P\mathbf{i}+Q\mathbf{j} \end{equation*} \end{definition} \begin{itemize} \item \begin{definition} Let $D$ denote a subset of space. \ A \textbf{vector field} on $D$ is a function $\mathbf{F}$ that assigns to each point $(x,y,z)$ in $D$ a three-dimensional vector $\mathbf{F}(x,y,z)$. \ In terms of its component functions, the vector field $\mathbf{F}$ is given by \begin{equation*} \mathbf{F}(x,y,z)=P(x,y,z)\mathbf{i}+Q(x,y,z)\mathbf{j}+R(x,y,z)\mathbf{k}% =\left\langle P\left( x,y,z\right) ,Q\left( x,y,z\right) ,R(x,y,z)\right\rangle \end{equation*}% or, for short, \begin{equation*} \mathbf{F}=P\mathbf{i}+Q\mathbf{j}+R\mathbf{k} \end{equation*} \end{definition} \begin{definition} If $f$ is a scalar function of two variables, its \textbf{gradient vector field} $\nabla f$ is defined by \begin{equation*} \nabla f=\frac{\partial f}{\partial x}\mathbf{i}+\frac{\partial f}{\partial y% }\mathbf{j} \end{equation*}% If $f$ is a scalar function of three variables, its \textbf{gradient vector field} $\nabla f$ is defined by \begin{equation*} \nabla f=\frac{\partial f}{\partial x}\mathbf{i}+\frac{\partial f}{\partial y% }\mathbf{j}+\frac{\partial f}{\partial z}\mathbf{k} \end{equation*} \end{definition} \begin{definition} $\mathbf{F}$ is a \textbf{conservative vector field} if there exists a scalar function $f$ such that $\nabla f=\mathbf{F}$. \ In this case $f$ is called a \textbf{potential function} for $\mathbf{F}$\textbf{.} \end{definition} \end{itemize} \paragraph{Example:} Let $\Phi \left( x,y,z\right) =xyz+3x^{4}y^{2}z^{3}.$ Then $\nabla \Phi =\left( yz+12x^{3}y^{2}z^{3}\right) \vec{i}+\left( xz+6x^{4}yz^{3}\right) \vec{j}+\left( xy+9x^{4}y^{2}z^{2}\right) $ \subsubsection{\protect\vspace{1pt}Divergence} A vector field is a vector-valued function. If \begin{equation*} \vec{F}(x,y,z)=\left[ p(x,y,z),q(x,y,z),r(x,y,z)\right] =p\vec{i}+q\vec{j}+r% \vec{k} \end{equation*}% is a vector field, then the scalar \begin{equation*} \nabla \cdot \vec{F}=\func{div}\vec{F}=\frac{\partial p}{\partial x}\left( a,b,c\right) +\frac{\partial q}{\partial y}\left( a,b,c\right) +\frac{% \partial r}{\partial z}\left( a,b,c\right) \end{equation*}% is the \emph{divergence} of $F$ at the point $\left( a,b,c\right) .$ \vspace{1pt} \subsubsection{\protect\vspace{1pt}Curl} If $\vec{F}(x,y,z)=\left( p(x,y,z),q(x,y,z),r(x,y,z)\right) =p\vec{i}+q\vec{j% }+r\vec{k}$ is a vector field, then the vector \begin{equation*} \nabla \times \vec{F}=\left( \frac{\partial r}{\partial y}-\frac{\partial q}{% \partial z},\frac{\partial p}{\partial z}-\frac{\partial r}{\partial x},% \frac{\partial q}{\partial x}-\frac{\partial p}{\partial y}\right) =\left( \frac{\partial r}{\partial y}-\frac{\partial q}{\partial z}\right) \vec{i}% +\left( \frac{\partial p}{\partial z}-\frac{\partial r}{\partial x}\right) \vec{j}+\left( \frac{\partial q}{\partial x}-\frac{\partial p}{\partial y}% \right) \vec{k} \end{equation*}% is called the \emph{curl} of $F$. \paragraph{Example:} Let $\vec{F}=xy\vec{i}+xz^{2}\vec{j}+ze^{x}\sin y\vec{k}.$ Find $\func{div}% \vec{F}$ and $\func{curl}\vec{F}.$ \begin{equation*} \nabla \cdot \vec{F}=\func{div}\vec{F}=y+e^{x}\sin y \end{equation*} \begin{eqnarray*} \nabla \times \vec{F} &=&\func{curl}\vec{F}=\left\vert \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ xy & xz^{2} & ze^{x}\sin y% \end{array}% \right\vert \\ &=&\left( ze^{x}\cos y-2xz\right) \vec{i}-ze^{x}\sin y\vec{j}+\left( z^{2}-x\right) \vec{k} \end{eqnarray*} \subsection{Line Integrals} We define the line integral as follows: A curve $C$ may be described in three dimensions via \qquad \begin{equation*} x=f(t);\qquad y=g(t);\qquad z=h(t)\text{ \ \ \ }a\leq t\leq b \end{equation*}% or% \begin{equation*} \overrightarrow{r}(t)=f(t)\overrightarrow{i}+g(t)\overrightarrow{j}+h(t)% \overrightarrow{k}\text{ \ \ }a\leq t\leq b \end{equation*} If \begin{equation*} \overrightarrow{F}(x,y,z)=P(x,y,z)\overrightarrow{i}+Q(x,y,z)\overrightarrow{% j}+R(x,y,z)\overrightarrow{k} \end{equation*} then \begin{eqnarray*} \int_{C}\overrightarrow{F}\cdot d\overrightarrow{r} &=&\int_{C}Pdx+Qdy+Rdz=% \int_{a}^{b}\overrightarrow{F}(f(t),g(t),h(t))\cdot \overrightarrow{r}% ^{\prime }(t)dt \\ &=&\int_{a}^{b}\left\{ P(f(t),g(t),h(t))f^{\prime }(t)+Q(f(t),g(t),h(t))g^{\prime }(t)+R(f(t),g(t),h(t))h^{\prime }(t)\right\} dt \end{eqnarray*} \paragraph{Example:} Evaluate $\int_{C}\overrightarrow{F}\bullet \overrightarrow{dr}$ , where $% \overrightarrow{F}=(x+2y)\overrightarrow{i}+(x^{2}-y^{2})\overrightarrow{j}$ and $C$ is the line segment joining $(0,0)$ and $(1,1)$. \vspace{1pt} \paragraph{SOLUTION} We have to parametrize $\overrightarrow{F}$ and $\overrightarrow{r}$ first. Since we are moving from $0$ to $1$ along the line $x=y,$ it behoves us to set $x=y=t$ as our parameter. Then \begin{equation*} \overrightarrow{F}=(t+2t)\overrightarrow{i}+(t^{2}-t^{2})\overrightarrow{j}% =3t\overrightarrow{i}. \end{equation*} \vspace{1pt} Next, \begin{eqnarray*} \overrightarrow{r} &=&x(t)\overrightarrow{i}+y(t)\overrightarrow{j}=t% \overrightarrow{i}+t\overrightarrow{j}. \\ \overrightarrow{r}^{\prime }\left( t\right) &=&\overrightarrow{i}+% \overrightarrow{j}. \end{eqnarray*} \vspace{1pt} Finally, \begin{equation*} \overrightarrow{F}\bullet \overrightarrow{r}^{\prime }\left( t\right) =3t% \overrightarrow{i}\cdot (\overrightarrow{i}+\overrightarrow{j})=3t \end{equation*} \vspace{1pt} Integrate now: \begin{equation*} \int_{C}\overrightarrow{F}\bullet \overrightarrow{dr}=\int_{0}^{1}3t\;dt=% \frac{3}{2}. \end{equation*} \subsubsection{Path Independence} Certain line integrals depend only on the integrand and endpoints $A$ and $B$% . Such integrals are called \emph{path independent} or are said to be \emph{% independent of the path}. \vspace{1pt}Often one must consider situations in which the path $C$ is a closed curve. Hence the starting point $A$ and ending point $B$ are the same. This is usually written as \begin{equation*} \oint_{C}\overrightarrow{F}\cdot d\overrightarrow{r}. \end{equation*} For plane curves we take the positive direction of $C$ so that the interior of the closed curve is always to the left as $C$ is traversed. \vspace{1pt} \vspace{1pt}\FRAME{dtbpF}{3.7377in}{2.6584in}{0pt}{}{}{lin3.gif}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 3.7377in;height 2.6584in;depth 0pt;original-width 361.375pt;original-height 255.9375pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/lin3.gif';file-properties "XNPEU";}} The following are equivalent: \vspace{1pt} $\int_{C}\overrightarrow{F}\cdot d\overrightarrow{r}$ is path independent $% \leftrightarrow $ there exists a $G$ such that $\overrightarrow{F}=\nabla G$ \qquad \qquad \qquad \qquad \qquad \qquad $\ \ \ \ \leftrightarrow \oint_{C}% \overrightarrow{F}\cdot d\overrightarrow{r}=0$ for any closed path $C$ \qquad \qquad \qquad \qquad \qquad \qquad $\ \ \ \ \leftrightarrow $ $\nabla \times \vec{F}=curl\vec{F}=0$ Note in two dimensions with $\vec{F}=P\left( x,y\right) \vec{i}+Q\left( x,y\right) \vec{j}$ the last condition becomes% \begin{equation*} Q_{x}=P_{y} \end{equation*} \paragraph{Example:} Let $\overrightarrow{F}=y^{2}z^{3}\overrightarrow{i}+2xyz^{3}\overrightarrow{% j}+(3xy^{2}z^{2}+z)\overrightarrow{k}$. \ Show that curl $\overrightarrow{F}=% \overrightarrow{0}$. Evaluate $\oint_{C}\overrightarrow{F}\bullet \overrightarrow{dr}$, where $C$ is the ellipse in $x,y,z$ -space given by \begin{equation*} \frac{x^{2}}{4}+\frac{y^{2}}{9}=1,z=3. \end{equation*} \paragraph{\protect\vspace{1pt}SOLUTION} \begin{equation*} \func{curl}\overrightarrow{F}=\nabla \times \overrightarrow{F}=\det \left[ \begin{array}{lll} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ y^{2}z^{3} & 2xyz^{3} & 3xy^{2}z^{2}+z% \end{array}% \right] =6xyz^{2}\overrightarrow{i}+2yz^{3}\overrightarrow{k}+3y^{2}z^{2}% \overrightarrow{j}-2yz^{3}\overrightarrow{k}-6xyz^{2}\overrightarrow{i}% -3y^{2}z^{2}\overrightarrow{j}=\overrightarrow{0}. \end{equation*} \vspace{1pt} We have shown that for the given $\overrightarrow{F}$, $\func{curl}\vec{F}=0. $ \ \ Therefore the integral is path-independent, so the integral around a closed curve is zero. Thus $\oint_{C}\overrightarrow{F}\bullet \overrightarrow{dr}=0.$ \vspace{1pt} \paragraph{Example:} Consider the \ $\int_{C}\overrightarrow{F}\cdot \overrightarrow{dr}$, where \ $\overrightarrow{F}=(2xyz+z^{2}y)\vec{i}+(x^{2}z+z^{2}x)\vec{j}% +(x^{2}y+2xyz)\vec{k}.$ Show that $\ \nabla \times \overrightarrow{F}=% \overrightarrow{0}.$ What does this tell you about \ $\oint_{C}% \overrightarrow{F}\cdot \overrightarrow{dr}$ , \ where \ $C$ \ is any closed curve? $\vspace{1pt}$ \paragraph{\protect\vspace{1pt}SOLUTION} \vspace{1pt}% \begin{eqnarray*} \nabla \times \overrightarrow{F} &=&curl\;\overrightarrow{F}=\left\vert \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ 2xyz+z^{2}y & x^{2}z+z^{2}x & x^{2}y+2xyz% \end{array}% \right\vert \\ &=&(x^{2}+2xz-x^{2}-2zx)\vec{i}-(2xy+2yz-2xy-2zy)\vec{j}% +(2xz+z^{2}-2xz-z^{2})\vec{k} \\ &=&\overrightarrow{0} \end{eqnarray*} \vspace{1pt} Then \ $\oint_{C}\overrightarrow{F}\cdot \overrightarrow{dr}=0$ \ for any closed curve \ $C.$ \ Or, equivalently, \ $\int_{C}\overrightarrow{F}\cdot \overrightarrow{dr}$ \ is independent of the path taken between two given points. \paragraph{\protect\vspace{1pt}Example} \vspace{1pt}Find a function $\ \Phi (x,y,z)$ \ such that \ $\nabla \Phi =% \overrightarrow{F},$ where $\vec{F}$ is the vector field above. $\vspace{1pt}$ \paragraph{\protect\vspace{1pt}SOLUTION} \vspace{1pt}$\overrightarrow{F}=\nabla \Phi =\dfrac{\partial \Phi }{\partial x}\vec{i}+\dfrac{\partial \Phi }{\partial y}\vec{j}+\dfrac{\partial \Phi }{% \partial z}\vec{k}\qquad $We set equal the corresponding components. $\dfrac{\partial \Phi }{\partial x}=2xyz+z^{2}y\qquad \Longrightarrow \qquad $% \begin{equation*} \Phi (x,y,z)=x^{2}yz+xyz^{2}+h(y,z) \end{equation*} \begin{equation*} \dfrac{\partial \Phi }{\partial y}=x^{2}z+z^{2}x+h_{y}=x^{2}z+z^{2}x \end{equation*} Thus $h=g\left( z\right) \qquad $ \begin{equation*} \dfrac{\partial \Phi }{\partial z}=x^{2}y+2xyz+g^{\prime }\left( z\right) =x^{2}y+2xyz \end{equation*} \vspace{1pt} $g\left( z\right) =C,$ a constant \vspace{1pt} Finally, we have \ \ \begin{equation*} \Phi (x,y,z)=x^{2}yz+xyz^{2}+C \end{equation*} \paragraph{Example:} Evaluate% \begin{equation*} \int_{C}\left( x+2y\right) dx+\left( x^{2}-y^{3}\right) dy \end{equation*} where $C$ consists of the segments from $\left( 1,1\right) $ to $\left( 3,1\right) $ and $\left( 3,1\right) $ to $\left( 3,-1\right) .$ Sketch $C.$ Solution: The path $C$ is shown below. \FRAME{dtbpFX}{3.0007in}{1.9993in}{0pt}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 3.0007in;height 1.9993in;depth 0pt;display "USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine "MuPAD";xmin "-5";xmax "5";xviewmin "0.9998";xviewmax "3.0002";yviewmin "-1.0002";yviewmax "1.0002";plottype 4;axesFont "Times New Roman,12,0000000000,useDefault,normal";numpoints 49;plotstyle "patch";axesstyle "normal";axestips TRUE;gridLines TRUE;xis \TEXUX{x};var1name \TEXUX{$x$};function \TEXUX{$\left( 3,1,3,-1\right) $};linecolor "black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";curveColor "[flat::RGB:0000000000]";curveStyle "Line";function \TEXUX{$\left( 1,1,3,1\right) $};linecolor "black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";curveColor "[flat::RGB:0000000000]";curveStyle "Line";VCamFile 'MXAH7W0A.xvz';}} Let $C_{1}$ be the segment from $\left( 1,1\right) $ to $\left( 3,1\right) $ and $C_{2}$ be the segment from $\left( 3,1\right) $ to $\left( 3,-1\right) . $ Then $C=C_{1}\cup C_{2}.$ \begin{eqnarray*} \int_{C}\left( x+2y\right) dx+\left( x^{2}-y^{3}\right) dy &=&\int_{C_{1}}\left( x+2y\right) dx+\left( x^{2}-y^{3}\right) dy+\int_{C_{2}}\left( x+2y\right) dx+\left( x^{2}-y^{3}\right) dy \\ &=&\int_{C_{1}}\left( x+2\left( 1\right) \right) dx+\int_{C_{2}}\left( \left( 3\right) ^{2}-y^{3}\right) dy \\ &=&\int_{1}^{3}\left( x+2\right) dx+\int_{1}^{-1}\left( 9-y^{3}\right) dy=8-18=-10 \end{eqnarray*} \vspace{1pt} \subsection{Green's Theorem} Theorem:\qquad\ Let $P(x,y)$ and $Q(x,y)$ be functions of two variables which are continuous and have continuous first partial derivatives in some rectangular region $H$ in the $x,y-$ plane. If $C$\ is a simple, closed, piecewise smooth curve lying entirely in $H$, and if $R$ is the bounded region enclosed by $C$, then \vspace{1pt}\qquad \qquad \qquad \qquad \qquad \begin{equation*} \oint_{C}\{P(x,y)dx+Q(x,y)dy\}=\int \int_{R}\left( \dfrac{\partial Q}{% \partial x}-\dfrac{\partial P}{\partial y}\right) dA \end{equation*} Note: Green's Theorem applies only to a closed curve. Corollary: Let $R$ be a bounded region in the $x,y-$ plane. Then the area of $R$ is given by \vspace{1pt}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \begin{equation*} A=\frac{1}{2}\oint_{C}(xdy-ydx)=\oint_{C}xdy=-\oint_{C}ydx \end{equation*} \paragraph{\protect\vspace{1pt}Example:} Evaluate \begin{equation*} \oint_{C}\left( 1+\tan x\right) dx+\left( x^{2}+e^{y}\right) dy \end{equation*} Where $C$ is the positively oriented boundary of the region $R$ enclosed by the curves $y=\sqrt{x},$ $x=1,$ and $y=0.$ Be sure to sketch $C.$ \emph{Solution:} The region enclosed by $C$ is shown below. \FRAME{dtbpFX}{3.0007in}{2.0001in}{0pt}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 3.0007in;height 2.0001in;depth 0pt;display "USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine "MuPAD";xmin "0";xmax "1.0";xviewmin "-0.00010000010002";xviewmax "1.00010000010002";yviewmin "-0.00010000010002";yviewmax "1.00010000010002";plottype 4;axesFont "Times New Roman,12,0000000000,useDefault,normal";numpoints 49;plotstyle "patch";axesstyle "normal";axestips FALSE;xis \TEXUX{x};var1name \TEXUX{$x$};function \TEXUX{$\sqrt{x}$};linecolor "black";linestyle 1;pointstyle "point";linethickness 2;lineAttributes "Solid";var1range "0,1.0";num-x-gridlines 49;curveColor "[flat::RGB:0000000000]";curveStyle "Line";rangeset"X";function \TEXUX{$\left( 1,1,1,0,0,0\right) $};linecolor "black";linestyle 1;pointstyle "point";linethickness 2;lineAttributes "Solid";curveColor "[flat::RGB:0000000000]";curveStyle "Line";VCamFile 'MXAH7X0C.xvz';}} We use Green's Theorem to evaluate the integral since $C$ is a closed curve.% \begin{eqnarray*} \oint_{C}\left( 1+\tan x\right) dx+\left( x^{2}+e^{y}\right) dy &=&\int \int_{R}\left( \frac{\partial \left( x^{2}+e^{y}\right) }{\partial x}-\frac{% \partial \left( 1+\tan x\right) }{\partial y}\right) dA \\ &=&\int_{0}^{1}\int_{0}^{\sqrt{x}}\left( 2x-0\right) dydx=2\int_{0}^{1}x^{% \frac{3}{2}}dx=\frac{4}{5} \end{eqnarray*} \paragraph{Example:} Verify that Green's Theorem is true for the line integral% \begin{equation*} \oint_{C}xydx+x^{2}dy \end{equation*} where $C$ is the triangle with vertices $\left( 0,0\right) ,\left( 1,0\right) ,$ and $\left( 1,2\right) .$ \emph{Solution:} The triangle is shown below. $\left( 0,0,1,0,1,2,0,0\right) $\FRAME{dtbpFX}{% 3.0007in}{1.9993in}{0pt}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 3.0007in;height 1.9993in;depth 0pt;display "USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine "MuPAD";xmin "-5";xmax "5";xviewmin "-0.0001";xviewmax "1.0001";yviewmin "-0.0002";yviewmax "2.0002";plottype 4;axesFont "Times New Roman,12,0000000000,useDefault,normal";numpoints 49;plotstyle "patch";axesstyle "normal";axestips FALSE;xis \TEXUX{x};var1name \TEXUX{$x$};function \TEXUX{$\left( 0,0,1,0,1,2,0,0\right) $};linecolor "black";linestyle 1;pointstyle "point";linethickness 2;lineAttributes "Solid";curveColor "[flat::RGB:0000000000]";curveStyle "Line";VCamFile 'MXAH7W0B.xvz';}} The boundary consists of $3$ line segments $C_{1}:0\leq x\leq 1,y=0;C_{2}:0\leq y\leq 2,x=1;C_{3}:y=2x,$ $x=1$ to $0.$ Thus% \begin{equation*} \oint_{C}xydx+x^{2}dy=\int_{0}^{1}0dx+\int_{0}^{2}\left( 1\right) dy+\int_{1}^{0}\left( 2x^{2}+2x^{2}\right) dx=\frac{2}{3} \end{equation*} Also% \begin{eqnarray*} \oint_{C}xydx+x^{2}dy &=&\iint_{R}\left( \frac{\partial \left( x^{2}\right) }{\partial x}-\frac{\partial \left( xy\right) }{\partial y}\right) dA \\ &=&\iint_{R}\left( 2x-x\right) dA \\ &=&\int_{0}^{1}\int_{0}^{2x}xdydx=\int_{0}^{2}\int_{\frac{y}{2}}^{1}xdxdy=% \frac{2}{3} \end{eqnarray*} \paragraph{Example:} Find the area of the region bounded by the hypocycloid with vector equation $% \qquad \qquad \qquad \qquad \qquad \qquad $% \begin{equation*} \vec{r}\left( t\right) =\cos ^{3}t\text{ }\vec{i}+\sin ^{3}t\text{ }\vec{j},% \text{ \ \ \ \ \ \ }0\leq t\leq 2\pi . \end{equation*} $\vspace{1pt}$\qquad \begin{equation*} A=\oint_{C}xdy=-\oint_{C}ydx=\dfrac{1}{2}\oint xdy-ydx \end{equation*} $\vspace{1pt}$ $\qquad $We have $x=\cos ^{3}t,dx=3\cos ^{2}t\left( -\sin t\right) dt,$ \ and \ $y=\sin ^{3}t,dy=3\sin ^{2}t\cos tdt.$ \ Using \ $A=\oint_{C}xdy:A=\oint_{C}xdy=\int_{0}^{2\pi }\cos ^{3}t\left( 3\sin ^{2}t\cos t\right) dt=\dfrac{3\pi }{8}.$ \ \vspace{1pt} \paragraph{Example: \ } Evaluate% \begin{equation*} \oint_{C}\left( 1+10xy+y^{2}\right) dx+\left( 6xy+5x^{2}\right) dy \end{equation*} 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*} \paragraph{Example:} Evaluate% \begin{equation*} \oint_{C}x^{2}ydx-xy^{2}dy \end{equation*} where $C$ is the circle $x^{2}+y^{2}=4$ with counterclockwise orientation$.$ Solution: Since the path is a closed curve, we may use Green's Theorem to evaluate the line integral. Thus \begin{eqnarray*} \oint_{C}x^{2}ydx-xy^{2}dy &=&\diint\limits_{x^{2}+y^{2}\leq 4}\left( \frac{% \partial \left( -xy^{2}\right) }{\partial x}-\frac{\partial \left( x^{2}y\right) }{\partial y}\right) dA \\ &=&\diint\limits_{x^{2}+y^{2}\leq 4}\left( -y^{2}-x^{2}\right) dA=-\int_{0}^{2\pi }\int_{0}^{2}r^{2}\cdot rdrd\theta \\ &=&-8\pi \end{eqnarray*} \paragraph{Example:} Verify Green's theorem for \vspace{1pt}% \begin{equation*} \oint\nolimits_{C}\left( 4x-2y\right) dx+\left( 2x+6y\right) dy \end{equation*} where\ $C$ \ is the ellipse \ $x=2\cos \theta ,$ \ $y=\sin \theta ,$ \ $% 0\leq \theta \leq 2\pi .$ \ (Recall that the area of the ellipse \ $\frac{% x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ \ is \ $\pi ab.$) $\vspace{1pt}$ \paragraph{\protect\vspace{1pt}SOLUTION} \vspace{1pt} For this ellipse, \ $a=2$ \ and \ $b=1$. \ Let \ $G$ \ be the interior of \ $% C$. \ Green's theorem states that the two integrals \ $\oint% \nolimits_{C}Pdx+Qdy$ \ and \ $\iint\nolimits_{G}\left( Q_{x}-P_{y}\right) dxdy$ \ are equal. \ We must verify this. \vspace{1pt} Since $\ Q_{x}=2$ \ and \ $P_{y}=-2$, \begin{equation*} \iint\nolimits_{G}\left( Q_{x}-P_{y}\right) dxdy=\iint\nolimits_{G}4dxdy=4\iint\nolimits_{G}dxdy=4(AreaofG)=4(\pi )(2)(1)=8\pi \end{equation*} \vspace{1pt} The ellipse is already parametrized by $\theta $. \ Since \ $dx=-2\sin \theta \,d\theta $ \ and $\ dy=\cos \theta \,d\theta $, \begin{eqnarray*} \oint\nolimits_{C}Pdx+Qdy &=&\oint\nolimits_{C}\left( 4x-2y\right) dx+\left( 2x+6y\right) dy \\ &=&\int_{0}^{2\pi }\left[ \left( 8\cos \theta -2\sin \theta \right) \left( -2\sin \theta \right) +\left( 4\cos \theta +6\sin \theta \right) \left( \cos \theta \right) \right] \,d\theta \\ &=&\int_{0}^{2\pi }\left[ -16\sin \theta \cos \theta +4\sin ^{2}\theta +4\cos ^{2}\theta +6\sin \theta \cos \theta \right] \,d\theta \\ &=&\int_{0}^{2\pi }\left[ 4-10\sin \theta \cos \theta \right] \,d\theta =8\pi \end{eqnarray*} \vspace{1pt} \vspace{1pt} \end{document} 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