%\documentclass[c]{beamer}

%\usecolortheme{whale}


\documentclass[xcolor=dvipsnames,envcountsect]{beamer} 
%\documentclass{beamer}
\beamertemplatenavigationsymbolsempty 
\usecolortheme[named=Blue]{structure} %change
%\usecolortheme{seahorse}
%\usecolortheme{rose}
 %default blue to brown
%\usetheme{Warsaw}
%\usetheme{CambridgeUS}
\useoutertheme{infolines}
%\usecolortheme{dolphin}
%\usefonttheme{serif}
\usetheme{Madrid}

%\addtocounter{framenumber}{-5} 
%\setbeamertemplate{footline}[default] 
\usefonttheme{professionalfonts} % using non standard fonts for beamer
%\usefonttheme{serif} % default family is serif


\usepackage{multicol}
%\usepackage[compatibility=false]{caption}
%\usepackage{subfig}
\usepackage{algcompatible}
\usepackage[section]{algorithm}
\usepackage{algorithmicx}
%\usepackage{algorithmic}
\usepackage{algpseudocode}

%\usepackage[caption=false]{subfig}
\usepackage{amsmath}
\DeclareMathOperator{\sech}{sech}
\DeclareMathOperator{\csch}{csch}

%\usetheme{Frankfurt}
%\usetheme{AnnArbor}
% % % % % % % % % % %
%\usepackage{fancyhdr}
%\usepackage{graphicx}
%\usepackage{caption}
%\usepackage{tikz,siunitx}
%
%\usetikzlibrary{arrows,decorations.pathmorphing,backgrounds,positioning,fit,petri}
%
%\usetikzlibrary{automata,arrows}
%\usepackage{subcaption}
%\usepackage{enumitem}
%\setlist{nolistsep}
%\usepackage{float}
% % % % % % % % % % %


\usepackage{extarrows}
\usepackage{verbatim}
\usepackage{bm}
\usepackage[all,cmtip]{xy}
\usepackage[framemethod=default]{mdframed}
\def\Ker{\textrm {Ker\,}}
\def\Im{\textrm{Im\,}}
\def\Z{\mathbb{Z}}
\def\N{\mathcal{N}}
\def\Z{\mathbb{Z}}
\def\F{\mathbb{F}}
\def\Aut{\mathrm{Aut}}
\def\rightiso{\xrightarrow{\simeq}}
\def\B{B}
\def\H{H}
\def\A{\mathcal{A}}
\def\Bb{\overline{B}}
\def\Rb{\overline{R}}
%\def\p{} 
\def\p{\pause}
\def\cl{\color{blue}}
\def\clg{\color{red}}
%\usefonttheme{structuresmallcapsserif}
\def\cl{\color{blue}}
\def\clg{\color{red}}



\setbeamertemplate{theorems}[numbered]



\newtheorem{proposition}{Proposition}[section]
\newtheorem*{conjecture}{Conjecture}
%\theoremstyle{definition} \newtheorem{algorithm}{Algorithm}[section]
\theoremstyle{definition} \newtheorem{remark}{Remark}[section]


\renewcommand{\algorithmicrequire}{\textbf{Input:}}
\renewcommand{\algorithmicensure}{\textbf{Output:}}
\renewcommand{\algorithmicprocedure}{\textbf{Procedure:}}
\renewcommand{\Procedure}{\textbf{Function:}}
\renewcommand{\EndProcedure}{\textbf{EndFunction:}}

\usepackage{ifthen}

% This is the file main.tex


\newcommand\uleft[2]{\,{}^{#1}\hspace*{-.1em}{#2}}
\def\idgroup{\mathbf{1}}
\newcommand\rdef[1]{\textit{\textcolor{blue}{#1}}}
\newcommand\mdef[1]{\textcolor{Brown}{#1}}

\newcommand\link[2]{\hyperlink{#2}{\beamergotobutton{#1}}}

\def\GAP{{\sf GAP} }
\def\HAP{{\sf HAP} }
\def\XMod{{\sf XMod}}
\def\C{\mathcal{C}}
\def\ZG{\mathbb{Z}G}
\input{rbg}
\def\minusvspace{\vspace*{-18pt}}
\def\xyscale{\xymatrixrowsep{0.25in} \xymatrixcolsep{0.25in}}

\def\xysmall{\xymatrixrowsep{0.15in} \xymatrixcolsep{0.15in}} 
\def\Ne{\mathcal{N}}
\def\CW{\textrm{CW}}
\parskip 0.1in

\input{def}

\def\xd{\vspace*{10pt}}
\def\ixd{\vspace*{3pt}}



 




\title[]{MA140-Engineering Calculus}
\author[ Lecture 30 ]{\large{ Lecture 30} 
\and %\\ \small{Supervisor: Dr. John Burns}
\\ 
 \includegraphics[scale=0.0]{Figures/lec14.jpg}}
%\date[June 2-4, 2016]{June 2-4, 2016}
%\institute[Uppsala University]{ }


\begin{document}
\begin{frame}
\titlepage
\end{frame}


\section{MA140}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
We begin by rotating the line segment $y=r$, from $x=A$ to $x=B$.\\
If we rotate this line segment $AB$ having length $\Delta x$ about the $x-$axis, we generate a cylinder with surface area $$2\pi y\Delta x=2\pi r\Delta x$$ This area is the same as that of a rectangle with side lengths $\Delta x$ and $2\pi y$ 
\begin{center}
			\includegraphics[scale=0.25]{Figures/lec2004.pdf}
		\end{center}
\end{frame}
\begin{frame}

{\textcolor{red}{Surface area:}}
We are interested in the surface generated by rotating the curve about the x-axis.
\begin{center}
			\includegraphics[scale=0.25]{Figures/lec2000.pdf}
		\end{center}
\end{frame}
\begin{frame}
Suppose that the arc length from $p$ to $Q$ is $\Delta S_{k}$, so the surface area of the typical band above is: 
$$\Delta S_{k} 2\pi f(x_{k})$$
So the surface area can be approximated by the following sum:
$$S\cong \sum ^{n}_{k=1} \Delta S_{k} 2\pi f(x_{k})$$
From the last lecture, we know that $$\Delta S_{k}=\Delta x_{k} \sqrt{1+(\frac{\Delta y_{k}}{\Delta x_{k}})^{2}}$$
So: 
$$S \cong \sum ^{n}_{k=1} (\Delta x_{k} \sqrt{1+(\frac{\Delta y_{k}}{\Delta x_{k}})^{2}}) 2\pi f(x_{k})$$

\end{frame}
\begin{frame}
Let $n \to \infty $ or $\Delta x_{k} \to 0$, then:
$$S=\lim_{n\to \infty} \sum ^{n}_{k=1} (\Delta x_{k} \sqrt{1+(\frac{\Delta y_{k}}{\Delta x_{k}})^{2}}) 2\pi f(x_{k})$$
As this is a Riemann sum: 
$$S=2\pi \int_{a}^{b} f(x)\sqrt{1+(\frac{dy}{dx})^{2}}dx$$
\end{frame}
\begin{frame}
\begin{example}
Find the area of the surface generated by revolving the curve $y=2\sqrt{x}$, $1\leq x\leq 2$, about the $x-$axis
\end{example}
\begin{center}
			\includegraphics[scale=0.25]{Figures/lec2001.pdf}
		\end{center}
\end{frame}
\begin{frame}
We evaluate the formula $$S=\int_{a}^{b} 2\pi y\sqrt{1+(\frac{dy}{dx})^{2}}$$
with: $$a=1, b=2, y=2\sqrt{x}, \quad \frac{dy}{dx}=\frac{1}{\sqrt{x}}$$ 
So: $$\sqrt{1+(\frac{dy}{dx})^{2}}=\sqrt{1+(\frac{1}{\sqrt{x}})^{2}}$$ $$=\sqrt{1+\frac{1}{x}}=\sqrt{\frac{x+1}{x}}=\frac{\sqrt{x+1}}{\sqrt{x}}$$
\end{frame}
\begin{frame}
So: $$S=\int_{1}^{2} 2\pi \cdot 2\sqrt{x} \cdot \frac{\sqrt{x+1}}{\sqrt{x}} dx=4\pi \int_{1}^{2} \sqrt{x+1} dx$$
$$=4\pi \cdot \frac{2}{3}(x+1)^{3/2}\Big ]^{2}_{1}=\frac{8\pi }{3}(3\sqrt{3}-2\sqrt{2})$$
\end{frame}
\begin{frame}
\begin{example}
A reflector is formed by rotating $y=\sqrt{x}$ between $x=0$ and $x=1$ about the $x-$axis. What is the surface area.
\end{example}
\begin{example}
Find the area of the surface generated by revolving the curve $y=\frac{x^{3}}{9}$ between $x=0$ and $x=2$  
\end{example}
\end{frame}

\begin{frame}
We evaluate the formula $$S=\int_{a}^{b} 2\pi y\sqrt{1+(\frac{dy}{dx})^{2}}$$
with: $$a=0, b=2, y=\frac{x^{3}}{9}, \quad \frac{dy}{dx}=\frac{x^{2}}{3}$$ 
So:  $$\sqrt{1+(\frac{dy}{dx})^{2}}=\sqrt{1+(\frac{x^{2}}{3})^{2}}$$ $$=\sqrt{1+\frac{x^{4}}{9}}=\sqrt{\frac{9+x^{4}}{9}}=\frac{\sqrt{9+x^{4}}}{3}$$
\end{frame}
\begin{frame}
So: $$S=\int_{0}^{2} 2\pi \cdot (\frac{x^{3}}{9}) \cdot \frac{\sqrt{9+x^{4}}}{3} dx=\frac{2\pi}{27} \int_{0}^{2} x^{3} \sqrt{9+x^{4}} dx$$
$$=\frac{2\pi}{27} \cdot \frac{(9+x^{4})^{3/2}}{6} \Big ]^{2}_{0}=\frac{98\pi}{77}$$
\end{frame}
\end{document}