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\title[]{MA140-Engineering Calculus}
\author[ ]{\large{ } 
\and %\\ \small{Supervisor: Dr. John Burns}
\\ 
 \includegraphics[scale=0.4]{Figures/lec14.jpg}}
%\date[June 2-4, 2016]{June 2-4, 2016}
%\institute[Uppsala University]{ }


\begin{document}


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

\begin{frame}

{\textcolor{red}{Find the volume of solids:}}
\begin{center}
			\includegraphics[scale=0.35]{Figures/1111.jpg}
		\end{center}
		\begin{center}
			\includegraphics[scale=0.35]{Figures/2222.jpg}
		\end{center}
\end{frame}
%\begin{frame}
%\begin{definition}
%The volume of a solid of known integrable cross-sectional area $A(x)$ from $x=a$ to $x=b$ is the integral of $A$ from $a$ to $b$, $$V=\int_{a}^{b} A(x) dx$$
%\end{definition}
%\begin{theorem}
%{\textcolor{red}{Calculating the volume of solids:}}
%\begin{itemize}
%\item[(1)] Sketch the solid and a typical cross-section.
%\item[(2)] Find a formula for $A(x)$, the area of a typical cross-section.
%\item[(3)] Find the limits of integration.
%\item[(4)] Integrate $A(x)$ using the Fundamental Theorem.  
%\end{itemize}
%\end{theorem}
%\end{frame}
%\begin{frame}
%\begin{example}
%A pyramid $3$ m high has a square base that is $3$ m on a side. Find
%the volume of the pyramid.
%\end{example}
%We draw the pyramid: 
%\begin{center}
%			\includegraphics[scale=0.28]{Figures/lec603.pdf}
%		\end{center}
%\end{frame}
%\begin{frame}
%The cross-section at $x$ is a square $x$ meters on a side, so its area is: $$A(x)=x^{2}$$ 
%The squares lie on the planes from $x=0$ to $x=b$. So the volume is $$V=\int_{0}^{3} A(x) dx=\int_{0}^{3} x^{2} dx=\frac{x^{3}}{3} \Big ]_{0}^{3}=9$$ 
%\end{frame}
\begin{frame}
The solid generated by rotating a plane region about an axis in its plane is called a solid of
revolution. To find the volume of a solid we need only
observe that the cross-sectional area $A(x)$ is the area of a disk of radius $R(x)$, the distance
of the planar region’s boundary from the axis of revolution. The area is then $$A(x)=\pi (radius)^{2}=\pi [R(x)]^{2}$$ 
So the definition of volume gives: $$V=\int_{a}^{b} A(x) dx=\int_{a}^{b} \pi [R(x)]^{2} dx$$
\end{frame}
\begin{frame}
\begin{example}
The region between the curve $y=\sqrt{x}$, $0\leq x\leq 4$, and the $x-$axis is revolved about the $x-$axis to generate a solid. Find its volume.
\end{example}
We draw figures showing the region, a typical radius, and the generated solid.
\begin{center}
			\includegraphics[scale=0.28]{Figures/lec604.pdf}
		\end{center}
\end{frame}
\begin{frame}
\begin{center}
			\includegraphics[scale=0.38]{Figures/lec605.pdf}
		\end{center}
\end{frame}
\begin{frame}
The volume is: $$V=\int_{a}^{b} \pi [R(x)]^{2} dx==\int_{0}^{4} \pi [\sqrt{x}]^{2} dx$$ 
$$=\pi\int_{0}^{4} x dx=\pi \frac{x^{2}}{2} \Big ]_{0}^{4}=\pi \frac{(4)^{2}}{2}=8\pi$$
\end{frame}
\begin{frame}
\begin{example}
The circle $$x^{2}+y^{2}=a^{2}$$
is rotated about the $x-$axis to generate a sphere. find its volume
\end{example}
We imagine the sphere cut into thin slices, \begin{center}
			\includegraphics[scale=0.26]{Figures/lec606.pdf}
		\end{center}
\end{frame}
\begin{frame}
the cross-sectional area at a typical point $x$ between $-a$ and $a$ is: $$A(x)=\pi y^{2}=\pi (a^{2}-x^{2})$$
Therefore, the volume is $$V=\int_{-a}^{a} A(x) dx=\int_{-a}^{a} \pi (a^{2}-x^{2}) dx$$
$$=\pi \Big [ a^{2}x-\frac{x^{3}}{3} \Big ]_{-a}^{a}=\frac{4}{3} \pi a^{3}$$  
\end{frame}
\begin{frame}
\begin{example}
Sketch the area between $x=2$ and $x=3$, under the curve $$y=\frac{1}{x-1}$$
Also find the volume of the solid if this area is rotated around the $x-$axis.
\end{example}
\end{frame}
\end{document}