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\title[]{MA140-Engineering Calculus}
\author[ Lecture 32 ]{\large{ Lecture 32 } 
\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}

 {\textcolor{red}{Centroid of a plane area:}}\\
 \begin{center}
			\includegraphics[scale=1.0]{Figures/lec7000.pdf}
		\end{center}

\end{frame}
\begin{frame}
Suppose the centroid of a typical strip is at $(x_{k},y_{k})$
Then: $$System \enspace mass\cong \sum_{k=1}^{n} \Delta m_{k}$$
The moments of the entire system about the two axes are: $$Moment \enspace about \enspace x-axis\cong \sum_{k=1}^{n} y_{k} \Delta m_{k}$$
$$Moment \enspace about \enspace y-axis\cong \sum_{k=1}^{n} x_{k} \Delta m_{k}$$


\end{frame}
\begin{frame}
The $x$-coordinate of the system's center of mass is defined to be:
$$\bar{x}=\frac{Moment \enspace about \enspace y-axis}{System \enspace mass}\cong \frac{\sum_{k=1}^{n} x_{k} \Delta m_{k}}{\sum_{k=1}^{n} \Delta m_{k}}$$
As $\delta=1$, $\Delta m_{k}=\Delta A_{k}=(f_{2}(x_{k})-f_{1}(x_{k}))\Delta x$. So:
$$\bar{x}\cong \frac{\sum_{k=1}^{n} x_{k} (f_{2}(x_{k})-f_{1}(x_{k}))\Delta x}{\sum_{k=1}^{n} (f_{2}(x_{k})-f_{1}(x_{k}))\Delta x}$$
The sums is a Riemann sum so when $\Delta x\to 0$ or $n\to \infty$, then $$\bar{x}= \frac{\int_{a}^{b} x (f_{2}(x)-f_{1}(x)) dx}{\int_{a}^{b} (f_{2}(x))-f_{1}(x)) dx}=\frac{\int_{a}^{b} x (f_{2}(x)-f_{1}(x)) dx}{A}$$
\end{frame}
\begin{frame}
The $y$-coordinate of the system's center of mass is defined to be:
$$\bar{y}=\frac{1}{2A}\int_{a}^{b}  \Big [(f_{2}(x))^{2}-(f_{1}(x))^{2}\Big ] dx$$
\begin{example}
Find the centroid of the area under the curve $y=\sqrt{x-2}$ between the lines $x=2$ and $x=5$
\end{example}
\end{frame}
\begin{frame}
$$Area=\int_{2}^{5} \sqrt{x-2} dx=\int_{2}^{5} (x-2)^{1/2} dx=\frac{(x-2)^{3/2}}{3/2}\Big ]_{2}^{5}$$ 
$$=\frac{2}{3}[(5-2)^{3/2}-(2-2)^{3/2}]=\frac{2}{3}3\sqrt{3}=2\sqrt{3}$$
the $x$ coordinate of the centroid is: $$\bar{x}=\frac{1}{A}\int_{2}^{5} x\sqrt{x-2} dx$$
Let $u=x-2$, so $x=u+2$ and $du=dx$ then: 
$$\int x\sqrt{x-2} dx=\int (u+2)\sqrt{u} du=\int u^{3/2}+2u^{1/2} du=\frac{u^{5/2}}{5/2}+2\frac{u^{3/2}}{3/2}+c$$ 
$$=\frac{2}{5}(x-2)^{5/2}+\frac{4}{3}(x-2)^{3/2}+c$$
\end{frame}
\begin{frame}
$$\bar{x}=\frac{1}{A}\int_{2}^{5} x\sqrt{x-2} dx=\frac{1}{2\sqrt{3}}\Big( \frac{2}{5}(x-2)^{5/2}+\frac{4}{3}(x-2)^{3/2}\Big ]_{2}^{5}\Big )=\frac{19}{5}$$
Also: 
$$\bar{y}=\frac{1}{2A}\int_{2}^{5} [f(x)]^{2} dx=\frac{1}{2A}\int_{2}^{5} (x-2) dx=\frac{1}{2A} (\frac{x^{2}}{2}-2x\Big ]_{2}^{5})$$ So $$\bar{y}=\frac{3\sqrt{3}}{8}$$
\end{frame}

\end{document}