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Cone Cylinder.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} {\large Cone Cylinder} \begin{align*} &\text{\bf Q6 a ii):}\\ \\ &\text{\underline {Using Formulae}}\\ &\text{The cone volume formula is}\quad V_{cone}=\pi r^2\frac{h}{3},\quad\text{where $h$ is the hight.}\\ &\text{In this question}~~V_{cone}=\pi r^2\frac{2x}{3}.\\ %&\text{From Pythagoras's Theorem,}\quad h=\sqrt{400-r^2},\quad\text{so}~~V_{cone}=\pi r^2\frac{\sqrt{400-r^2}}{3}.\\ &\text{The volume of the cylendar at the bottom}~~V_{cyl}=\pi r^2 x.\\ &\because r^2=20^2-(2x)^2=400-4x^2.\\ &\therefore~\text{the total volume}~~V=V_{cone}+V_{cyl}=\pi r^2\frac{2x}{3}+\pi r^2 x =\pi r^2 x\left(\frac{2}{3}+1\right)\\ &\qquad\qquad=\frac{5}{3}\pi(400-4x^2)x=\frac{20}{3}\pi(100x-x^3).\\ \\ \\ &\text{\underline {Using Integration}}\\ &\text{In fact, the volume formulae are derived using integration anyway, so you may go straight to treat it as}\\ &\text{a graph and integrate to find the volume of the solid formed by rotating the graph about the x-axis.}\\ \\ &\text{Let us use the same diagram we drew in the lesson, with $u$ as the horizontal axis, and $R$ as the radius at $u$.}\\ &\frac{R}{u}=\frac{r}{2x},\quad\text{so}~~R=\frac{ru}{2x}.\quad R^2=\frac{r^2}{(2x)^2}u^2. \quad\text{($r$ and $x$ are constants. Only $u$ is varying here.)}\\ %=\frac{400-4x^2}{4x^2}u=\left(\frac{100}{x^2}-1\right)u \\ &V =\int_{0}^{2x}\pi R^2~du+\int_{2x}^{3x}\pi r^2~du\\ &=\int_{0}^{2x}\pi \frac{r^2}{(2x)^2}u^2~du+\pi r^2\int_{2x}^{3x}~du\\ &=\pi\frac{r^2}{(2x)^2}\int_{0}^{2x}u^2~du+\pi r^2\int_{2x}^{3x}~du\\ &=\pi r^2\left(\frac{1}{(2x)^2}\left[\frac{u^3}{3}\right]_{0}^{2x}+\Big[u\Big]_{2x}^{3x}\right)\\ &=\pi r^2\left(\frac{1}{(2x)^2}\left[\frac{(2x)^3}{3}\right]+\Big[3x-2x\Big]\right)\\ &=\pi r^2\left(\frac{2x}{3}+x\right)\\ &=\frac{5}{3}\pi r^2 x\\ &=\frac{5}{3}\pi(400-4x^2)x\\ &=\frac{20}{3}\pi(100x-x^3).\\ \end{align*} \end{document}