<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\setlength\oddsidemargin{0pt} \setlength\evensidemargin{0pt}
\setlength\topmargin{0pt}
\begin{document}
\parindent=0pt
{\bf Exam Question

Topic: Volume of Revolution}

Find the volume of revolution obtained by rotating the region in
the $x$-$y$ plane bounded by the lines $x=0,y=0,x=1$ and the curve
$y=\mathrm{e}^x$ about (i) the line $y=3$, (ii) the line $x=-2$.

Give your answer in terms of e, and also as an approximation
correct to 3 decimal places, using your calculator.
 \vspace{0.5in}

{\bf Solution}


(i)
\begin{eqnarray*}
V &amp;=&amp;\pi\int_0^1[3^2-(3-\mathrm{e}^x)^2]\,
dx=\pi\int_0^1(6\mathrm{e}^x-\mathrm{e}^{2x})\, dx\\ &amp;=&amp;
\pi\left[6\mathrm{e}^x-\mathrm{e}^{2x}/2\right]_0^1 =
\pi[6\mathrm{e}-\mathrm{e}^2/2-6+1/2]\\&amp;=&amp;
\frac{\pi}{2}[12\mathrm{e}+\mathrm{e}^2-11]=22.353\ \mathrm{(3\
d.p.)}
\end{eqnarray*}

(ii)
\begin{eqnarray*}
V &amp;=&amp;2\pi\int_0^1(x+2)\mathrm{e}^x\, dx
=\left[2\pi(x+2)\mathrm{e}^x\right]_0^1-\int_0^12\pi\mathrm{e}^x\,
dx \\ &amp;=&amp;
\left[2\pi(x+2)\mathrm{e}^x\right]_0^1-\left[2\pi\mathrm{e}^x\right]_0^1
= 2\pi(2\mathrm{e}-1)=27.867\ \mathrm{(3\ d.p.)}
\end{eqnarray*}




\end{document}
</pre></body></html>