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{\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 $y=-2$, (ii) the line $x=3$.

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[(\mathrm{e}^x+2)^2-2^2]\,
dx=\pi\int_0^1(\mathrm{e}^{2x}+4\mathrm{e}^x)\, dx\\ &amp;=&amp;
\pi\left[\frac{\mathrm{e}^{2x}}{2}+4\mathrm{e}^x\right]_0^1 =
\pi\left[\frac{\mathrm{e}^2}{2}+4\mathrm{e}-\frac{1}{2}-4\right]\\&amp;=&amp;
\pi\left[\frac{\mathrm{e}^2}{2}+4\mathrm{e}-\frac{9}{2}\right]=31.628\
\mathrm{(3\ d.p.)}
\end{eqnarray*}

(ii)
\begin{eqnarray*}
V &amp;=&amp;2\pi\int_0^1(3-x)\mathrm{e}^x\, dx
=2\pi\left[(3-x)\mathrm{e}^x\right]_0^1+2\pi\int_0^1\mathrm{e}^x\,
dx
\\ &amp;=&amp;
2\pi(2\mathrm{e}-3)+2\pi(\mathrm{e}-1)=2\pi(3\mathrm{e}-4)=26.106\
\mathrm{(3\ d.p.)}
\end{eqnarray*}




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