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{\bf Exam Question

Topic: Surface of Revolution}

Find the area of the curved surface obtained by rotating the graph
of $y=\sqrt{x},$ between $x=0$ and $x=1,$ about the line $y=0.$

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{\bf Solution}

\begin{eqnarray*}
\mathrm{Area}\ &=& 2\pi\int_0^1 y\,
\sqrt{1+\left(\frac{dy}{dx}\right)^2}\, dx = 2\pi\int_0^1
\sqrt{x}\, \sqrt{1+\left(\frac{1}{2\sqrt{x}}\right)^2}\, dx\\ &=&
2\pi\int_0^1 \sqrt{1+\textstyle{\frac{1}{4}}}\, dx
=\textstyle{\frac{4\pi}{3}}\left[\left(x+\textstyle{\frac{1}{4}}\right)^{3/2}\right]_0^1\\
&=&\textstyle{\frac{4\pi}{3}}\left[\left(\textstyle{\frac{5}{4}}\right)^{3/2}
-\left(\textstyle{\frac{1}{4}}\right)^{3/2}\right]
=\textstyle{\frac{\pi}{6}}\left(5\sqrt{5}-1\right).
\end{eqnarray*}

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