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{\bf Exam Question

Topic: Volume of Revolution}

Let $R$ denote the region in the first quadrant of the $x$-$y$
plane bounded by the $x$-axis, the $y$-axis, and the curve
$y=1-x^2.$

(i) Find the volume of the solid obtained by rotating $R$ around
the line $x=3.$

(ii) Find the total surface area of the solid obtained by rotating
$R$ around the $y$-axis.

 \vspace{0.5in}

{\bf Solution}

\begin{eqnarray*}
\mathrm{(i)}\ \ V&=&2\pi\int_0^1(3-x)(1-x^2)\,
dx=2\pi\int_0^1(3-x-3x^2+x^3)\, dx\\
&=&2\pi\left(3-\frac{1}{2}-1+\frac{1}{4}\right)=\frac{7\pi}{2}
\end{eqnarray*}

\begin{eqnarray*}
\mathrm{(ii)}\ \ S&=&2\pi\int_0^1 x\sqrt{1+4x^2}\, dx +\pi\\
&=&2\pi\left[\frac{1}{12}\left(1+4x^2\right)^{3/2}\right]_0^1+\pi
=\frac{\pi}{6}\left(5^{3/2}-1\right)+\pi\\ &=&
\frac{\pi}{6}\left(5^{3/2}+5\right)=\frac{5\pi}{6}\left(\sqrt{5}+1\right)
\end{eqnarray*}





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