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\noindent\underline{Honors HW3} (due 13 February) \\
We looked at \textbf{volumes} of solids of revolution in class. In this assignment, we will look at \textbf{surface area} of solids of revolution.
From the text, Section 2.4, p. 279, the surface area of a solid of revolution formed by rotating the curve $f(x)$ above $[a,b]$ around the $x$-axis is given by
\[ \mathrm{SurfaceArea} = 2\pi \displaystyle\int_a^b f(x) \sqrt{1+(f'(x))^2} \mathrm{d}x.\]
\begin{enumerate}[1.)]
\item Find the surface area of the solid of revolution obtained by rotating the curve $f(x)=x^3$ lying above $[0,1]$ around the $x$-axis.
\item Find the surface area of the solid of revolution obtained by rotating the curve $f(x)=\sqrt{9-x^2}$ lying above the interval $[-2,2]$ around the $x$-axis.
\item Set up \textbf{but do not evaluate} an integral to compute the surface area of the solid of revolution obtained by rotating the curve $f(x)=\sin(2x)$ lying above $\left[ 0, \dfrac{\pi}{8} \right]$ about the $x$-axis.
\item Use software (my suggestion: WolframAlpha) to numerically solve the integral that you found in Problem 3.
\end{enumerate}
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