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\noindent\underline{Honors HW5} (due 4 March) \\
In HW3 you considered \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
\[ \mathrm{SurfaceArea} = 2\pi \displaystyle\int_a^b f(x) \sqrt{1+(f'(x))^2} \mathrm{d}x.\]

In this project, you will look at ``Gabriel's horn" which is a surface defined by rotating a curve around the $x$-axis. This surface is interesting because it has finite volume but infinite surface area (so you can fill up its volume with paint, but you cannot paint it itself).
\begin{enumerate}[1.)]
\item Use the washer method to find the volume of the surface bounded by the curves $\dfrac{1}{x}$, $y=0$, and $x=1$ (note: this will have an infinite domain). You should arrive at a finite volume.
\item Use the surface area formula to \textbf{write down the integral} you would compute to find the surface area of Gabriel's horn.
\item Plot the curves $y=\dfrac{1}{x}$ and $y=\dfrac{1}{x} \sqrt{1+\dfrac{1}{x^4}}$ in Desmos (or your favorite plotting utility). Which curve is on ``top" over the interval $[1,\infty)$?
\item You should have found in 3.) that $\dfrac{1}{x} \sqrt{1+\dfrac{1}{x^4}} > \dfrac{1}{x}$. Using this fact and the general idea that if $f(x) \geq g(x)$, then $\displaystyle\int f(x) \mathrm{d}x \geq \displaystyle\int g(x) \mathrm{d}x$ (``more area"), conclude that the integral for surface area is greater than an integral which results in $\infty$. This proves the surface area of Gabriel's horn is infinite.
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