# Show the surface area of revolution of $e^{-x}$ is finite

I need to show that the surface area of revolution of $e^{-x}$ is finite when the region from $x=0$ to $x=\infty$ is rotated about the x-axis.

I tried using the surface area formula, but got stuck on the integration, so I thought that maybe there was a trick to show that it was infinite without actually finding the definite integral.

This is where I got with my integration:

\begin{align} SA &= 2\pi\, \int^b_a\, f(x)\sqrt{1+f'(x)^2}\, \mathrm{d}x \\ &= 2\pi\, \int^\infty_0\, e^{-x}\sqrt{1+e^{-2x}}\, \mathrm{d}x \\ &= \lim_{t \to \infty} 2\pi\, \int^t_0\, e^{-x}\sqrt{1+e^{-2x}}\, \mathrm{d}x \\ \end{align}

\begin{align} \mathrm{let\,\,} u &= e^{-x}.\\ \mathrm{d}u &= -e^{-x}\, \mathrm{d}x\\ \mathrm{d}x &= \frac{-1}{e^{-x}} \,\mathrm{d}u\\ \mathbf{But:}\, u &= e^{-x}, \,\mathrm{so}\\ \mathrm{d}x &= \frac{-1}{u} \mathrm{d}u. \end{align} When $x=0$, $u=1$ and when $x=t$, $u=e^{-t}$. However, as $t \to \infty$, $u \to 0$. (Is this a bad assumption to make here?) So,

\begin{align} SA &= -2\pi\int_1^0 \sqrt{1+u^2}\, \mathrm{d}u. \end{align}

This looks like it could be a nice enough integral, but I can't immediately see how to do it, and Wolfram|Alpha seems to use magic (and $\sinh^{-1}$, which is almost the same thing... =P)

Remember that I have only been asked to show the surface area is finite. I don't actually have to find the surface area. (But if you know a method that doesn't involve too many advanced tricks, I'm happy to hear it.)

Can we just say that the area under that curve is obviously finite, because it is the area bounded by a continuous curve between $x=0$ and $x=1$? Should we perhaps say that this area is less than the area under $-2\pi \int_1^0 1+x^2\,\mathrm{d}x$, which is obviously finite, and thus this area itself is finite? What is the best way to answer this?

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We want to calculate the surface area $S(M)$ from $0$ to $M$, and see what happens to $S(M)$ as $M$ gets large. You had done all the necessary work. We have $$S(M)=\int_0^M 2\pi e^{-x}\sqrt{1+e^{-2x}}\,dx.$$ Note that $\sqrt{1+e^{-2x}}\le \sqrt{2}$ in our interval. So the function we are integrating is between $0$ and $2\sqrt{2}\pi e^{-x}$.
The integral $$\int_0^\infty 2\sqrt{2}\pi e^{-x}\,dx$$ is an upper bound for $S(M)$, so $$\int_0^\infty 2\pi e^{-x}\sqrt{1+e^{-2x}}\,dx$$ converges.
Remark: There are various ways to find explicit expressions for the integral. The most pleasant is through a hyperbolic function substitution. Quite a bit uglier, but doable, is your substitution, followed by $u=\tan t$.
Your substitution process that leads to $\int_0^1 \sqrt{1+u^2}\,du$ can be justitified. The part that made you justifiably nervous can be dealt with by noting that for $S(M)$ we are integrating from $u=e^{-M}$ to $1$.
And there is no issue of convergence for $\int_0^1 2\pi \sqrt{1+u^2}\,du$. We are integrating a continuous function on a bounded interval.