Compute the surface area swept out if the graph $y = e^{-x}$, for $0 \leq x \leq 1$, is revolved about the $x$-axis.
The formula for the surface area is
$$S = 2\pi\int_{a}^{b} f(x) \sqrt{1 + f'(x)^{2}}\, dx.$$
So, $$ S = 2\pi \int_{0}^{1} e^{-x} \sqrt{1 + e^{-2x}}\, dx. $$
My professor recommended that I use trig substitution from this point. I tried letting $e^{-x} = \tan\theta$, but I think my limits of integration would be undefined [$-\ln(\tan 0)$]. Just this starting point would be helpful.
I also tried letting $u = -e^{-x}$, but it didn't really help.
Letting $u=e^{-x}$, we have $$\int_0^1 e^{-x} \sqrt{1+e^{-2x}} dx=-\int_1^{1/e}\sqrt{1+u^2} du=\int_{1/e}^1\sqrt{1+u^2} du$$
This last integral can be done with a trig substitution, or can be looked in a table of standard integrals. The limits of integration shouldn't cause any problems.