surface area of $y = e^{-x}, x \geq 1$, about $x$-axis Find the surface area generated by rotating $y = e^{-x}, x \geq 1$ about the $x$-axis or state that the integral diverges.
I have the equation set up, but when I change the bounds, I end up with a lower bound of $\tan(e^{-1})$. Help!
 A: I start from the integral you mentioned in the comments, for it was correct:
$2\pi \int_{1}^{\infty}e^{-x}\sqrt{1+e^{-2x}}dx$
Make the substitution $u=e^{-x}$:
$=-2\pi \int_{u=1/e}^{0}\sqrt{1+u^{2}}du$
Now, substitute: $u=tan(\theta)$, giving $du=sec^{2}(\theta)$:
$=-2\pi \int_{\theta=tan^{-1}(1/e)}^{0}\sqrt{1+tan^{2}(\theta)}sec^{2}(\theta)$
This substitution is valid because the range of $tan(\theta)$ is all real numbers, so it can take on any value which u would have.
Now, using $tan^{2}(\theta)+1=sec^{2}(\theta)$ and canceling out the square root:
$=-2\pi \int_{tan^{-1}(1/e)}^{0}sec^{3}(\theta)d\theta$
This is a really famous integral so I won't do it by hand (it'll be on a google search in a second). It evaluates to:
$\int sec^{3}(\theta)d\theta=\frac{1}{2}sec(\theta)tan(\theta)+\frac{1}{2}ln|sec(\theta)+tan(\theta)|$
I then distribute the $2\pi$, and use the (-) to reverse the order of the bounds. Plugging in $tan^{-1}(1/e)$ gives terms with $sec(tan^{-1}(1/e))$. Using the same trig identity as before (solve for secant) shows that $sec(tan^{-1}(1/e))=\sqrt{(1/e)^{2}+1}$.
All in all, that gives:
$\frac{\pi}{e} \sqrt{(1/e)^{2}+1}+\frac{\pi}{e}ln|\sqrt{(1/e)^{2}+1}+1/e|$
