Calculate the surface area generated by rotating

Calculate the surface area generated by rotating the curve $y=e^{-x}$ about the $x$-axis for $x>0$.

Step 1: Set up

\[y'=-e^{-x} \implies 1+(y')^2=1+e^{-2x}\]

\[S=2\pi\int_0^\infty e^{-x}\sqrt{1+e^{-2x}}\,dx\]

Step 2: Substitute $u=e^{-x}$

$du=-e^{-x}dx$; the limits $x=0,\infty$ become $u=1,0$:

\[S=2\pi\int_0^1\sqrt{1+u^2}\,du\]

Step 3: Evaluate

\[S=2\pi\left[\frac{u\sqrt{1+u^2}}2+\frac12\ln\left(u+\sqrt{1+u^2}\right)\right]_0^1 = 2\pi\left[\frac{\sqrt2}2+\frac12\ln(1+\sqrt2)\right]\]

\[\boxed{S=\pi\left(\sqrt2+\ln(1+\sqrt2)\right)}\]

Step 4: Numerical value

\[S\approx\pi(1.414+0.881)\approx2.295\pi\approx7.21\]

Step 5: Interpretation

Even though $x$ ranges over the infinite interval $[0,\infty)$, the surface area is finite — the curve decays fast enough that the horn-shaped surface it generates has finite area, exactly $\pi(\sqrt2+\ln(1+\sqrt2))$, half the value obtained in Problem 120 for one full arch of $\sin x$ (a different curve whose arc-length integrand happens to reduce to the same standard form).