Answer
$$
x= t+ e^{t}, \quad y= e^{-t}, \quad 0 \leq t \leq \pi / 1
$$
The surface area of the curve, which is obtained by rotating the given curve about the x-axis, is
$$
S=\int 2 \pi y d s=\int_{0}^{\pi / 2} 2 \pi e^{-t} \sqrt{ 1+2e^{t}+e^{2t}+e^{-2t} } d t \approx 10.6705
$$
Work Step by Step
$$
x= t+ e^{t}, \quad y= e^{-t}, \quad 0 \leq t \leq \pi / 1
$$
then
$$
d x / d t=1+e^{t} , \quad d y / d t = -e^{-t}
$$
so
$$
\begin{aligned}\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2} &=(1+e^{t}) ^{2} + (-e^{-t}) ^{2} \\
&=1+2e^{t}+e^{2t}+e^{-2t}
\end{aligned}
$$
Thus the surface area the curve, which is obtained by rotating the given curve about the x-axis, is
$$
S=\int 2 \pi y d s=\int_{0}^{\pi / 2} 2 \pi e^{-t} \sqrt{ 1+2e^{t}+e^{2t}+e^{-2t} } d t \approx 10.6705
$$
