Answer
The surface area is $S \simeq 22.943$
Work Step by Step
We have
$x\left( t \right) = t$, ${\ \ \ }$ $x'\left( t \right) = 1$,
$y\left( t \right) = {{\rm{e}}^t}$, ${\ \ \ }$ $y'\left( t \right) = {{\rm{e}}^t}$.
By Eq. (4) of Theorem 3, the surface obtained by rotating $c\left( t \right)$ about the $x$-axis for $0 \le t \le 1$ has surface area:
$S = 2\pi \mathop \smallint \limits_0^1 {{\rm{e}}^t}\sqrt {1 + {{\rm{e}}^{2t}}} {\rm{d}}t$
By changing of variable, we write $u = {{\rm{e}}^t}$. So, $du = {{\rm{e}}^t}dt$. Thus, the integral becomes
$S = 2\pi \mathop \smallint \limits_1^{\rm{e}} \sqrt {1 + {u^2}} {\rm{d}}u$
Let $u = \sinh v$. So, $du = \cosh vdv$. Since ${\cosh ^2}v = 1 + {\sinh ^2}v$, so $\cosh v = \sqrt {1 + {{\sinh }^2}v} $. Thus,
$S = 2\pi \mathop \smallint \limits_{{{\sinh }^{ - 1}}1}^{{{\sinh }^{ - 1}}{\rm{e}}} {\cosh ^2}v{\rm{d}}v$
Using the identity given on page 412: ${\cosh ^2}v = \frac{1}{2}\left( {\cosh 2v + 1} \right)$, we get
$S = \pi \mathop \smallint \limits_{{{\sinh }^{ - 1}}1}^{{{\sinh }^{ - 1}}{\rm{e}}} \left( {\cosh 2v + 1} \right){\rm{d}}v$
$S = \pi \left( {\frac{1}{2}\sinh 2v + v} \right)|_{{{\sinh }^{ - 1}}1}^{{{\sinh }^{ - 1}}{\rm{e}}}$
$S = \pi \left( {\frac{1}{2}\sinh \left( {2{{\sinh }^{ - 1}}{\rm{e}}} \right) - \frac{1}{2}\sinh \left( {2{{\sinh }^{ - 1}}1} \right) + {{\sinh }^{ - 1}}{\rm{e}} - {{\sinh }^{ - 1}}1} \right)$
$S \simeq 22.943$
