Answer
The surface area is $S \simeq 28.8472$.
Work Step by Step
We have
$x\left( t \right) = t$, ${\ \ \ }$ $x'\left( t \right) = 1$,
$y\left( t \right) = \sin t$, ${\ \ \ }$ $y'\left( t \right) = \cos t$.
Note that the curve is a sine curve: $y=\sin x$. By symmetry the surface area obtained by rotating $c\left( t \right)$ about the $x$-axis for $0 \le t \le 2\pi $ is twice the surface area for $0 \le t \le \pi $. Thus, by Eq. (4) of Theorem 3, the surface area is
$S = 4\pi \mathop \smallint \limits_0^\pi \sin t\sqrt {1 + {{\cos }^2}t} {\rm{d}}t$
Write $u=\cos t$. So, $du = - \sin tdt$. Thus, the integral becomes
$S = - 4\pi \mathop \smallint \limits_1^{ - 1} \sqrt {1 + {u^2}} {\rm{d}}u$.
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,
$S = - 4\pi \mathop \smallint \limits_{{{\sinh }^{ - 1}}1}^{{{\sinh }^{ - 1}}\left( { - 1} \right)} {\cosh ^2}v{\rm{d}}v$
Using the identity given on page 412: ${\cosh ^2}v = \frac{1}{2}\left( {\cosh 2v + 1} \right)$, we get
$S = - 2\pi \mathop \smallint \limits_{{{\sinh }^{ - 1}}1}^{{{\sinh }^{ - 1}}\left( { - 1} \right)} \left( {\cosh 2v + 1} \right){\rm{d}}v$
$S = - 2\pi \left( {\frac{1}{2}\sinh 2v + v} \right)|_{{{\sinh }^{ - 1}}1}^{{{\sinh }^{ - 1}}\left( { - 1} \right)}$
$S = - 2\pi \left( {\frac{1}{2}\sinh \left( {2{{\sinh }^{ - 1}}\left( { - 1} \right)} \right) - \frac{1}{2}\sinh \left( {2{{\sinh }^{ - 1}}1} \right) + {{\sinh }^{ - 1}}\left( { - 1} \right) - {{\sinh }^{ - 1}}1} \right)$
$S \simeq 28.8472$.
