Answer
The surface area is $S = \frac{{64\pi }}{3}$.
Work Step by Step
We have
$x\left( t \right) = t - \sin t$, ${\ \ \ }$ $x'\left( t \right) = 1 - \cos t$,
$y\left( t \right) = 1 - \cos t$, ${\ \ \ }$ $y'\left( t \right) = \sin t$.
The $x$-interval for one arch of the cycloid is $0 \le x \le 2\pi $ as is shown in Figure 25 of Exercise 97 (Section 12.1). The corresponding $t$-interval is $0 \le x \le 2\pi $.
By Eq. (4) of Theorem 3, the surface obtained by rotating $c\left(t\right)$ about the $x$-axis for $0 \le x \le 2\pi $ has surface area:
$S = 2\pi \mathop \smallint \limits_0^{2\pi } \left( {1 - \cos t} \right)\sqrt {{{\left( {1 - \cos t} \right)}^2} + {{\left( {\sin t} \right)}^2}} {\rm{d}}t$
$S = 2\pi \mathop \smallint \limits_0^{2\pi } \left( {1 - \cos t} \right)\sqrt {2\left( {1 - \cos t} \right)} {\rm{d}}t$
$S = 2\sqrt 2 \pi \mathop \smallint \limits_0^{2\pi } {\left( {1 - \cos t} \right)^{3/2}}{\rm{d}}t$
Since $2{\sin ^2}\frac{t}{2} = 1 - \cos t$, we get
$S = 2\sqrt 2 \pi \left( {{2^{3/2}}} \right)\mathop \smallint \limits_0^{2\pi } {\sin ^3}\frac{t}{2}{\rm{d}}t$
$S = 8\pi \mathop \smallint \limits_0^{2\pi } {\sin ^3}\frac{t}{2}{\rm{d}}t$.
Let $u = \frac{t}{2}$. So, $du = \frac{1}{2}dt$. The integral becomes
$S = 16\pi \mathop \smallint \limits_0^\pi {\sin ^3}u{\rm{d}}u$.
Note that this integral has been evaluated in Example 1 of Section 8.2 (page 397), that is:
$\smallint {\sin ^3}x{\rm{d}}x = \frac{{{{\cos }^3}x}}{3} - \cos x + C$
So,
$S = 16\pi \left( {\frac{{{{\cos }^3}x}}{3} - \cos x} \right)|_0^\pi = 16\pi \left( { - \frac{1}{3} + 1 - \frac{1}{3} + 1} \right) = \frac{{64\pi }}{3}$.
