Answer
The surface area is $S = \frac{6}{5}\pi $.
Work Step by Step
We have
$x\left( t \right) = {\cos ^3}t$, ${\ \ \ }$ $x'\left( t \right) = - 3{\cos ^2}t\sin t$,
$y\left( t \right) = {\sin ^3}t$, ${\ \ \ }$ $y'\left( t \right) = 3{\sin ^2}t\cos t$.
By Eq. (4) of Theorem 3, the surface obtained by rotating $c\left(t\right)$ about the $x$-axis for $0 \le t \le \frac{\pi }{2}$ has surface area:
$S = 2\pi \mathop \smallint \limits_0^{\pi /2} {\sin ^3}t\sqrt {{{\left( { - 3{{\cos }^2}t\sin t} \right)}^2} + {{\left( {3{{\sin }^2}t\cos t} \right)}^2}} {\rm{d}}t$
$S = 2\pi \mathop \smallint \limits_0^{\pi /2} {\sin ^3}t\sqrt {9{{\cos }^4}t{{\sin }^2}t + 9{{\sin }^4}t{{\cos }^2}t} {\rm{d}}t$
$S = 6\pi \mathop \smallint \limits_0^{\pi /2} {\sin ^3}t\sqrt {{{\sin }^2}t{{\cos }^2}t\left( {{{\cos }^2}t + {{\sin }^2}t} \right)} {\rm{d}}t$
$S = 6\pi \mathop \smallint \limits_0^{\pi /2} {\sin ^4}t\cos t{\rm{d}}t$
Let $u=\sin t$. So, $du = \cos tdt$.
$S = 6\pi \mathop \smallint \limits_0^1 {u^4}{\rm{d}}u = 6\pi \frac{1}{5}{u^5}|_0^1 = \frac{6}{5}\pi $.
