Answer
The surface area is $S = \sqrt 2 \pi $.
Work Step by Step
We have
$x\left( t \right) = {\sin ^2}t$, ${\ \ \ }$ $x'\left( t \right) = 2\sin t\cos t$,
$y\left( t \right) = {\cos ^2}t$, ${\ \ \ }$ $y'\left( t \right) = - 2\cos t\sin t$.
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:
$S = 2\pi \mathop \smallint \limits_0^{\pi /2} \left( {{{\cos }^2}t} \right)\sqrt {{{\left( {2\sin t\cos t} \right)}^2} + {{\left( { - 2\cos t\sin t} \right)}^2}} {\rm{d}}t$
$S = 4\sqrt 2 \pi \mathop \smallint \limits_0^{\pi /2} \sin t{\cos ^3}t{\rm{d}}t$
Write $u = \cos t$. So, $du = - \sin tdt$. Thus,
$S = - 4\sqrt 2 \pi \mathop \smallint \limits_1^0 {u^3}{\rm{d}}u = - \sqrt 2 \pi {u^4}|_1^0 = \sqrt 2 \pi $.
