Answer
The surface area is $S = \frac{\pi }{6}\left( {5\sqrt 5 - 1} \right)$
Work Step by Step
We have
$x\left( t \right) = {t^2}$, ${\ \ \ }$ $x'\left( t \right) = 2t$,
$y\left( t \right) = t$, ${\ \ \ }$ $y'\left( t \right) = 1$.
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:
$S = 2\pi \mathop \smallint \limits_0^1 t\sqrt {{{\left( {2t} \right)}^2} + 1} {\rm{d}}t = 2\pi \mathop \smallint \limits_0^1 t\sqrt {1 + 4{t^2}} {\rm{d}}t$
By changing of variable, we write $u=1+4t^2$. So, $du = 8tdt$. Thus, the integral becomes
$S = \frac{\pi }{4}\mathop \smallint \limits_1^5 \sqrt u {\rm{d}}u = \frac{\pi }{4}\cdot\frac{2}{3}{u^{3/2}}|_1^5 = \frac{\pi }{6}\left( {{5^{3/2}} - 1} \right) = \frac{\pi }{6}\left( {5\sqrt 5 - 1} \right)$
