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