## Calculus (3rd Edition)

The surface area is $S = m\sqrt {1 + {m^2}} \pi {A^2}$
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: $S = 2\pi \mathop \smallint \limits_0^A mt\sqrt {1 + {m^2}} {\rm{d}}t$ $S = 2\pi m\sqrt {1 + {m^2}} \mathop \smallint \limits_0^A t{\rm{d}}t = 2\pi m\sqrt {1 + {m^2}} \frac{1}{2}{t^2}|_0^A$ $S = m\sqrt {1 + {m^2}} \pi {A^2}$.