Answer
General formula for the length of a helix:
$s = \sqrt {{{\left( {2\pi NR} \right)}^2} + {h^2}} $
Work Step by Step
In Exercise 21, we obtain the parametrization of a helix of radius $R$ and height $h$ that makes $N$ complete turns, given by
${\bf{r}}\left( t \right) = \left( {R\cos \left( {\frac{{2\pi Nt}}{h}} \right),R\sin \left( {\frac{{2\pi Nt}}{h}} \right),t} \right)$ ${\ \ }$ for $0 \le t \le h$
The derivative is
${\rm{r}}'\left( t \right) = \left( { - \frac{{2\pi N}}{h}R\sin \left( {\frac{{2\pi Nt}}{h}} \right),\frac{{2\pi N}}{h}R\cos \left( {\frac{{2\pi Nt}}{h}} \right),1} \right)$
So,
$||{\rm{r}}'\left( t \right)|{|^2} = {\left( { - \frac{{2\pi N}}{h}R\sin \left( {\frac{{2\pi Nt}}{h}} \right)} \right)^2}$
$ + {\left( {\frac{{2\pi N}}{h}R\cos \left( {\frac{{2\pi Nt}}{h}} \right)} \right)^2} + {1^2}$
$||{\rm{r}}'\left( t \right)|{|^2} = {\left( {\frac{{2\pi NR}}{h}} \right)^2} + 1$
Thus, the length of the helix is
$s = \mathop \smallint \limits_0^h \sqrt {{{\left( {\frac{{2\pi NR}}{h}} \right)}^2} + 1} {\rm{d}}t = h\sqrt {{{\left( {\frac{{2\pi NR}}{h}} \right)}^2} + 1} $
$s = h\sqrt {\frac{{{{\left( {2\pi NR} \right)}^2} + {h^2}}}{{{h^2}}}} = \sqrt {{{\left( {2\pi NR} \right)}^2} + {h^2}} $
