Answer
$$L = 2\sqrt {10} \pi $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = 3\cos t{\bf{i}} + 3\sin t{\bf{j}} + t{\bf{k}};\,\,\,\,\,\,\,\,\,\,\,0 \leqslant t \leqslant 2\pi \cr
& \cr
& {\text{Calculate }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {3\cos t{\bf{i}} + 3\sin t{\bf{j}} + t{\bf{k}}} \right] \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {3\cos t} \right]{\bf{i}} + \left[ {3\sin t} \right]{\bf{j}} + \left[ t \right]{\bf{k}} \cr
& {\bf{r}}'\left( t \right) = - 3\sin t{\bf{i}} + 3\cos t{\bf{j}} + {\bf{k}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \left\| { - 3\sin t{\bf{i}} + 3\cos t{\bf{j}} + {\bf{k}}} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( { - 3\sin t} \right)}^2} + {{\left( {3\cos t} \right)}^2} + {{\left( 1 \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {9{{\sin }^2}t + 9{{\cos }^2}t + 1} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {9\left( {{{\sin }^2}t + {{\cos }^2}t} \right) + 1} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {10} \cr
& \cr
& {\text{Find the arc length using }}L = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& {\text{for the interval }}\,0 \leqslant t \leqslant 2\pi ,\,\,{\text{ }}a = 0{\text{ and }}b = 2\pi . \cr
& L = \int_0^{2\pi } {\sqrt {10} } dt \cr
& \cr
& {\text{Integrate and evaluate}} \cr
& L = \sqrt {10} \left[ t \right]_0^{2\pi } \cr
& L = \sqrt {10} \left( {2\pi - 0} \right) \cr
& L = 2\sqrt {10} \pi \cr} $$
