Find the arc length $s$ of the logarithmic spiral represented by $r=f(\theta)=e^{3\theta }$ from $\theta = 0$ to $\theta =2.$

Solution We use the arc length formula $s=\int_{\alpha }^{\beta } \sqrt{\,r^{2}+\left( \dfrac{dr}{d\theta }\right) ^{2}}\,d\theta$ with $r=e^{3\theta }$. Then $\dfrac{dr}{d\theta }=3e^{3\theta }$ and $$\begin{eqnarray*} s&=&\int_{0}^{2}\sqrt{(e^{3\theta })^{2}+(3e^{3\theta })^{2}}\,d\theta =\int_{0}^{2}\sqrt{10e^{6\theta }}d\theta =\sqrt{10}\int_{0}^{2}e^{3\theta }\,d\theta\\ &=& \sqrt{10}\left[\dfrac{e^{3\theta}}{3}\right]^2_0 =\frac{\sqrt{10}}{3}(e^{6}-1) \end{eqnarray*}$$