
Let us now transform this expression into polar coördinates by means
of the substitutions
\begin{align*} % Not running text into the display
x &= \rho \cos \theta,\quad y = \rho \sin \theta. \\
\intertext{\indent Then}
dx &= -\rho \sin \theta\, d\theta + \cos \theta\, d\rho, \\
dy &=  \rho \cos \theta\, d\theta + \sin \theta\, d\rho,
\end{align*}
and we have
\begin{align*}
[dx^2 + dy^2]^{\frac{1}{2}}
  &= [(-\rho \sin \theta\, d\theta + \cos \theta\, d\rho)^2
    + ( \rho \cos \theta\, d\theta + \sin \theta\, d\rho)^2]^{\frac{1}{2}} \\
  &= [\rho^2\, d\theta^2 + d\rho^2]^{\frac{1}{2}}.
\end{align*}

If the equation of the curve is
\begin{align*}
\rho &= f(\theta), \\
\intertext{then}
d\rho &= f'(\theta)\, d\theta = \frac{d\rho}{d\theta}\, d\theta.
\end{align*}

Substituting this in the above differential expression, we get
\[
\left[\rho^2 + \left(\frac{d\rho}{d\theta}\right)^2\right]^{\frac{1}{2}}\, d\theta.
\]

If then $\alpha$ and~$\beta$ are the limits of the independent variable~$\theta$
corresponding to the limits in~$(A)$ and~$(B)$, p.~374, % [** F1: Hard-coded page number]
we get the \textbf{formula for the length of the arc},
\[
\tag{A} % Okay if tag is on the right
s = \int_\alpha^\beta \left[\rho^2 + \left(\frac{d\rho}{d\theta}\right)^2\right]^{\frac{1}{2}}\, d\theta,
\]
where $\rho$ and~$\dfrac{d\rho}{d\theta}$ in terms of~$\theta$ must be substituted
from the equation of the given curve.
% [** Illustration]

In case it is more convenient to use $\rho$ as the independent variable,
and the equation is in the form
\begin{align*}
\theta &= \phi(\rho), \\
\intertext{then}
d\theta &= \phi'(\rho)\, d\rho = \frac{d\theta}{d\rho}\, d\rho.
\end{align*}

Substituting this in
\[
[\rho^2\, d\theta^2 + d\rho^2]^{\frac{1}{2}}
\]
gives
\[
\left[\rho^2 \left(\frac{d\theta}{d\rho}\right)^2 + 1\right]^{\frac{1}{2}}\, d\rho.
\]

Hence if $\rho_1$ and~$\rho_2$ are the corresponding limits of the independent
variable~$\rho$, we get the \textbf{formula for the length of the arc},
\[
\tag{B}
s = \int_{\rho_1}^{\rho_2} \left[\rho^2 \left(\frac{d\theta}{d\rho}\right)^2 + 1\right]^{\frac{1}{2}}\, d\rho,
\]
where $\dfrac{d\theta}{d\rho}$ in terms of~$\rho$ must be substituted from the equation of the
given curve.
