1295. Calculate the length of the circumference

$$
x^{2}+y^{2}=R^{2}
$$
Solution. Differentiating the given equation, we find $x d x+y d y=0$, hence $\frac{d y}{d x}=-\frac{x}{y}$.

Using formula (1), we find the length of the arc $AB$, i.e., $\frac{1}{4}$ of the circumference (Fig. 83):

$$
\begin{aligned}
& \frac{1}{4} s=\int_{0}^{R} \sqrt{1+\frac{x^{2}}{y^{2}}} d x=\int_{0}^{R} \frac{\sqrt{x^{2}+y^{2}}}{y} d x= \\
& =\int_{0}^{R} \frac{R}{\sqrt{R^{2}-x^{2}}} d x=\left.R\left(\arcsin \frac{x}{R}\right)\right|_{0} ^{R}=\frac{1}{2} \pi R
\end{aligned}
$$

from which

$$
s=2 \pi R
$$

In this problem, formula (2) could also have been applied.