#### Answer

$$2\pi$$

#### Work Step by Step

Since
\begin{aligned}
x^{2}+y^{2}&=1 \\
y&=\sqrt{1-x^{2}} \\
y^{\prime} &=-\frac{x}{\sqrt{1-x^{2}}}
\end{aligned}
Then
\begin{aligned}
s&=2 \int_{-1}^{1} \sqrt{1+\left(-\frac{x}{\sqrt{1-x^{2}}}\right)^{2}} d x \\
&=2 \int_{-1}^{1} \sqrt{1+\frac{x^{2}}{1-x^{2}}} d x \\
&=2 \int_{-1}^{1} \sqrt{\frac{x^{2}+1-x^{2}}{1-x^{2}}} d x \\
&=2 \int_{-1}^{1} \sqrt{\frac{x^{2}+1-x^{2}}{1-x^{2}}} d x \\
&=2 \int_{-1}^{1} \frac{1}{\sqrt{1-x^{2}}} d x\\
&=2\sin^{-1}x\bigg|_{-1}^{1}\\
&=2\pi
\end{aligned}