Example 2.61. Calculate the surface area formed by the rotation of the arc of the circle $x^{2}+y^{2}=16(y>0)$ over the segment $-1 \geqslant x \geqslant 1$ around the $O X$ axis.
Solution. We will use formula (2.30).

$$
\begin{gathered}
S=2 \pi \int_{-1}^{1} \sqrt{16-x^{2}} \sqrt{1+\left(-x / \sqrt{16-x^{2}}\right)^{2}} d x= \\
=2 \pi \int_{-1}^{1} 4 d x=16 \pi
\end{gathered}
$$