Example 2.63. Calculate the surface area formed by the rotation around the polar axis of the lemniscate $\rho=a \sqrt{\cos 2 \varphi}$ over the interval $0 \leqslant \varphi \leqslant \pi / 4$.
Solution. The general view of the rotating curve is shown in Fig. 2.10. We will use formula (2.34):

$$
\begin{aligned}
& \sqrt{\rho^{2}+\left(\frac{d \rho}{d \varphi}\right)^{2}}=\sqrt{a^{2} \cos 2 \varphi+\left(\frac{a \sin 2 \varphi}{\sqrt{\cos 2 \varphi}}\right)^{2}}=\frac{a}{\sqrt{\cos 2 \varphi}} \\
S= & \left|2 \pi \int_{0}^{\pi / 4}\left[a \sqrt{\cos 2 \varphi} \sin \varphi \frac{a}{\sqrt{\cos 2 \varphi}}\right]\right| d \varphi=\left|2 \pi a^{2} \int_{0}^{\pi / 4} \sin \varphi d \varphi\right|= \\
= & \left|2 \pi a^{2}(-\cos \varphi)\right|_{0}^{\pi / 4}=\left|2 \pi a^{2}\left(\frac{\sqrt{2}}{2}-1\right)\right|=\pi a^{2}(2-\sqrt{2})
\end{aligned}
$$