It can be shown (by using a variant of Formula 10.2.7 on page 665 of Stewart's text) that the area of the surface generated by rotating the polar curve $$r=f(\theta),\;\; a \le \theta \le b$$ where $f'$ is continuous and $0 \le a < b \le \pi$ about the line $\theta=\pi/2$ is $$S=\int_{a}^{\,b} {2\pi r \cos(\theta) \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}}\,d\theta$$

Use the above formula to find the surface area generated by rotating the lemniscate $r^2=\cos(2\theta)$ about the line $\theta=\pi/2.$

Surface area =