Volume of Solid Generated by Revolving Lemniscates
To find the volume of the solid generated by revolving the lemniscates $r^2=a^2\cos{2\theta}$ about the initial line, we can use the method of cylindrical shells.

Step 1: Setting Up the Integral
1. The volume element of a cylindrical shell is given by $dV=2\pi rh \,dr$, where $r$ is the radius, $h$ is the height, and $dr$ is the thickness of the shell.
2. The radius of the shell is $r=a\sqrt{\cos{2\theta}}$.
3. The height of the shell is $h=r \,d\theta$.
4. The limits of integration for $r$ are from $0$ to $a\sqrt{\cos{2\theta}}$.
5. The limits of integration for $\theta$ are from $0$ to $\pi/4$.

Step 2: Calculating the Volume
1. The volume of the solid is given by $V=\int_{0}^{\pi/4} \int_{0}^{a\sqrt{\cos{2\theta}}} 2\pi a\sqrt{\cos{2\theta}} \cdot a\sqrt{\cos{2\theta}} \,dr \,d\theta$.
2. Simplify the integral to get $V=2\pi a^2 \int_{0}^{\pi/4} \cos{2\theta} \,d\theta$.
3. Integrate with respect to $\theta$ to get $V=2\pi a^2 [\frac{1}{2}\sin{2\theta}]_{0}^{\pi/4}$.
4. Evaluate the integral to get $V=\pi a^2$.
Therefore, the volume of the solid generated by revolving the lemniscates $r^2=a^2\cos{2\theta}$ about the initial line is $\pi a^2$.