Solution:

Given curve: $r^{2}=a^{2} \cos 2 \theta \quad$ is lemniscate.

The density at any point is proportional to the square of dist. From the pole.

Distance from the pole =r

$\quad \therefore$ Density $\propto r^{2}$

$\therefore$ Density $=k . r^{2}$

The mass of the lemniscate is given by,

$\mathbf{M}=\int_{\theta_{1}}^{\theta_{21}} \int_{r_{1}}^{r_{2}}$ density $r \ d r \ d \theta$

$\therefore \quad \mathbf{M}=\mathbf{4} \times \int_{0}^{\frac{\pi}{4}} \int_{0}^{a \sqrt{\cos 2 \theta}} k . r^{2} . r \ d r \ d\theta$

$=4 k \times \int_{0}^{\frac{\pi}{4}}\left[\frac{r^{4}}{4}\right]_{0}^{a \sqrt{\cos 2 \theta}} \mathrm{d} \theta$

$=k \times \int_{0}^{\frac{\pi}{4}} a^{4} \cdot \cos ^{2} 2 \theta \cdot d \theta$

We can solve this definite integral by beta function.

$\quad$ Put $2 \theta=t \Rightarrow 2 d \theta=d t$

Limits $\left[0, \frac{\pi}{2}\right]$

$\therefore \mathrm{M}=k a^{4} \int_{0}^{\frac{\pi}{2}} \cos ^{2} t \cdot \frac{d t}{2}$

$\begin{aligned}=& \frac{k a^{4}}{2} \times \frac{1}{2} \beta\left(\frac{1}{2}, \frac{3}{2}\right) \\ \therefore \mathbf{M}=\frac{k a^{4} \pi}{8} \end{aligned}$