How to find this mass? Let $R$ be the region in the first quadrant of the plane bounded by the lemniscates of the following equations:


*

*$\rho^2=4\cos(2\theta)$,

*$\rho^2=9\cos(2\theta)$,

*$\rho^2=4\sin(2\theta)$, and

*$\rho^2=9\sin(2\theta)$.


Find the mass of this portion of plane if the density at the point $P(\rho,\theta)$ is $\delta(\rho, \theta)=16\rho$.
 A: The mass integral will involve the double integral for area in polar coordinates with an embedded density function:
$$m \ = \ \int_{\theta_1}^{\theta_2} \int_{\rho_{inner}}^{\rho_{outer}}\ \delta (\rho , \theta) \ \ \rho \ d\rho  \ \ d\theta \ . $$
Since the density function is $ \ \delta (\rho , \theta) \ = \ 16 \ \rho \ ,  $ this integral becomes
$$m \ = \ \int_{\theta_1}^{\theta_2} \int_{\rho_{inner}}^{\rho_{outer}}\ 16\rho^2 \ \ d\rho  \ \ d\theta \ . $$
Some of the hard work is going to be finding the limits of integration.  In the diagram below, the lemniscates (which, in full,  look like figure-eights or "infinity signs") intersect to form a region with four vertices.  The angles are found from solving the various combinations of $ \ a \cos 2 \theta \ = \ b \sin 2 \theta \ $ ; two vertices are on the ray $ \ \tan 2 \theta = 1 \ \Rightarrow \ \theta = \frac{\pi}{8} \ $ , one at $ \ \tan 2 \alpha = \frac{4}{9} \ , $ and the fourth at $ \ \tan 2 \beta = \frac{9}{4} \ . $  Despite appearances, the region is not symmetrical about $ \ \theta = \frac{\pi}{8} \ , $ so we need to split the integral into 
$$m \ = \ \int_{\alpha}^{\pi / 8} \int_{\rho_{inner}}^{\rho_{outer}}\ 16\rho^2 \ \ d\rho  \ \ d\theta \ \ + \ \  \int^{\beta}_{\pi / 8} \ \int_{\rho_{inner}}^{\rho_{outer}} \ 16\rho^2 \ \ d\rho  \ \ d\theta \ . $$
The red curves are the cosine-lemniscates and the blue ones, the sine-lemniscates.  For the first integral, the radial integral runs from $ \ 2 \ \sqrt{\cos 2 \theta} \  $ to $ \ 3 \ \sqrt{\sin 2 \theta} \  $ , while for the second, the radial integration goes from  $ \ 2 \ \sqrt{\sin 2 \theta} \  $ to $ \ 3 \ \sqrt{\cos 2 \theta} \ . $  So what you need to carry out is
$$m \ = \ \int_{\alpha}^{\pi / 8} \int_{2  \sqrt{\cos 2 \theta}}^{3  \sqrt{\sin 2 \theta}}\ 16\rho^2 \ \ d\rho  \ \ d\theta \ \ + \ \  \int^{\beta}_{\pi / 8} \ \int_{2  \sqrt{\sin 2 \theta}}^{3  \sqrt{\cos 2 \theta}} \ 16\rho^2 \ \ d\rho  \ \ d\theta \ . $$
This will be a bit unpleasant since the radicals won't just "square away" and none of the angular limits of integration have "nice" values...

A: Hint:
The formula for finding the mass bounded by a region in the xy-plane is given by
$$ m = \int\int_{D}\delta(x,y)dA=\int\int_{D}\delta(x,y)dydx,$$
where $\delta(x,y)$ is the density.
