<s>[INST] Mass of a Lamina of region $R=\{(x,y):(x-2)^2 +y^2 \le4 \: \land \: (x-1)^2 + y^2 \ge 1 \: \land \: y \ge-x\}$ The region $R$ occupied by the lamina is given by the following:
$$
R=\{(x,y):(x-2)^2 +y^2 \le4 \: \land \: (x-1)^2 + y^2 \ge 1 \: \land \: y \ge-x\}
$$
Assuming the density function $\rho(x,y)$ is continuous on $R$, I setup an iterated double integral that gives the mass of the lamina $M$ as follows:
$$
M= \int_{0}^{2} \int_{\sqrt{1-(x-1)^{2}}}^{\sqrt{4-(x-2)^{2}}} \rho(x,y) \, dy \,dx + \\ \int_{1}^{2} \int_{-x}^{\sqrt{1-(x-1)^{2}}} {\rho(x,y) \, dy \,dx +
\int_{2}^{4} \int_{-\sqrt{4-(x-2)^{2}}}^{\sqrt{4-(x-2)^{2}}} \rho(x,y) \, dy \,dx
}$$
Is this correct? Is there a shorter way to compute for the mass of the lamina?
 [/INST] As mentioned in comments, your work is correct except an incorrect sign in the upper bound of the second integral. But it is much simpler in polar coordinates.
$(x-1)^2 + y^2 = 1 \implies x^2 + y^2 = 2x$
$(x-2)^2 + y^2 = 4 \implies x^2 + y^2 = 4x$
Both circles form in two quadrants to the right of y-axis. So in polar coordinates, $x = r \cos\theta, y = r\sin\theta, - \pi/2 \le \theta \le \pi/2$
As the lamina is bound between two circles, bounds of $r$ is $2 \cos\theta \le r \le 4 \cos\theta$
As $~y \ge - x, ~\tan\theta \ge - 1$ and the bounds of $\theta$ is $ - \pi/4 \le \theta \le \pi/2$
So, the below integral gives the mass of the lamina -
$ \displaystyle \int_{-\pi/4}^{\pi/2} \int_{2\cos\theta}^{4\cos\theta} \rho(r \cos\theta, r\sin\theta) \cdot r ~ dr ~ d\theta$
</s>