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?

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$
