Can you do it geometrically?

Yes, it can be solved by geometry.

Let $(0,0)$ , $(0,1)$ , $(2,0)$ , and the point of intersection of the two circles be $O$ , $A$ , $B$ , and $P$ . The quadrilateral $AOBP$ is a kite with radii of the two circles as sides. Let $\angle BAO = \alpha$ , then $\angle PAO = 2 \alpha$ and $\angle ABO = \beta$ , then $\angle PBO = 2 \beta$ . We note that $\alpha = \tan^{-1} 2$ and $\beta = \tan^{-1} \frac 12$ .

The area of intersection is given by:

$\begin{aligned} \mathfrak E & = \text{Area of radius-1 segment OP} + \text{Area of radius-2 segment OP} \\ & = \text{Area of sector } AOP - \text{Area of } \triangle AOP + \text{Area of sector } BOP - \text{Area of } \triangle BOP \\ & = 1^2 \left(\alpha - \sin \alpha \cos \alpha \right) + 2^2 \left(\beta - \sin \beta \cos \beta \right) \\ & = (1.107148718 - 0.4) + 4(0.463647609-0.4) \\ & = 0.707148718 + 0.254590436 \\ & = 0.961739154 \end{aligned}$

$\implies \lfloor 1000 \mathfrak E \rfloor = \boxed{961}$

Let us take a calculus approach to this area. The circles in question can be modeled as:

$x^2 + (y-1)^2 = 1;$

$(x-2)^2 + y^2 = 4$

which intersect in the points $(x,y) = (0,0); (\frac{4}{5},\frac{8}{5})$ . These two points lie along the line $y=2x$ , which allows us to perform the following integration:

$A = \int_{0}^{4/5} \sqrt{4-(x-2)^2} - 2x dx + \int_{0}^{8/5} \sqrt{1-(y-1)^2} - \frac{y}{2} dy$ ;

or $\frac{x-2}{2}\sqrt{4-(x-2)^2} + 2\arcsin(\frac{x-2}{2}) - x^2 |_{0}^{4/5} + \frac{y-1}{2}\sqrt{1-(y-1)^2} + \frac{1}{2}\arcsin(y-1) - \frac{y^2}{4}|_{0}^{8/5} \approx 0.9617$

Thus, $\lfloor 1000A \rfloor = \boxed{961}.$