Area of the shaded region 
The numbers are to identify the circles
I've came out with this list of 4 inequalities(1 each circle), but I don't know if this is the best method to calculate it, neither how to solve it:
$(x+\frac{d}{2})^2+y^2\geq r_1^2 \\
(x-\frac{d}{2})^2+y^2\geq r_2^2 \\
(x+\frac{d}{2})^2+y^2\leq r_{1'}^2 \\
(x-\frac{d}{2})^2+y^2\leq r_{2'}^2$
The radius of the big circles can't be smaller than the small's.
The center of $1'$ is equal to the center of $1$, and the same with 2s.
 A: Disclamer: This is a brute force approach and should not be used unless all else fails. A geometric solution is probably possible.
As per the suggestion by Andrew in the comments, we can use Green's Theorem to find the area. Green's Theorem states that for any curve $C$ in the $xy$-plane which bounds region $R$ and vector field $\vec{F}=\langle M,N\rangle$ which is differentiable in $R$,
$$\oint_C \vec{F}\cdot d\vec{r} = \iint_R \left( \frac{\partial M}{\partial y}+\frac{\partial N}{\partial x} \right)dA. $$
To find the area of the region in the question, we just need a vector field where the integrand in the RHS is $1$. One such field is $\vec{F} = \langle 0,x\rangle$. We want to break $C$ (the boundary of the region whose area we are interested in) into four parts belonging to the four circles. Let $2k$ be the distance between the centers of the circles at the left and right side. The equations of these circles are
$$ (x\pm k)^2+y^2=R^2 \quad \text{and}\quad (x\pm k)^2+y^2=r^2. $$
We need to parametrise the curves, and we will do so in terms of $y$. Solving for $x$ in each curve we get
$$x=\pm k + \sqrt{R^2-y^2} \quad \text{and} \quad x=\pm k +\sqrt{r^2-y^2}$$
respectively. Let us cover $C$ counterclockwise starting from the bottom-most point, the intersection of the two big circles. Let us write $y_0$ for the $y$-coordinate of this point, $y_1$ for that of the intersection of a big and small circle, and $y_2$ for that of the intersection of the two small circles. Then the LHS can be written as:
$$\begin{split}
\oint_C Mdx + Ndy & = \oint_C xdy\\
& = \int_{y_0}^{y_1}k+\sqrt{R^2-y^2}dy + \int_{y_1}^{y_2} k+\sqrt{r^2-y^2}dy + \int_{y_2}^{y_1}-k+\sqrt{r^2-y^2}dy + \int_{y_1}^{y_0}-k+\sqrt{R^2-y^2}dy\\
& =^{\hspace{-0.3cm}*} 2k(y_2-y_0) +\int_{y_0}^{y_1}\sqrt{R^2-y^2}dy + \int_{y_1}^{y_2} \sqrt{r^2-y^2}dy - \int_{y_1}^{y_2}\sqrt{r^2-y^2}dy - \int_{y_0}^{y_1}\sqrt{R^2-y^2}dy\\
& = 2k(y_2-y_0).
\end{split}$$
(Note that the expression after $=^{\hspace{-0.3cm}*}$ is obtained by isolating the $k$ from each integral and combining them after integrating.) It remains to find $y_0$ and $y_2$, which is quite easy. The intersections of the two big and small circles respectively clearly occur when $x=0$, so
$$ y_0^2=R^2-k^2\implies y_0=-\sqrt{R^2-k^2} $$
and
$$ y_2^2=r^2-k^2\implies y_2=-\sqrt{r^2-k^2}. $$
Combining these, we get that the area of the desired region is $2k(y_2-y_0)=2k\left(\sqrt{R^2-k^2}-\sqrt{r^2-k^2}\right)$. Since $2k=d$, this is the same as
$$\mathrm{Area} = d\left( \sqrt{R^2-\frac{1}{4}d^2}+\sqrt{r^2-\frac{1}{4}d^2} \right).$$
