I want to calculate the area of the shaded region with just the radii and without any angles, I know we can calculate the area of the quadrilateral made by connecting the circles centers and subtract the area of the circles sectors, but without any angles or diagonal I think we can't calculate the area of the quadrilateral, but is there any other way of calculating the area of the shaded region, or at least can we get an interval of the area?
Note: I've phrased my answer using a counter-clockwise arrangement of circles, so $C_1$ is opposite $C_3$ and $C_2$ opposite $C_4$. I later realized this doesn't match the indices in your figure.
You have a one-parameter family of possible solutions. Just assume the center of one circle, $C_1$ is at $(0,0)$ and the center of the opposite circle $C_3$ is at $(t,0)$. You know that you need $t\ge r_1+r_3$ since the two circles don't overlap. You can also see that if you make $t$ too large the two circles will be too far apart, causing the other two circles $C_2$ and $C_4$ either to overlap or even unable to touch both $C_1$ and $C_3$ at the same time. So start with a value of $t$ slightly larger than $r_1+r_3$ when your try to imagine or draw this.
Now let's work out coordinates for the other two centers. For $C_2$ we want to meet two conditions so that it touches $C_1$ and $C_3$:
\begin{align*} x_2^2+y_2^2&=\left(r_1+r_2\right)^2\\ \left(x_2-t\right)^2+y_2^2&=\left(r_2+r_3\right)^2 \end{align*}
Subtract the two equations from one another and you get a linear equation for $x_2$ which you can use to express $x_2$ in terms of $t$. Plug that into the first equation to determine $y_2$ and pick the negative sign for the square root. Do the same computation for $C_4$ but pick the positive sign there. That way you get the circles $C_1$ through $C_4$ arranged in a counter-clockwise order which is the mathematical convention.
Now is the point where you can work out the maximal value for $t$. That's the point where $C_2$ touches $C_4$:
$$\left(x_2-x_4\right)^2+\left(y_2-y_4\right)^2=\left(r_2+r_4\right)^2$$
Since your $y_2$ and $y_4$ contain square roots, this will be a bit messy. I'd expect that by squaring both sides of the equation twice, sorting square roots on one side and everything else on the other side before each squaring, you should be able to get a degree 4 equation in $t$ for this, or more specifically a quadratic equation in $t^2$ because negative values of $t$ represent pretty much the same situation. I haven't tried this using the square roots, but I have done some computatations on polynomials using resultants to eliminate variables and come up with this:
$$0 = (r_2 + r_4)^2 t^4 \\ -2 (r_1^2 r_2^2 - 2 r_1 r_2^2 r_3 + r_2^2 r_3^2 + 2 r_1^2 r_2 r_4 + 4 r_1 r_2^2 r_4 + 4 r_1 r_2 r_3 r_4 + 4 r_2^2 r_3 r_4 + 2 r_2 r_3^2 r_4 + r_1^2 r_4^2 + 4 r_1 r_2 r_4^2 - 2 r_1 r_3 r_4^2 + 4 r_2 r_3 r_4^2 + r_3^2 r_4^2) t^2 \\ + (r_1 - r_3)^2 (r_1^2 r_2^2 - 2 r_1 r_2^2 r_3 + r_2^2 r_3^2 + 2 r_1^2 r_2 r_4 + 8 r_1 r_2^2 r_4 + 12 r_1 r_2 r_3 r_4 + 8 r_2^2 r_3 r_4 + 2 r_2 r_3^2 r_4 + r_1^2 r_4^2 + 8 r_1 r_2 r_4^2 + 16 r_2^2 r_4^2 - 2 r_1 r_3 r_4^2 + 8 r_2 r_3 r_4^2 + r_3^2 r_4^2)$$
Solve quadratic equation to get two candidates for $t^2$, then take the positive square roots of these to get two candidates for $t$. The smallest one that is greater than $r_1+r_3$ should be the one, but to be honest I don't have a good intuition as to what the other value represents, so do some experiments to be sure.
Now that you know the range of $t$ you can look at the shaded area. You'd have formulas which give you all the center coordinates as functions of $t$. You can get the area expressed in terms of that, although the expressions would be big and ugly and transcendental. You could even compute the derivative of that area function, and find it's roots to determine where the area is extremal. Anyways keeping the permissible range for $t$ in mind, of course.
Take the area at any extremal points within the permissible range, and also the area you get at each endpoint of the permissible range. These define the interval of achievable areas.
I would do the above with heavy use of computer algebra systems. And I'd not expect any formulas simple enough that I'd paste them here.