Electrostatic force on a test charge by uniformly charged annular disc

[YK0001]

Homework Statement

A uniformly charged disk with the total charge $Q$, has an exterior radius of $b_{1}$ and the disc is an annular disc with the interior radius of $b_{0}$.
A test charge $q_{t}$ has been placed $\overrightarrow{A}\text{meters}$ away from the center of the disk $\overrightarrow{O}$.
$$\overrightarrow{O} =\begin{bmatrix}0\\0\end{bmatrix}\text{ and }\overrightarrow{A} =\begin{bmatrix}x\\y\end{bmatrix}$$
Calculate the electrostatic force (scalar magnitude with direction or a vector quantity) on $q_{t}$ caused by the disk.

Relevant Equations

Coulomb's law: $\overrightarrow{F}_{12} = \frac{kq_{1}q_{2}}{\overrightarrow{r_{12}}^{2}}$
Volume by revolving area bound by a function: $A(x) = \pi(F(x)^2)\text{ and }V = \int_{a}^{b}A(x)dx\text{ or } Y = \int_{c}^{d}A(y)dy$

Since we know that $Q = \text{Total charge uniformly distributed in the disc}$ and the area of the disc is just $\pi((b_1-b_0)^2)$.
We can define $dQ_{\tiny{|}}$ as the charge of an infinitesimally small point on the disk, and $Q_{\tiny{|}}$ as the chage of an infinitesimally small cross sectional line from the disc, such that $Q = \pi(\int dQ_{\tiny{|}})^2$. We can then calculate the electrostatic force on $q_t$ from $dQ_{\tiny{|}}$ to be $dF_{\tiny{|}} = \frac{kQ_{\tiny{|}}q_t}{(r)^2} = \frac{kQ_{\tiny{|}}q_t}{x^2+b^2-2xbcos(\theta)}$, then we can conclude that: $$F_{\tiny{|}} = \int\frac{kdQ_{\tiny{|}}q_{t}}{x^2+b^2-2xbcos(\theta)} = kq_{t}\int\frac{dQ_{\tiny{|}}}{x^2+b^2-2xbcos(\theta)}$$.
This is where the problem arises, I don't know what to integrate the right side to respect with, I've hypothesized that it's with respect to $Q_{\tiny{|}}$ so it looks like $\int [...] dQ_{\tiny{|}}$ because that is the infinitesimally small term already present on that side of the equation, but I am unsure if this is the correct assumption to make, I'm also thinking that there are 3 changing variables in that integral (x, b, theta (theta not really, since we can get that in terms of x and b), but I'm not sure what that integral needs to be).

I have drawn a diagram showing the disc and have attached that as well, the blue shaded areas are charged.

 

[kuruman]

Since we know that $Q = \text{Total charge uniformly distributed in the disc}$ and the area of the disc is just $\pi((b_1-b_0)^2)$ . . .

. . . we can define a surface charge $~\sigma =\dfrac{Q}{\pi((b_1-b_0)^2)}.$ Then a small area element $dA$ has charge $dq=\sigma ~dA=\dfrac{Q}{\pi((b_1-b_0)^2)}dA.$ Can you express area element $dA$ in terms of spatial coordinates over which you can integrate?

As a matter of clarity, here $Q$ is the given total charge on the disk and a has a constant value. If you wish to subdivide it into small elements, you should given then a different symbol as I have done. Note that $Q=\int_0^Qdq.$ Your expression $Q = \pi(\int dQ_{\tiny{|}})^2$ makes no sense at all.