What is the mathematical rationale for these two lines in this question?
$$f_{r,\theta}(r,\theta)=\dfrac2\pi f_r(r),\quad r\in(0,R],\quad \theta\in\left(0,\dfrac\pi2\right),$$
then $$f_z(z)=\dfrac{\,\mathrm dF_z(z)}{\,\mathrm dz} = \dfrac2\pi\int\limits_z^R f_r(r)\dfrac{\mathrm dr}{\sqrt{r^2-z^2}}.$$
For the second line (above), specifically, how was $z$ made the lower limit i.e the transformation to $z$ from $\arccos\dfrac zr$ to derive $\dfrac{\mathrm dr}{\sqrt{r^2-z^2}}$?
Edit: The original transformed function is $$f_r(r)= \frac{2r}{R^{2}}$$
Also $z = r \cos \theta$
I have repeated the diagram here for ease of navigation.
I see question a and question b are a bit close, but I haven't been able to map them to this case yet.
An alternative approach. The problem as stated is a disguise for defining $F(z)$ the fraction of the quadrant between $0$ and $z$ and its derivative $f(z)$. That section of the quadrant is made of two pieces, a triangle with base $z$ and hypotenuse $R$ (area $=T$), and a sector of the circle between that hypotenuse and a vertical radius (area $=S)$. The total area of the quadrant is $A=\frac{\pi R^2}{4}$. The other areas are $T=\frac{z\sqrt{R^2-z^2}}{2}$ and $S=\frac{R^2}{2}(\frac{\pi}{2}-arcos{\frac{z}{R}})$. Thus $F(z)=\frac{S+T}{A}$. To get $f(z)$, we need $\frac{dS}{dz}=\frac{R^2}{2\sqrt{R^2-z^2}}$ and $\frac{dT}{dz}=\frac{R^2-2z^2}{2\sqrt{R^2-z^2}}$ Combining all this to get $f(z)=\frac{4\sqrt{R^2-z^2}}{\pi R^2}$.
The problem as stated sets up the system in polar coordinates and then asks to get $f(z)$ by integrating over $r, \theta$ to get the area fraction $F(z)$.