A segment is removed from a semi circle what is the pdf of the remaining area? Question: 
Given the illustration below, the cdf of  Z is 
\begin{equation}
Pr(Z\leq z) = \frac{\text{exp}(-\lambda A_z)-\text{exp}(-N/2)}{1-\text{exp}(-N/2)}
\end{equation}
How could one Prove that
\begin{equation}
f_z(Z) = \frac{2\lambda \sqrt{R^2-z^2}}{1-\text{exp}(-N/2)} \text{exp}[-\lambda A_z]                         
\end{equation}
for $0<z\leq R$
where $A_z$ = $R^2 [\cos^{-1} (z/R)- (z/R)\sqrt{1-(z/R)^2}]$
Challenge: 
The first problem I have is understanding how $A_z$ was derived, then, how the numerator of the first equation of the pdf was gotten. The closest clue I have to $A_z$ is this question and this one, but here, the height of the segment here is unknown.
In case it might interest someone, $N = \lambda \pi R^2$ and $\lambda$ represents the density of a two dimensional Poisson Point Process

 A: I understand that you have already recognised that this problem can be solved by using the identity relating the PDF of a distribution to the CDF: $ f_Z(z) = F_Z'(z)$,
which for your particular CDF is equivalent to:
$$ f_Z(z) = \frac{-\lambda A'(z) \exp(-\lambda A(z))}{1- \exp(-N/2)}.$$
So it remains to calculate $A(z)$, and then use this to derive $A'(z)$.

Calculating A(z)
In the following I refer to my diagram (unfortunately not as well produced as your graphic, but you should get the point).
Let $\theta_z$ denote the angle made between $A \hat O B$, using the standard trigonometric identify for the cosine we have that 
$$\theta_z = \cos^{-1}(z/R).$$
Therefore the area of the circular wedge with angle $2\theta_z$ (i.e. the wedge bounded by the lines $OB$ and $OC$) is
\begin{aligned}W_z &= \pi R^2 \times \frac{2\theta_z}{2 \pi} \\
& =  R^2 \theta_z \\
& = R^2 \cos^{-1}(z/R).
\end{aligned}
We can now obtain $A(z)$ by subtracting from $W_z$ the area, $T_z$, of the triangle made out by the lines $OBCO$, which is equal to the area of the rectangle with side lengths equal to those of the line segments $OA$ and $AB$. 
Since $A$ has coordinates $(z,0)$ the line segment $OA$ has length $z$; we can calculate the coordinates  of $B$ by relying on the formula $x^2 + y^2 = R^2$, which gives coordinates $(z, \sqrt{R^2 - z^2})$, and so the length of $AB$ is $\sqrt{R^2 - z^2}$. Combining these we get
$$T_z = z \sqrt{R^2 - z^2}$$,
and further more
\begin{aligned}
A(z) &= W_z - T_z \\
& = R^2 \cos^{-1}(z/R) - z \sqrt{R^2 - z^2} \\
& = R^2 \left( \cos^{-1}(z/R) - (z/R) \sqrt{1 - (z/R)^2} \right),
\end{aligned}
which is as you gave in the question statement.

Calculating $A'(z)$
This now follows from appealing to standard identities for derivatives. If we denote 
$$g(x) = \cos^{-1}(x) - x \sqrt{1 - x^2},$$
then we have $A(z) = R^2 g(z/R)$, and so using the chain rule:
\begin{aligned}
A'(z) & = R^2 \frac{d}{dz} g(z/R)  \\
& = R^2 g'(z/R) \frac{d}{dz} \left(\frac{z}{R}\right) \\
& = R g'(z/R).
\end{aligned}
So concentrating on $g(x)$ we differentiate this using the identity for the derivative of $\cos^{-1}$, as well as the product rule to handle the second term to obtain:
$$g'(x) = - \frac{1}{\sqrt{1-x^2}} - \frac{1-2x^2}{\sqrt{1-x^2}}, $$
where the first term above is contributed from $\cos^{-1}$, and the second is from the product term. Rearranging this we have
\begin{aligned}
g'(x) &= -2\frac{1 - x^2}{\sqrt{1-x^2}} \\
& = -2 \sqrt{1 - x^2}.
\end{aligned}
And so we have
\begin{aligned}
A'(z) &= Rg'(z/R) \\
& = -2R \sqrt{1 - (z/R)^2} \\
& = -2 \sqrt{R^2 - z^2}.
\end{aligned}
Substituting this into the original formula provided for $f_Z(z)$ gives the desired answer.
