Interior Area of Circle of Circles 
I am looking for the area of the white region interior to a set of circles with radius A, oriented on the edge of a larger circle with radius B, spaced apart from each other with distance C. You can assume that C is less than 2 times A, so that each smaller circle overlaps with its neighbors. To clarify, I am looking for the area of the circle with radius b, not covered by small circles.
 A: 
It's more convenient 
to use the number of circles $n$
instead of the distance between their centers.
Then
\begin{align}
\phi&=\tfrac\pi{n}
,
\end{align}
the distance between the centers $|C_iC_{i+1}|=2\,R\sin\tfrac\phi2$
and
the total area $S$ of the interior region 
consists of $n$ areas $S_p$ of petals $P_1OP_2$.
\begin{align}
S&=n\,S_p
,\\
S_p
&=2\,S_{\triangle OC_0P_1}-S_{\mathrm{seg}\,C_0P_1P_2}
\\
&=|OC_0|\cdot|P_1F|-\tfrac\theta2\,|C_0P_1|^2
\\
&=R\cdot r_a\sin\tfrac\phi2-\tfrac\theta2\,r^2
\tag{1}\label{1}
.
\end{align}
In $\triangle P_1OF$, 
$\triangle P_1FC_0$,
$\triangle P_1C_0G$
\begin{align}
\angle C_0GO&=\tfrac\pi2
,\\
\angle GOC_0&=\tfrac\phi2=\tfrac\pi{n}
,\\
\angle FP_1O&=\tfrac\pi2-\tfrac\phi2
,\\
\angle FC_0P_1&=\tfrac\theta2
,\\
\angle C_0P_1F&=\tfrac\pi2-\tfrac\theta2
,\\
\angle GP_1C_0&
=\tfrac\phi2+\tfrac\theta2
=\tfrac\pi{n}+\tfrac\theta2
,\\
|C_0G|&=R\sin\tfrac\phi2=R\sin\tfrac\pi{n}
,
\end{align}
\begin{align}
\sin\angle GP_1C_0&=
\sin(\tfrac\pi{n}+\tfrac\theta2)
=\tfrac{|C_0G|}{r}
=\tfrac{R}r\,\sin\tfrac\pi{n}
,\\
\tfrac\theta2&=
\arcsin\left(\tfrac{R}r\,\sin\tfrac\pi{n}\right)-\tfrac\pi{n}
,\\
\sin\tfrac\theta2&=
\tfrac{R}{r}\,\sin\tfrac\pi{n}\,
\left(
\cos\tfrac\pi{n}-\sqrt{\tfrac{r^2}{R^2}-\sin^2\tfrac\pi{n}}
\right)
,\\
r_a&=\frac{r}{\sin\tfrac\pi{n}}\,\sin\tfrac\theta2
\\
&=
R\,
\left(
\cos\tfrac\pi{n}-\sqrt{\tfrac{r^2}{R^2}-\sin^2\tfrac\pi{n}}
\right)
.
\end{align}
Finally,
\begin{align}
S(n,R,r)&=
\pi\,r^2+
n\cdot \left(
R^2\sin(\tfrac\pi{n})\,
\left(
\cos(\tfrac\pi{n})-\sqrt{\tfrac{r^2}{R^2}-\sin^2(\tfrac\pi{n})}
\right)
-r^2\,
\arcsin\left(\tfrac{R}r\,\sin(\tfrac\pi{n})\right)
\right)
.
\end{align}
Edit
An example with (more-or-less) nice expression for the area. 
Let $n=6$, $R=\sqrt3$, $r=1$.
Then 
\begin{align}
r_a&=\sqrt3\left(\tfrac{\sqrt3}2-\sqrt{\tfrac13-\tfrac14} \right)
=\tfrac32-\sqrt{1-\tfrac34}=1
,\\
\tfrac\theta2&=\arcsin(\tfrac{\sqrt3}2)-\tfrac\pi6=\tfrac\pi6
,\\
S(6,\sqrt3,1)
&=
\pi+6\cdot
\left(
\tfrac32\,
\left(
\tfrac{\sqrt3}2
-\sqrt{\tfrac{1}{3}-\tfrac{1}{4}}
\right)
-1\cdot
\arcsin\tfrac{\sqrt3}2
\right)
\\
&=
\pi+
9\,
\left(
\tfrac{\sqrt3}2
-\tfrac{\sqrt3}6
\right)
-6\cdot
\tfrac\pi3
\\
&=
3\sqrt3-\pi\approx 2.05
.
\end{align}
In the image below four grid cells represent one square unit:

A: The area within the intersection of two circles is called a lens; when the circles each have radius $a$ and their centres are distance $d$ apart, the area of the lens is
$$L = a^2\pi-2a^2\arctan\left(\frac{d}{\sqrt{4a^2-d^2}}\right)-\frac{d}{2}\sqrt{4a^2-d^2}$$
Suppose there are $n$ small circles. Then the total area is the area of the circles minus the area of the lenses, i.e.:
$$n(\pi a^2 - L)$$
A: Below is an example figure with $n=8$ small circles

Given $n$ small circles, the white area can be determined as the area of the large circle with radius $B$ minus the areas of the $n$ green sectors with angle $\alpha_1$ and the $n$ blue sectors with angle $\alpha_2$, i.e.: $$\text {Area} = \pi B^2- nA^2\frac{(\alpha_1 + \alpha_2)}{2}$$
The problem now is to find $\alpha_1$ and $\alpha_2$. Finding $\alpha_1$ is easy using the Cosine Rule, as we have an isosceles triangle with sides $A$ and base $C$, which gives $$\cos \alpha_1 = 1- \frac{C^2}{2A^2}$$
Finding $\alpha_2$ is harder. But we can use the triangle with the angle $\alpha_3$ shown above. We know that $2\alpha_3 = \frac{2\pi}{n}$ and hence $\alpha_3 = \frac{\pi}{n}$. We also know two sides of the triangle and can therefore use the Sine Rule to find $\alpha_4$: $$\frac{\sin \alpha_3}{A}=\frac{\sin \alpha_4}{B}$$
which gives $$\sin \alpha_4= \frac{B}{A} \sin \alpha_3$$
Hence $$\alpha_4 =\pi-\text {arcsin} (\frac{B}{A} \sin \alpha_3)$$ and therefore $$\alpha_2 = 2(\pi-\alpha_3-\alpha_4)$$
I suspect there is a simpler answer to this question, but I couldn't immediately find it.
