Finding area of a part of a circle I have the values of $L$, $R$ and $W$ in the picture below. 
The circle is drawn though the center of the rectangle. And the circle will always intersect the rectangle.
How can I find the area of the part in grey for any such rectangle and circle of this type?

 A: Here are some hints.
The area of a sector (a piece of the pie) with central angle $\theta$ is $\pi R^2 \theta/(2\pi)$.
From your diagram, let's say that $\theta$ is the angle between $B$ and $C$.  Call the center of the circle $O$.  Then the area of triangle $BOC$ is
$$A(BOC) = R\sin (\theta/2) \cdot R \cos (\theta/2).$$
Taking the difference of these two will get you the area of the white cap.
To calculate the center angle, drop a perpendicular from the center of the circle to the right side of the rectangle.  Call the point of intersection $E$.  Then triangle $EOA$ is a right triangle with one of the angles $\theta/2$.  The adjacent side has length $w/2$, and the hypotenuse is $R$.  This makes
$$\sin (\theta/2) = \frac{w/2}{R}.$$
You can use the Pythagorean theorem to find the other side, and then find $\cos(\theta/2)$.
To find $\theta$ directly, then, just use an inverse trig function:
$$\sin^{-1} (\sin (\theta/2)) = ...$$
Can you take it from here?
A: Since $ A,B,C,D $ are sketched as distinct points,evaluate
$$ 4 \int_0^{ w/2} \sqrt{ R^2- x^2} dx $$
as a  standard integral.
