Finding area of a part of a circle

Question

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?

Answer 1

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?

Answer 2

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.