# Finding the area of a portion of circle

How to find the area of green region? Angle between centre $A$ of first circle and the radius is given as alpha. Distance between $A$ and $B$ is $d$. Radius of two circles is $R$.

Can the area be found as area of two circles-area of first circle? Is there any simple way to find other than using integrals?

It is possible to find the "middle" area, and therefore the shaded area, without calculus. Let the top point of the top triangle be $P$, and the bottom point of the bottom triangle be $Q$.

Consider the sector that goes $PAQ$, and then around the circle back to $P$.

We know the angle of that sector (it is twice your given angle), and we know the radius, so we know the area.

Double that. This is the area of our "middle" region, except that the areas of the two triangles have been counted twice.

So calculate twice the area of the sector described earlier, and subtract the sum of the areas of the triangles. These areas are not hard to find.

There is a lot of calculations, so here is the outline.

Divide the intersection into two halves by drawing a vertical line. Each half is a sector minus a triangle. If you call $S$ the area of the sector and $\Delta$ the area of the triangle, then the intersection area is $2(S-\Delta)$. So your answer is $$\pi R^2 - 2 S + 2 \Delta$$

If $\theta$ is the angle subtended by the points of intersection at the center then $$S = \frac{\theta}{2} R^2$$ and $$\Delta = \frac{1}{2} R^2 \sin(\theta)$$

You probably know the relationship: $$S-\Delta = \frac{1}{2} R^2 (\theta - \sin(\theta))$$

I will let you find the $\theta$ from the given data.