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In the diagram below (from here fig.2, page.5) the enclosed area between two circles (shaded area) has been indicated $a_{t+\delta_{t}}$. Can anyone help me how can I compute this? is it true? $(0<d<2r)$.


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I've inserted the diagram (I think I got the right one), but it's not immediately clear to me from that diagram what the radii of the circles are. Is $d$ the distance between their centers? – Isaac Feb 5 '12 at 6:40
Good Work, @Isaac – user21436 Feb 5 '12 at 6:44
up vote 3 down vote accepted

edit You gave the expression $$\pi r^{2} - \int_{-\frac{d}{2}}^{\frac{d}{2}}\sqrt{{r}^{2}-x^{2}}dx ;(0\leq d \leq 2r).$$ Near as I can tell, the integral corresponds to the shaded region below, with the left and right edges set to match the pieces of the circle needed for the area you're trying to find.

integral area region

I suspect that there should be a factor of 2 in front of the integral, so that it gets the area above and below the axis:

double the integral

Then, when we take $$\pi r^{2} - 2\int_{-\frac{d}{2}}^{\frac{d}{2}}\sqrt{{r}^{2}-x^{2}}dx,$$ we'll be subtracting the area of that region from the area of the whole circle, leaving the shaded area shown below, which is equivalent to the original area you wanted.

circle minus the integral

To compute $\int\sqrt{r^2-x^2}dx$, make the substitution $x=r\sin\theta$ ($dx=r\cos\theta d\theta$), to get $$\begin{align} \int\sqrt{r^2-x^2}dx&=\int\sqrt{r^2-r^2\sin^2\theta}\;r\cos\theta d\theta \\ &=r^2\int\cos^2\theta d\theta \\ &=\frac{1}{2}r^2\left(\theta+\frac{1}{2}\sin2\theta\right) \\ &=\frac{1}{2}r^2\left(\arcsin\frac{x}{r}+\frac{x}{r}\frac{\sqrt{r^2-x^2}}{r}\right) \\ &=\frac{1}{2}\left(r^2\arcsin\frac{x}{r}+x\sqrt{r^2-x^2}\right). \end{align}$$

my original answer:

basic diagram

In general the technique that I'd use to find the area of such an overlap is to split it into two regions, divided by the chord common to the two circles.

Each piece is called a circular segment. The way I'd find the area of a circular segment is to find the area of the corresponding circular sector and subtract off the triangle.

specific diagram focusing on one circle

Here, I'm focusing on the purple circular segment. Together, the circular segment and the lighter purple triangle form a sector of the circle. The area of a sector of a circle is a fraction of the circle determined by the central angle. The area of the triangle we need to subtract off can be found using the formula $\frac{1}{2}r^2\sin\theta$, where $\theta$ is the central angle in the circle. Subtracting these gives the area of the circular segment.

Adding the two circular segment areas will give the area of the overlap of the two circles.

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@Issac: Thank you for your explanation. However, the overlap area of two circles (shaded area) has been given as: $$\pi r^{2} - \int_{-\frac{d}{2}}^{\frac{d}{2}}\sqrt{{r}^{2}-x^{2}}dx ;(0\leq d \leq 2r).$$ Is it true? If it is true, how can be computed in cartezian coordinate? (The radius of the circles are equal$(r_{1}=r_{2}=r)$ and d is the distance between centers.) – sorena Feb 5 '12 at 7:29
@sorena: I've edited my answer to hopefully better address your question. I think there should be a factor of 2 in front of the integral, as explained in my edit. I also described how to compute the integral. – Isaac Feb 5 '12 at 7:51
@Issac: Thank you very much for your helpful information and useful answer. – sorena Feb 5 '12 at 8:48
@Issac:Excusme me, for the final question, if I have $\theta \in \{0,\pi\}$ how can be the interval of integral be as $\int_{-\infty}^{\infty}f(\theta)d\theta$. Is it true? – sorena Feb 5 '12 at 8:50
@sorena: If you're talking about $\int\cos^2\theta d\theta$ resulting from the substitution $x=r\sin\theta$, the bounds of the original integral were $x=-\frac{d}{2}$ and $x=\frac{d}{2}$ where $0\le\frac{d}{2}\le r$, so bounds correspond to $\theta=-\arcsin\frac{d}{2r}$ and $\theta=\arcsin\frac{d}{2r}$, both of which are in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. – Isaac Feb 5 '12 at 8:56

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