how can I obtain enclosed area between two circles in cartesian coordinates? 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? $(0<d<2r)$.

 A: 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.

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:

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.

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:

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.

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.
