# Area of a circular segment.

See the picture below:

How can I calculate the area in black, using no handy formulas which will give me the answer if I plug in the right values? I had the idea to take $\displaystyle \int_{0.5r}^{r}$, but the problem is I don't know which function to take the integral of.

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That function would be $l(x)=2\sqrt{r^2-x^2}$, but a geometric approach should be quicker. – Carlos Eugenio Thompson Pinzón Oct 16 '13 at 19:14
@CarlosEugenioThompsonPinzón What would the geometric approach be? – Phaptitude Oct 16 '13 at 19:20

If $\theta$ is the angle of the arc (in radians), the area of the sector is $\frac12\theta\cdot R^2$, and the area of the triangle is $2\frac12R\sin\frac\theta2R\cos\frac\theta2=\frac12R^2\sin\theta$.
The area of the segment is therfor: $A=\frac12R^2(\theta-\sin\theta)$. Now, we have that $\cos\frac\theta2=\frac12$ which means that $\frac\theta2=\frac13\pi$ and $\theta=\frac23\pi$. Replacing: $$A=\frac12R^2(\frac23\pi-\sqrt3)=(\frac13\pi-\frac12\sqrt3)R^2$$
Since the circle is given by $x^2+y^2=R^2$, the function to integrate is $\sqrt{R^2-x^2}$. The area is twice that integral.