# Formula for area of a circular segment

Consider the circle through 3 points $$(0, 0)$$, $$(d, 0)$$, $$(d/2,d/2\cdot p)$$ where $$d>0$$ and $$p>0$$.

I need a formula for the area of the region in the upper half-space of $$\mathbb{R}^2$$ enclosed by the x-axis and the circle. I tried to write down a parametrization of the region over the unit square in such a way that one parameter parametrizes the line between $$(0, 0)$$ and $$(d, 0)$$ as well as the circular arc boundary to the region while the other parameter draws lines between these.

When integrating the determinant of the jacobian I ended up with $$\frac{d^2\cdot\left(p^3-p+(p^2+1)^2 \arctan(p)\right)}{8\cdot p^2}$$

but I'm not sure it is correct, because the determinant was a huge expression and I couldn't tell how to argue that it would not change sign over the unit square. I know the form correctly is $$d^2f(p)$$, but is $$f(p)$$ correct?

• Could you edit in the calculation that obtained your result? If you did handle it incorrectly, we'll likely spot a mistake. – J.G. Oct 21 '19 at 13:36
• Did you check your result with the formula given at en.wikipedia.org/wiki/Circular_segment#Area ? – Aretino Oct 21 '19 at 17:26

The area above the $$x$$-axis and under the circle is just the area difference between the circle sector of angle $$2\theta$$ and an overlapping isosceles triangle of base length $$d$$. The radius $$r$$ of the circle can be obtained from,

$$r^2 = \left( \frac d2 \right)^2+\left(r-\frac{pd}{2}\right)^2$$

which gives

$$r= \frac{1+p^2}{4p}d\tag{1}$$

The angle $$\theta$$ satisfies,

$$\tan \frac{\theta}{2} = \frac{\frac d2}{2r-\frac{pd}{2}} = p\tag{2}$$

The area difference described above is,

$$A = \frac12(2\theta)r^2 - \frac12 d\left(r-\frac{pd}{2}\right)$$

Substitute (1) and (2) into above expression,

$$A=\frac{d^2}{8p^2} \left[(p^2-1)p+(p^2+1)^2 \tan^{-1}(p)\right]$$

which confirms that your result is correct. Note that the result is only valid for $$0.

For $$p>1$$ the formula holds unchanged: