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?

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

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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 $0<p \le 1$.

For $p>1$ the formula holds unchanged:


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