Area of a rectangle in a Poincare half-plane

Question

Let (-1,1), (-1,2), (1,1), and (1,2) be points in the Poincaré half-plane and consider the rectangle formed by joining the points with geodesics. I want to find the area of this rectangle using calculus instead of using angle excess, i.e. like one would use integration to find the area of a certain domain in $\mathbb{R}^2$. I think I would have to use the metric $ d\hat{s}^2 = \frac{ds^2}{y^2}$ where ${ds}^2 = {dx}^2 + {dy}^2$ but that is as far as I got. What would a solution look like?

Answer 1

I suppose we have to compute the area delimited by the segments $AB$ and $CD$, and the arcs of circles centered in origin $\overset\frown{AC}$ and $\overset\frown{BD}$ in the following picture:

Then $x$ runs in the interval $[-1,1]$, and because the two circles have equations $x^2 +y^2 =2$ and $x^2+y^2=5$, for each fixed $x$ the variable $y$ runs between $\sqrt{2-x^2}$ and $\sqrt{5-x^2}$. Then the integral to be computed is: $$ \begin{aligned} J &= \int_{-1}^1dx\int_{\sqrt{2-x^2}}^{\sqrt{5-x^2}}\frac {dy}{y^2} = \int_{-1}^1dx\left[-\frac 1y\right]_{\sqrt{2-x^2}}^{\sqrt{5-x^2}} = \int_{-1}^1 \frac {dx}{\sqrt{2-x^2}} - \int_{-1}^1 \frac {dx}{\sqrt{5-x^2}} \\ &=\left[ \arcsin\frac x{\sqrt2} - \arcsin\frac x{\sqrt5} \right]_{-1}^1 =2\arcsin\frac 1{\sqrt 2}-2\arcsin\frac1{\sqrt 5} =\frac\pi2-2\arcsin\frac1{\sqrt 5} \ . \end{aligned} $$

$\square$

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And indeed, we expect as result the "defect" value, $2\pi$ minus sum of the internal angles. The two internal angles in $A,C$ are $\pi/4$. Together they add to $\pi/2$. There are further two blue angles with value $\pi/2$ to be subtracted. And there remain the two pink angles, also to be subtracted. So the area is: $$ 2\pi - 3\cdot\frac \pi2-2\arctan\frac 12=\frac \pi2 -2\arcsin\frac 1{\sqrt 5} \ . $$