# Hyperbolic Area

• May 4th 2008, 10:59 AM
hercules
Hyperbolic Area
Find the hyperbolic area of the euclidean rectangle with vertices (0,1) , (0,3),(5,3),(5,1).

the area of a hyperbolic triangle is equal to its Defect. Now any hyperbolic polygon can be divided into hyperbolic triangles to get the area. But with this euclidean rectangle two sides are not geodesics. Will an inversion work? Is there another method to find the Hyperbolic area of a euclidean figure?

Thank You.
• May 4th 2008, 01:05 PM
ThePerfectHacker
Quote:

Originally Posted by hercules
Find the hyperbolic area of the euclidean rectangle with vertices (0,1) , (0,3),(5,3),(5,1).

the area of a hyperbolic triangle is equal to its Defect. Now any hyperbolic polygon can be divided into hyperbolic triangles to get the area. But with this euclidean rectangle two sides are not geodesics. Will an inversion work? Is there another method to find the Hyperbolic area of a euclidean figure?

Thank You.

Draw a diagnol, which in this case is a curved geodesic. You end up with 2 hyperbolic triangles. Add up the areas.
• May 4th 2008, 01:27 PM
hercules
Quote:

Originally Posted by ThePerfectHacker
Draw a diagnol, which in this case is a curved geodesic. You end up with 2 hyperbolic triangles. Add up the areas.

That was my original thought but the bottom and top sides of the rectangle are not geodesics. Is that still considered a hyperbolic triangle?
• May 5th 2008, 07:35 PM
hercules
Quote:

Originally Posted by ThePerfectHacker
Draw a diagnol, which in this case is a curved geodesic. You end up with 2 hyperbolic triangles. Add up the areas.

I'm still trying to figure this out. Drawing the diagonal doesn't work because of the reason i mentioned above. All lines of the hyperbolic triangle must be geodesics. A possible route to the solution is using integrals (double at that). I'm horrible with those. Going to review all of calculus this summer then definitely i will be a great helper on this forum......But i need help now .
• May 5th 2008, 07:41 PM
ThePerfectHacker
Quote:

Originally Posted by hercules
I'm still trying to figure this out. Drawing the diagonal doesn't work because of the reason i mentioned above. All lines of the hyperbolic triangle must be geodesics. A possible route to the solution is using integrals (double at that). I'm horrible with those. Going to review all of calculus this summer then definitely i will be a great helper on this forum......But i need help now .

I misunderstood your question. I see that you have an Euclidean triangle, not a hyberbolic one.

Let $\displaystyle R = \{ (a,b)\in\mathbb{R}^2| 0\leq a\leq 5\mbox{ and }1\leq b\leq 3\}$.

Then $\displaystyle a(R) = \iint_R \frac{dA}{y^2} = \int_0^5 \int_1^3 \frac{1}{y^2} ~ dy~dx$