Computing area of equilateral triangle on a sphere

tainted
Messages
28
Reaction score
0

Homework Statement



Suppose T is an equilateral triangle on the sphere of radius R = 1. Let \alpha denote the angle at any of the three vertices’s of the triangle. (Recall that 3\alpha > n.) Use the result of the last problem on the previous homework and the inclusion - exclusion principle (together with an orange and a knife) to compute the area of T .

Homework Equations



The result to the last problem on the previous homework is A = \alpha2R^2

The Attempt at a Solution


I assumed that all angle on the equilateral triangle where 90 degrees or \frac{\pi}{2}; therefore making the volume equal to 1/8 that of the whole sphere

So I did
A = \alpha2R^2 where A is the area of T
A = \frac{\pi}{2}2R^2
A = \pi*R^2 That would be the area of 1/4 of the sphere overall, but because I am taking the area of an equilateral triangle, I took half of that to get
A = \frac{\pi}{2}R^2
A = \frac{\pi}{2}*1
A = \frac{\pi}{2}

Would that be correct? I just kind of picked 90 degrees or \frac{\pi}{2} for \alpha, but I assume it could be anything between 60 and up to 90 degrees which would change my answer. How do I know which angle to pick?
 
Last edited:
Physics news on Phys.org
I am concerned that my answer will be completely different if I used an angle other than 90 degrees. Also I do not know what an orange and knife have to do with this problem.
 
You are supposed to do it for ANY angle using inclusion/exclusion. Look at the pictures here: http://planetmath.org/AreaOfASphericalTriangle.html Just look at the pictures, ignore the solution and try to work it out for yourself.
 
Last edited by a moderator:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top