Surface area of spherical section delineated by 2 perpendicular circular planes/central angles The problem concerns visible area based on a field of view from the center of a sphere.
I was never taught spherical trigonometry so even basic terminology is hard. After trying to figure out the problem for a few hours by myself I know some terms but don't fully understand them.
From the center of a sphere, based on two central angles (in degrees) that fall along two circular, perpendicular planes (x and y in what would be spherical coordinates I think).
How large is the visible surface area based on one horizontal and one vertical angle? As in, for a person who can turn left-right/up-down 360°?
This interactive Wolfram Demonstration has solutions for spherical triangles but those need 3 input parameters and I don't know enough about them to tell if they're even the right shape to cover all cases. It glitches on some inputs, too.
Basic examples:
- hemisphere: would result from a full 360° horizontal/left-right angle and a 90° vertical/up-down angle; this could be represented by a "spherical triangle" which itself has all 3 angles at 180°.
- semi-hemisphere: 180° horizontal, 90° vertical or a spherical triangle that has 90°, 90° and 180° angles.
Those are the simplest subsections of a sphere where the area are whole multiples of π*r^2.  
What's the formula that would allow the calculation of this surface area from these two central angles, for cases like 45°/270°, 180°/10°, any other number?
 A: I think the "perpendicular circular plans/central angles" means the longitude and lantitude angles when you treat the sphere as our earth.  The question is asking the surface area when 2 sets of longitude and latitude angles are given.  In order words we are going to find the area on earth given the sets of angles.
For surface area we have a formula below using spherical coordinates :
$$x = rcos\theta cos\phi$$
$$y = rsin\theta cos\phi$$
$$z = rsin\phi$$
Where $\theta$ is the longitude and $\phi$ is the latitude.
$$S = \iint\sqrt{EG - F^2} d\theta d\phi$$
Where
$$E = (\frac{\partial x}{\partial\theta})^2 + (\frac{\partial y}{\partial\theta})^2 + (\frac{\partial z}{\partial\theta})^2$$
$$G = (\frac{\partial x}{\partial\phi})^2 + (\frac{\partial y}{\partial\phi})^2 + (\frac{\partial z}{\partial\phi})^2$$
$$F = \frac{\partial x}{\partial\theta}\frac{\partial x}{\partial\phi} + \frac{\partial y}{\partial\theta}\frac{\partial y}{\partial\phi} + \frac{\partial z}{\partial\theta}\frac{\partial z}{\partial\phi}$$
Hence the surface area S is
$$S = r^2\int_{\theta_1}^{\theta_2}\int_{\phi_1}^{\phi_2}cos{\phi}d{\phi}d{\theta} = r^2(\theta_2 - \theta_1)(sin{\phi_2} - sin{\phi_1})$$
Take $\theta $ from 0 to $\frac{\pi}{2}$ and $\phi$ from 0 to $\frac{\pi}{2}$ we have the area of $\frac{\pi r^2}{2}$ which is one eighth is a sphere of $4\pi r^2$
