How to calculate arbitrary area on surface of sphere?

Question

I am trying to calculate the solid angle subtended by arbitrary-shaped loops on a sphere's surface.

First, I parametrize circular loops by:

$$\theta(t,k_{x0},r) = k_{x_0} + r \cos(t);$$ $$\phi(t,k_{y0},r) = k_{y_0} + r \sin(t);$$

where $0\leq t\leq2\pi$, and $k_{x_0}$, $k_{y_0}$ define the loop's center. So, we can say that this step draws out a circular loop on the $\theta/\phi$ plane.

Then I project these onto the sphere's surface using spherical coordinates, as follows:

$$x(\theta,\phi)=r \cos{\theta}\sin{\phi}, y(\theta,\phi)=r \sin{\theta}\sin{\phi}, z(\theta,\phi)=r \cos{\phi}$$

How do I go about calculating the surface area within these $(x,y,z)$ loops on the surface? This will allow me to calculate the solid angle I need.

The solid angle is given by: $$\Omega= \iint_S \frac{\hat{r}\cdot\hat{n}}{r^2} \, \mathrm{d}\Sigma = \iint_{\mathcal{R}}\sin \theta \, \mathrm{d} \theta \, \mathrm{d} \phi=\frac{\textrm{spherical surface area}}{r^2}$$

I tried using various types of RegionMeasures to calculate this area (such as defining the area within the loop on the sphere as a Region, and by varying the radius from 0 to r, calculating the length of each loop in between and summing it all up), but I feel like I am missing a simple answer to my problem. Maybe what I am missing is a way to somehow map my arbitrary loops into a appropriate integration bounds for $\Omega$, but I tried to avoid this by resorting to Mathematica.

So far, I found the following post most useful: Integrate to calculate enclosed area

Thanks in advance for your time!

Note: I am parametrizing these loops in a peculiar way because I am trying to investigate a physics problem where the functions $x(\theta,\phi),y(\theta,\phi),z(\theta,\phi)$ will be different, and make my loops twist and turn. The ultimate goal is to find the solid angle in these cases, but I wanted to start with the sphere.

Answer 1

A simple solution here is be to use the Geo functionality, in particular GeoArea, which can compute the area of any polygon on the surface of the sphere (or of an ellipsoid of revolution). There is also the primitive GeoCircle, consisting of the points on the spheroid at a given distance from a given center.

Note that in the Geo functionality we use degrees, not radians.

For example suppose your disk has a center with the following latitude and longitude, radius and is on a sphere of this radius:

centerLat = Quantity[30, "AngularDegrees"];
centerLon = Quantity[-40, "AngularDegrees"];
geodiskRadius = Quantity[1, "Meters"];
sphereRadius = Quantity[1, "Meters"];

Then we can compute the area in stereoradians as

QuantityMagnitude[GeoArea[GeoDisk[{centerlat, centerlon}, geodiskRadius], GeoModel -> sphereRadius], "Meters"^2]
(* 8.89791 *)

You can also compute the area of any spherical polygon (with edges being great circles). For example this spherical triangle covers an octant of the full surface:

geoTriangle = Polygon[GeoPosition[{{0, 0}, {90, 0}, {0, 90}}]];

Therefore its area is 4Pi / 8:

QuantityMagnitude[GeoArea[geoTriangle, GeoModel -> sphereRadius], "Meters"^2]
(* 1.5708 *)

As I said, these functions can work on ellipsoids of revolution. Use GeoModel -> {a, b} where a and b are the equatorial and polar radiuses respectively.

Answer 2

v = {0, 0, 1};
r = Sin[Pi/4];
R = ImplicitRegion[
   {x^2 + y^2 + z^2 == 1, v.{x, y, z} >= Sqrt[1 - r^2]},
   {{x, -1.1, 1.1}, {y, -1.1, 1.1}, {z, -1.1, 1.1}}
   ];
Show[
 DiscretizeRegion[R],
 Graphics3D[Sphere[]]
 ]
Area[R]

1.8403

Answer 3

Another approach, while slow, is to use RegionIntersection with geometry primitives. For example:

Area @ RegionIntersection[
    Cylinder[{{0, 0, 0}, {0, 0, 1}}, 1/Sqrt[2]],
    Sphere[{0,0,0}, 1]
]
% //N

-(-2 + Sqrt[2]) 𝜋

1.8403

in agreement with @Henrik's answer.