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For example, assuming the Earth is a sphere of radius 6371 km, the surface area of the arctic (north of the Arctic Circle, at latitude 66.56° as of August 2016 [7]) is 2π ⋅ 6371 2 | sin 90° − sin 66.56° | = 21.04 million km 2 (8.12 million sq mi), or 0.5 ⋅ | sin 90° − sin 66.56° | = 4.125% of the total surface area of the Earth.
A sphere of radius r has area element = . This can be found from the volume element in spherical coordinates with r held constant. [9] A sphere of any radius centered at zero is an integral surface of the following differential form: + + =
If the radius of the sphere is denoted by r and the height of the cap by h, the volume of the spherical sector is =.. This may also be written as = (), where φ is half the cone angle, i.e., φ is the angle between the rim of the cap and the direction to the middle of the cap as seen from the sphere center.
The two effects exactly cancel each other out. In the extreme case of the smallest possible sphere, the cylinder vanishes (its radius becomes zero) and the height equals the diameter of the sphere. In this case the volume of the band is the volume of the whole sphere, which matches the formula given above.
For example, one sphere that is described in Cartesian coordinates with the equation x 2 + y 2 + z 2 = c 2 can be described in spherical coordinates by the simple equation r = c. (In this system—shown here in the mathematics convention—the sphere is adapted as a unit sphere, where the radius is set to unity and then can generally be ignored ...
Thus, the segment volume equals the sum of three volumes: two right circular cylinders one of radius a and the second of radius b (both of height /) and a sphere of radius /. The curved surface area of the spherical zone—which excludes the top and bottom bases—is given by =.
The sphere's radius is taken as unity. For specific practical problems on a sphere of radius R the measured lengths of the sides must be divided by R before using the identities given below. Likewise, after a calculation on the unit sphere the sides a, b, and c must be multiplied by R.
Fix a plane through the center of the ball. Let r denote the distance between a point in the plane and the center of the sphere, and let θ denote the azimuth. Intersecting the n-ball with the (n − 2)-dimensional plane defined by fixing a radius and an azimuth gives an (n − 2)-ball of radius √ R 2 − r 2.