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where S n − 1 (r) is an (n − 1)-sphere of radius r (being the surface of an n-ball of radius r) and dA is the area element (equivalently, the (n − 1)-dimensional volume element). The surface area of the sphere satisfies a proportionality equation similar to the one for the volume of a ball: If A n − 1 (r) is the surface area of an (n ...
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 set of points drawn from a uniform distribution on the surface of a unit 2-sphere, generated using Marsaglia's algorithm. To generate uniformly distributed random points on the unit -sphere (that is, the surface of the unit -ball), Marsaglia (1972) gives the following algorithm.
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 V = 2 π r 3 3 ( 1 − cos φ ) , {\displaystyle V={\frac {2\pi r^{3}}{3}}(1-\cos \varphi )\,,} where φ is half the cone aperture angle, i.e., φ is the angle between the rim of the cap and the ...
A sphere (top), rotational ellipsoid (left) and triaxial ellipsoid (right) The volume of a sphere of radius R is . Given the volume of a non-spherical object V, one can calculate its volume-equivalent radius by setting = or, alternatively:
The volume of the unit ball in Euclidean -space, and the surface area of the unit sphere, appear in many important formulas of analysis. The volume of the unit n {\\displaystyle n} -ball, which we denote V n , {\\displaystyle V_{n},} can be expressed by making use of the gamma function .
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 =.
An approximation for the volume of a thin spherical shell is the surface area of the inner sphere multiplied by the thickness t of the shell: [2] V ≈ 4 π r 2 t , {\displaystyle V\approx 4\pi r^{2}t,}