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where V is the volume of a sphere and r is the radius. S A = 4 π r 2 {\displaystyle SA=4\pi r^{2}} where SA is the surface area of a sphere and r is the radius.
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.
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.
The volume can be computed without use of the Gamma function. As is proved below using a vector-calculus double integral in polar coordinates, the volume V of an n-ball of radius R can be expressed recursively in terms of the volume of an (n − 2)-ball, via the interleaved recurrence relation:
The n-sphere of unit radius centered at the origin is denoted S n and is often referred to as "the" n-sphere. The ordinary sphere is a 2-sphere, because it is a 2-dimensional surface which is embedded in 3-dimensional space. In topology, the n-sphere is an example of a compact topological manifold without boundary.
For n = 2, in a 2-dimensional plane , "balls" according to the L 1-norm (often called the taxicab or Manhattan metric) are bounded by squares with their diagonals parallel to the coordinate axes; those according to the L ∞-norm, also called the Chebyshev metric, have squares with their sides parallel to the coordinate axes as their boundaries.
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:
Lines, L. (1965), Solid geometry: With Chapters on Space-lattices, Sphere-packs and Crystals, Dover. Reprint of 1935 edition. A problem on page 101 describes the shape formed by a sphere with a cylinder removed as a "napkin ring" and asks for a proof that the volume is the same as that of a sphere with diameter equal to the length of the hole.