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The surface-area-to-volume ratio has physical dimension inverse length (L −1) and is therefore expressed in units of inverse metre (m −1) or its prefixed unit multiples and submultiples. As an example, a cube with sides of length 1 cm will have a surface area of 6 cm 2 and a volume of 1 cm 3. The surface to volume ratio for this cube is thus
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.
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 ...
The ratio of the volume of a sphere to the volume of its circumscribed cylinder is 2:3, as was determined by Archimedes. The principal formulae derived in On the Sphere and Cylinder are those mentioned above: the surface area of the sphere, the volume of the contained ball, and surface area and volume of the cylinder.
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.
The sphere has the smallest surface area of all surfaces that enclose a given volume, and it encloses the largest volume among all closed surfaces with a given surface area. [11] The sphere therefore appears in nature: for example, bubbles and small water drops are roughly spherical because the surface tension locally minimizes surface area.
Defined by Wadell in 1935, [1] the sphericity, , of an object is the ratio of the surface area of a sphere with the same volume to the object's surface area: = where is volume of the object and is the surface area.
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,}