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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:
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 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 angle, i.e., φ is the angle between the rim of the cap and the direction ...
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 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, 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 ...
When a regular dodecahedron is inscribed in a sphere, it occupies more of the sphere's volume (66.49%) than an icosahedron inscribed in the same sphere (60.55%). [10] The resulting of both spheres' volumes initially began from the problem by ancient Greeks, determining which of two shapes has a larger volume: an icosahedron inscribed in a ...