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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.
On the sphere, he showed that the surface area is four times the area of its great circle. In modern terms, this means that the surface area is equal to: =. The result for the volume of the contained ball stated that it is two-thirds the volume of a circumscribed cylinder, meaning that the volume is
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:
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,}
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 ...
In all of these arrangements each sphere touches 12 neighboring spheres, [2] and the average density is π 3 2 ≈ 0.74048. {\displaystyle {\frac {\pi }{3{\sqrt {2}}}}\approx 0.74048.} In 1611, Johannes Kepler conjectured that this is the maximum possible density amongst both regular and irregular arrangements—this became known as the Kepler ...
where C is the circumference of a circle, d is the diameter, and r is the radius.More generally, = where L and w are, respectively, the perimeter and the width of any curve of constant width.
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 =.