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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.
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 ball in n dimensions is called a hyperball or n-ball and is bounded by a hypersphere or (n−1)-sphere. Thus, for example, a ball in the Euclidean plane is the same thing as a disk, the area bounded by a circle. In Euclidean 3-space, a ball is taken to be the volume bounded by a 2-dimensional sphere. In a one-dimensional space, a ball is a ...
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 =.
Hart (2009) [3] states that the "volume of a spherical wedge is to the volume of the sphere as the number of degrees in the [angle of the wedge] is to 360". Hence, and through derivation of the spherical wedge volume formula, it can be concluded that, if V s is the volume of the sphere and V w is the volume of a given spherical wedge,
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
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,}