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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.
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: + + =
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:
For example, one sphere that is described in Cartesian coordinates with the equation x 2 + y 2 + z 2 = c 2 can be described in spherical coordinates by the simple equation r = c. (In this system—shown here in the mathematics convention—the sphere is adapted as a unit sphere, where the radius is set to unity and then can generally be ignored ...
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 sphere's radius is taken as unity. For specific practical problems on a sphere of radius R the measured lengths of the sides must be divided by R before using the identities given below. Likewise, after a calculation on the unit sphere the sides a, b, and c must be multiplied by R.
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.
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 ...